Trigonometric functions. Attnom Academy

M5.3.1 Chapter 1

Radians

Is there a better way to measure an angle than in degrees?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A An arc as long as the radius cuts an angle of 45° at the centre.
You picked A. Read on. At the end, the chapter comes back to your guess.
B An arc as long as the radius cuts an angle of 90° at the centre.
You picked B. Read on. At the end, the chapter comes back to your guess.
C An arc as long as the radius cuts an angle of about 57° at the centre.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Change between degrees and radians
Find arc length and sector area using radians

Degrees come from an old habit of cutting a circle into 360 parts. Nothing about a circle makes 360 special. A radian is a unit that comes from the circle itself, and it makes the formulas for circles and for trigonometry short. From here on, most of the mathematics you meet uses radians.

Take a circle with radius r. Cut a piece of string of length r and bend it along the arc of the circle. The angle at the centre that the string covers is one radian. An arc as long as the radius cuts off one radian.

The arc has the same length as the radius, so the angle at the centre is 1 radian. The ticks show the two equal lengths.

This works for a circle of any size. A bigger circle has a longer radius and a longer arc, in the same ratio. Two radii of arc cut off 2 radians, and three radii cut off 3 radians. So an angle in radians is the number of radii that fit along its arc.

Definition. An angle θ at the centre of a circle has size θ = s ÷ r in radians. Here s is the length of the arc it cuts off and r is the radius. It is a plain number, so it has no unit. We may write "rad" to make it clear.

Now the whole circle. Its circumference is 2πr, which is 2π radii. So the full turn is 2π radians. Half a turn is π radians. This is the link between the two units: 2π radians = 360°, so π radians = 180°.

A quarter turn is π/2, a half turn is π, three quarters is 3π/2 and a full turn is 2π. The point at 60° is at angle π/3, and the point at 135° is at angle 3π/4.

One radian is 180° ÷ π, which is about 57.3°. To change units, use the fact that π radians = 180°. To change degrees to radians, multiply by π/180. To change radians to degrees, multiply by 180/π.

The angles you will use most. Learn them as fractions of π: 30° is one sixth of 180°, so it is π/6.

Worked example

Change 150° to radians. Then change 7π/4 radians to degrees.

We use π radians = 180°. So 1° is π/180 radians.
Multiply 150 by π/180. This gives 150π/180.
Cancel by 30. 150 ÷ 30 = 5 and 180 ÷ 30 = 6.
So 150° = 5π/6 radians.
For the other way, 1 radian is 180/π degrees. Multiply 7π/4 by 180/π. The π cancels.
We are left with 7 × 180 ÷ 4. Since 180 ÷ 4 = 45, this is 7 × 45 = 315.
So 7π/4 radians = 315°.

Answer: 150° = 5π/6 radians and 7π/4 radians = 315°.

Radians are usually written with no unit. If an angle has a degree sign, it is in degrees. If it has no sign, such as θ = 2 or x = π/3, it is in radians. From now on, an angle with no sign is in radians.

Arc length. The angle θ is the fraction θ/2π of a full turn. So the arc is that fraction of the circumference 2πr. The 2π cancels, and we are left with a short result: s = rθ. This also follows straight from the definition θ = s/r.

Sector area. A sector is a slice of a circle between two radii. It is the fraction θ/2π of the whole circle, which has area πr². So A = (θ/2π) × πr² = ½r²θ. Because s = rθ, we can also write A = ½rs.

Arc length s = rθ. Sector area A = ½r²θ. Perimeter of the sector = 2r + rθ, because it has two straight sides as well as the arc. These need θ in radians.

Compare with degrees. If θ is in degrees, the arc is (θ/360) × 2πr and the sector is (θ/360) × πr². The extra π/180 sits in every formula. Radians remove it. That is why radians are used in calculus and in physics.

Worked example

A sector has radius 8 cm and angle 0.75 radians. Find its arc length and its area.

The angle is in radians, so s = rθ and A = ½r²θ both apply.
Arc length: s = 8 × 0.75 = 6 cm.
Area: A = ½ × 8² × 0.75. Since 8² = 64, ½ × 64 = 32.
Then 32 × 0.75 = 24. The area is 24 cm².
Check with A = ½rs: ½ × 8 × 6 = 24. This agrees.

Answer: Arc length 6 cm and area 24 cm².

The sector of the worked example: radius 8 cm and θ = 0.75 radians.

Worked example

A sector has radius 6 cm and arc length 9 cm. Find its angle in radians, its area and its perimeter.

We know r and s, so use θ = s ÷ r.
θ = 9 ÷ 6 = 1.5 radians.
Area: A = ½rs, since s is given. A = ½ × 6 × 9 = 27 cm².
Check with A = ½r²θ: ½ × 36 × 1.5 = 18 × 1.5 = 27. This agrees.
Perimeter: two radii and the arc. P = 6 + 6 + 9 = 21 cm.

Answer: θ = 1.5 radians, area 27 cm² and perimeter 21 cm.

Radius 6 cm and arc 9 cm. The angle is 9 ÷ 6 = 1.5 radians, about 86°.

A segment is the part of a circle between a chord and the arc. Its area is the sector minus the triangle formed by the two radii and the chord. The triangle has area ½r² sin θ, from the rule area = ½ab sin C that you know. So the segment has area ½r²θ − ½r² sin θ = ½r²(θ − sin θ). Use radian mode for sin θ.

Worked example

A chord AB of a circle with radius 10 cm cuts an angle of 1.4 radians at the centre. Find the area of the segment between the chord and the minor arc.

Segment = sector − triangle. The angle is in radians, so the sector is ½r²θ.
Sector: ½ × 10² × 1.4 = 50 × 1.4 = 70 cm².
Triangle: ½r² sin θ = 50 × sin 1.4. The calculator is in radian mode, and sin 1.4 = 0.98545.
So the triangle is 50 × 0.98545 = 49.27 cm².
Segment = 70 − 49.27 = 20.73 cm². To 3 significant figures this is 20.7 cm².
Check: the formula ½r²(θ − sin θ) = 50 × (1.4 − 0.98545) = 50 × 0.41455 = 20.73. This agrees.

Answer: The segment has area 20.7 cm².

The hatched segment is the sector with the triangle taken away.

Worked example

A sector has radius 5 cm and perimeter 16 cm. Find its angle in radians and its area.

The perimeter is two radii plus the arc: 16 = 5 + 5 + s.
So the arc is s = 16 − 10 = 6 cm.
The angle is θ = s ÷ r = 6 ÷ 5 = 1.2 radians.
The area is A = ½rs = ½ × 5 × 6 = 15 cm².
Check with A = ½r²θ: ½ × 25 × 1.2 = 12.5 × 1.2 = 15. This agrees.

Answer: θ = 1.2 radians and the area is 15 cm².

Radius 5 cm and perimeter 16 cm. Take away the two radii, and 6 cm of arc is left.

Many exam questions join a sector to a triangle. The tangent to a circle at A is perpendicular to the radius OA. So if the tangent meets OB, extended, at T, triangle OAT has a right angle at A. Then AT = r tan θ and OT = r ÷ cos θ, with the calculator in radian mode.

Worked example

Sector OAB has radius 6 cm and angle 0.8 radians. The tangent at A meets OB, extended, at T. Find the perimeter and the area of the region bounded by AT, TB and the arc AB.

Triangle OAT has a right angle at A, with OA = 6 cm and angle AOT = 0.8 radians.
AT = 6 tan 0.8 = 6 × 1.0296 = 6.178 cm.
OT = 6 ÷ cos 0.8 = 6 ÷ 0.6967 = 8.612 cm. So TB = OT − OB = 8.612 − 6 = 2.612 cm.
The arc AB = rθ = 6 × 0.8 = 4.8 cm.
The perimeter is AT + TB + arc AB = 6.178 + 2.612 + 4.8 = 13.59 cm, which is 13.6 cm.
The area is triangle OAT − sector OAB = ½ × 6 × 6.178 − ½ × 6² × 0.8.
This is 18.53 − 14.4 = 4.13 cm².

Answer: Perimeter 13.6 cm and area 4.13 cm²

The angle θ is 0.8 radians. The point B lies on OT, 6 cm from O. The hatched region is triangle OAT with the sector OAB taken away.

For a small angle θ in radians, sin θ is almost θ, tan θ is almost θ, and cos θ is almost 1 − θ²/2. We write sin θ ≈ θ, tan θ ≈ θ and cos θ ≈ 1 − θ²/2. At θ = 0.1: sin 0.1 = 0.0998, tan 0.1 = 0.1003 and cos 0.1 = 0.9950, while 1 − 0.1²/2 = 0.995. These hold only in radians, and only for small angles.

See it three ways

With things

Wrap a piece of string, as long as the radius, round the edge of a round tin lid. Mark where it ends. The angle from the centre of the lid to the two ends of the string is one radian. Two lengths of string give two radians.

As a picture

In words

The size of an angle in radians is the length of its arc divided by the radius.

In symbols

θ = s ÷ r, so s = rθ. For a full turn, s = 2πr, so θ = 2π.

Try it

Measure a radian with string

You need: a round lid or plate, a piece of thread, a ruler, a pencil.

Measure the radius of the lid, from its centre to its edge, and cut a thread that long.
Lay the thread along the edge of the lid. Mark the two ends of the thread on the edge.
Mark the middle of the lid. Draw a line from it to each mark.
Look at the angle between the two lines. It is about 57°, a little less than a sixth of the circle.

What you will see: The angle is the same whatever the size of the lid. A little more than six pieces of thread, 6.28 of them, go round the whole edge.

If you cannot try it, this is what happens.

A wrong idea many people have

The arc is s = rθ, and I can put the angle in degrees.

Why it feels right: The angle is the same angle, whatever unit we use, so it seems any unit should work. A sector of 60° looks like it has θ = 60.

The test that breaks it: Take r = 3 cm and θ = 60°. The rule would give s = 3 × 60 = 180 cm, but the whole circle has a circumference of only 2π × 3 = 18.85 cm. An arc cannot be longer than the circle.

What is true: The formulas s = rθ and A = ½r²θ hold only for θ in radians, because a radian is defined by arc ÷ radius. Change degrees to radians first: 60° = π/3, and then s = 3 × π/3 = π = 3.14 cm.

A wrong idea many people have

The perimeter of a sector is rθ.

Why it feels right: The arc is the curved edge, and that edge feels like the whole outline of the sector.

The test that breaks it: Draw a sector with radius 6 cm and θ = 1.5. The arc rθ is 9 cm. But the outline also has the two straight sides, each 6 cm.

What is true: The perimeter is 2r + rθ. For this sector, 6 + 6 + 9 = 21 cm. Read each question to see whether it asks for the arc or for the perimeter.

A wrong idea many people have

The calculator mode does not matter for sin and cos, because a sine is just a number.

Why it feels right: The buttons are the same in every mode, and sin 30 gives the same kind of answer.

The test that breaks it: Type sin 1. In degree mode it shows 0.0175. In radian mode it shows 0.8415. The number 1 means 1° in one mode and 1 radian, about 57.3°, in the other. Type sin 30: degree mode gives 0.5 and radian mode gives −0.988.

What is true: Set the calculator to radian mode (often shown as RAD or R) whenever the angle is in radians. Check with sin(π/6): it must show 0.5. If it shows 0.009, the calculator is still in degree mode.

Key words

radian: the angle at the centre of a circle that cuts off an arc equal in length to the radius. About 57.3°.
arc: a part of the edge of a circle.
sector: the slice of a circle between two radii and an arc.
segment: the part of a circle between a chord and the arc it cuts off.
chord: a straight line joining two points on a circle.

Back to your guess

The answer is C: An arc as long as the radius cuts an angle of about 57° at the centre.

The angle is about 57.3°. The whole circle has an arc of 2π radii, which is a little over 6 radii, and 360° ÷ 6.28 is about 57.3°. This angle is called one radian, and it is the natural unit for angles.

Keep it

Copy this into your exercise book.

π rad = 180°. Degrees to radians: × π/180. Radians to degrees: × 180/π. s = rθ, A = ½r²θ, perimeter 2r + rθ. Radians only. Calculator in RAD mode.

Draw: A sector with radius r, arc s and angle θ, with the three formulas beside it.

M5.3.2 Chapter 2

The unit circle and any angle

How can I find sin 150° or cos 225° when the angle does not fit in a right angled triangle?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A sin 150° is the same as sin 30°.
You picked A. Read on. At the end, the chapter comes back to your guess.
B sin 150° does not exist, because a triangle cannot have an angle of 150° and a right angle.
You picked B. Read on. At the end, the chapter comes back to your guess.
C sin 150° is negative, because 150° is more than 90°.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use the unit circle to find sin and cos of any angle
Find the sign of sin, cos and tan in each quarter

In a right angled triangle, sin, cos and tan are ratios of sides. That works only for angles under 90°, because a triangle with a right angle cannot hold another angle that big. But a wheel, a swing or a sound wave turns through any angle. We need sin and cos for every angle.

The unit circle is a circle with radius 1 and its centre at the origin. Start at the point (1, 0). Turn round the circle through an angle θ, anticlockwise for positive θ. Call the point you reach P.

The radius OP has length 1. Its end P has x coordinate cos θ and y coordinate sin θ.

For an acute angle, the radius OP is the hypotenuse of a right angled triangle with hypotenuse 1. The side next to θ is the x coordinate of P and the side opposite θ is its y coordinate. So cos θ = adjacent ÷ 1 = x, and sin θ = opposite ÷ 1 = y.

This gives a definition that works for every angle. Definition. For any angle θ, the point P reached on the unit circle has coordinates P = (cos θ, sin θ). So cos θ is the x coordinate and sin θ is the y coordinate. And tan θ = sin θ ÷ cos θ, which is the gradient of OP, whenever cos θ is not 0.

Because x and y lie between −1 and 1, we always have −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. A negative angle turns clockwise. After a full turn of 360° (2π radians) the point is back where it started, so sin(θ + 360°) = sin θ and cos(θ + 360°) = cos θ.

The axes divide the plane into four quadrants, numbered anticlockwise from the top right. In each quadrant x and y have a fixed sign, so sin θ and cos θ have a fixed sign. This gives the signs of all three ratios.

The signs of sin, cos and tan in each quadrant. The sign of tan is the sign of sin divided by the sign of cos.

Each quadrant names what is positive there. Start in the fourth quadrant and go anticlockwise: Cos, All, Sin, Tan. The first letters spell CAST.

Angles in the other quadrants give the same numbers as acute angles, with a sign. The acute angle between the radius OP and the x axis is the reference angle. It is θ in the first quadrant, 180° − θ in the second, θ − 180° in the third and 360° − θ in the fourth.

The ray at 150° makes 30° with the negative x axis. The ray at 225° makes 45° with it. These acute angles are the reference angles.

The point at 150° is the mirror image of the point at 30° in the y axis. It has the same height, so sin 150° = sin 30°. Its x coordinate has the opposite sign, so cos 150° = −cos 30°. Reflection gives every other quadrant the same way.

P at 30° and Q at 150° are reflections in the y axis. They have the same y coordinate, so sin 150° = sin 30°. The x coordinates are opposites.

Rule. To find sin, cos or tan of any angle, follow four steps. First find the quadrant. Then find the reference angle. Then take the exact or calculator value for that angle. Last, give it the sign for the quadrant. For angles on the axes, read the point itself: it is (1, 0) at 0°, (0, 1) at 90°, (−1, 0) at 180° and (0, −1) at 270°.

Reflections and turns give results worth learning. For every x: sin(−x) = −sin x, cos(−x) = cos x and tan(−x) = −tan x where tan x is defined. Also sin(π − x) = sin x and cos(π − x) = −cos x. And sin(π + x) = −sin x and cos(π + x) = −cos x. Last, sin(π/2 − x) = cos x and cos(π/2 − x) = sin x. In degrees, replace π by 180°.

Each is a mirror image or a half turn of the point at x on the unit circle. A full turn, 2π or 360°, brings the point back, so sin(x + 2nπ) = sin x and cos(x + 2nπ) = cos x for every integer n. For an angle of more than one turn, or a negative angle, take away or add whole turns first.

The exact values for 30°, 45° and 60° come from two triangles you know: half of an equilateral triangle, and half of a square.

Half of an equilateral triangle with side 2. So sin 30° = 1/2, cos 30° = √3/2 and tan 30° = 1/√3 = √3/3.

Half of a square with side 1. So sin 45° = 1/√2 = √2/2, cos 45° = √2/2 and tan 45° = 1.

The exact values to learn. At 90° the point is (0, 1), so cos 90° = 0 and tan 90° = 1/0 has no value.

Because tan θ = sin θ ÷ cos θ, tan θ is not defined when cos θ = 0. That happens at 90°, 270°, and every angle 180° apart from these. In radians these are π/2, 3π/2, and so on. Near these angles tan θ becomes very large, positive on one side and negative on the other.

Worked example

Find the exact values of sin 150°, cos 150° and tan 150°.

150° lies between 90° and 180°, so it is in the second quadrant.
The reference angle is 180° − 150° = 30°.
We know sin 30° = 1/2, cos 30° = √3/2 and tan 30° = √3/3.
In the second quadrant sin is positive, cos is negative and tan is negative.
So sin 150° = 1/2, cos 150° = −√3/2 and tan 150° = −√3/3.
Check: tan 150° = sin 150° ÷ cos 150° = (1/2) ÷ (−√3/2) = −1/√3 = −√3/3.

Answer: sin 150° = 1/2, cos 150° = −√3/2, tan 150° = −√3/3.

Worked example

Given that cos θ = −5/13 and that 180° < θ < 270°, find sin θ and tan θ.

The point P = (cos θ, sin θ) lies on the unit circle, so x² + y² = 1.
Here x = −5/13, so x² = 25/169 and y² = 1 − 25/169.
Write 1 as 169/169. Then the top is 169 − 25 = 144.
So y² = 144/169. The square roots are 13 on the bottom and 12 on the top.
So y = 12/13 or y = −12/13, because both have a square of 144/169.
θ is in the third quadrant, where sin is negative. So sin θ = −12/13.
Then tan θ = sin θ ÷ cos θ = (−12/13) ÷ (−5/13) = 12/5.

Answer: sin θ = −12/13 and tan θ = 12/5.

The point P is in the third quadrant, so both of its coordinates are negative: cos θ = −5/13 and sin θ = −12/13.

Worked example

Find the exact values of cos(7π/6), sin(−π/3) and tan(5π/4).

Change each to degrees if that helps. 7π/6 = 210°, −π/3 = −60° and 5π/4 = 225°.
First cos 210°. It is in the third quadrant. The reference angle is 210° − 180° = 30°.
cos 30° = √3/2, and cos is negative in the third quadrant. So cos(7π/6) = −√3/2.
Next sin(−60°). A turn of 60° clockwise reaches the fourth quadrant, where sin is negative.
The reference angle is 60° and sin 60° = √3/2. So sin(−π/3) = −√3/2.
Last tan 225°. It is in the third quadrant, with reference angle 225° − 180° = 45°.
tan 45° = 1, and tan is positive in the third quadrant. So tan(5π/4) = 1.

Answer: cos(7π/6) = −√3/2, sin(−π/3) = −√3/2 and tan(5π/4) = 1.

Worked example

Find the exact values of cos(25π/6) and sin(−750°).

25π/6 is more than a full turn. Write 25π/6 = 4π + π/6, where 4π is two full turns.
So cos(25π/6) = cos(π/6) = √3/2.
For −750°, add two full turns, 720°. Then −750° + 720° = −30°.
So sin(−750°) = sin(−30°) = −sin 30° = −1/2.

Answer: cos(25π/6) = √3/2 and sin(−750°) = −1/2

See it three ways

With things

A child on a wheel that turns steadily goes up and down. At the start the child is level with the centre. After a quarter turn the child is at the top, after a half turn level again, and after three quarters at the bottom.

As a picture

In words

sin θ is the height of the point above the x axis, and cos θ is how far it is to the right of the y axis. Both are negative when the point is below or to the left.

In symbols

P = (cos θ, sin θ). At θ = 150°, P = (−√3/2, 1/2), so sin 150° = 1/2 and cos 150° = −√3/2.

Try it

Measure sin and cos on a big circle

You need: paper, a ruler, a protractor, a pencil.

Draw an x axis and a y axis that cross in the middle of the paper. Mark 10 cm from the crossing on each side. Let 10 cm stand for 1.
Draw a circle of radius 10 cm with its centre at the crossing.
With the protractor, mark 150° from the right hand end of the x axis, turning anticlockwise. Draw the radius to that mark.
From the end of the radius, measure down to the x axis, and measure across to the y axis. Divide each length by 10. The one across is negative, because it is to the left.

What you will see: The height is about 0.5 and the distance across is about −0.87. These are sin 150° and cos 150°. Try 30° and see the same height.

If you cannot try it, this is what happens.

A wrong idea many people have

sin 150° = −1/2, because 150° is bigger than 90° and sin goes negative there.

Why it feels right: The cosine went negative after 90°, and the sine seems to follow it.

The test that breaks it: The point at 150° is above the x axis, so its height sin 150° is positive. The graph of the point turning round the circle only goes below the x axis after 180°.

What is true: sin is positive in the first and second quadrants, from 0° to 180°. Use the quadrant, not whether the angle is above 90°.

A wrong idea many people have

sin 150° = sin 60°, because 150° = 90° + 60°.

Why it feels right: You may subtract 90° as a habit from a right angled triangle, which has an angle of 90°.

The test that breaks it: Look at the point at 150° and its mirror at 30°. Both are at height 1/2. But sin 60° = √3/2, which is a different height.

What is true: The reference angle is the acute angle with the x axis, not with the y axis. In the second quadrant it is 180° − θ. So the reference angle is 30°.

A wrong idea many people have

tan 90° = 1 ÷ 0 = 0.

Why it feels right: Dividing 1 by a small number gives a big number, and dividing by nothing seems to give nothing.

The test that breaks it: Try 1 ÷ 0.1 = 10, 1 ÷ 0.01 = 100, 1 ÷ 0.001 = 1000. The smaller the bottom, the bigger the answer. It does not settle at 0.

What is true: Division by 0 has no value, so tan 90° is not defined. The same is true at 270° and at every angle 180° from these.

Key words

unit circle: the circle with radius 1 and its centre at the origin.
quadrant: one of the four regions into which the axes cut the plane, numbered 1 to 4 anticlockwise from the top right.
reference angle: the acute angle between the radius OP and the x axis.

Back to your guess

The answer is A: sin 150° is the same as sin 30°.

sin 150° = sin 30° = 1/2. A triangle cannot show it, but a circle can. We turn through 150° round a circle of radius 1 and read the height of the point we reach. That height is the same as at 30°.

Keep it

Copy this into your exercise book.

P = (cos θ, sin θ) on the unit circle. Signs: first All, second Sin, third Tan, fourth Cos. Size from the reference angle. tan θ = sin θ ÷ cos θ, none when cos θ = 0.

Draw: A circle with axes, four quadrants marked A, S, T, C, and one radius with its reference angle.

M5.3.3 Chapter 3

Trig graphs and their changes

What do the graphs of sin, cos and tan look like, and how do they change when the equation changes?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The graph of y = sin 2x is twice as tall as y = sin x.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The graph of y = sin 2x repeats twice as often as y = sin x.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The graph of y = sin 2x is the same as the graph of y = 2 sin x.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Sketch y = 2 sin 3x and give its period and amplitude
Sketch y = tan x and mark its asymptotes

Imagine the point P going round the unit circle at a steady speed. Its height above the x axis is sin x, where x is the angle turned. If we plot that height against x, we get a wave. The same wave describes tides, sound, a swing and the mains electricity in a wall. In this chapter x is in radians.

The graphs of y = sin x (solid) and y = cos x (dashed) from 0 to 2π. Both lie between −1 and 1.

Read the main facts from the picture. The sine wave starts at 0, climbs to 1 at π/2, comes back to 0 at π, falls to −1 at 3π/2, and returns to 0 at 2π. Then it repeats for ever, in both directions. The cosine wave is the same shape, starting at its peak, 1.

Three words describe a wave. The period is the length of x after which the graph repeats. The amplitude is the greatest height of the wave above its middle line. The midline is the horizontal line halfway between the greatest and least values. For y = sin x and y = cos x the period is 2π, the amplitude is 1 and the midline is the x axis.

The graph of y = tan x is different. Since tan x = sin x ÷ cos x, it is not defined when cos x = 0. That is at x = π/2, 3π/2, −π/2 and so on, at every odd multiple of π/2. Near those values tan x grows without limit. The vertical lines there are asymptotes: the graph gets as close as we like but never touches them.

The graph of y = tan x. The dashed lines are the asymptotes x = −π/2 and x = π/2. The graph repeats every π, and it passes through the origin.

For tan x, the period is π, not 2π. The graph crosses the x axis at 0, π, 2π and so on. It rises from the left of each pair of asymptotes to the right, and it takes every value from very negative to very positive. It has no greatest or least value, so it has no amplitude.

Now the changes. Put a number a in front: y = a sin x. Every height is multiplied by a, so the wave is stretched up and down by a factor of a. The amplitude is |a|. If a is negative, the wave is turned upside down as well.

The number in front stretches the graph up and down. y = 2 sin x has amplitude 2 and the same period 2π.

Put a number b next to x: y = sin bx. Now x is multiplied by b before the sine is taken. The wave goes through its cycle b times as fast. So it is squeezed sideways by a factor of b, and its period is 2π ÷ b. For y = sin 2x the period is π, and for y = sin 3x it is 2π/3. The amplitude does not change.

The number next to x squeezes the graph sideways. y = sin 2x has period π: it completes 2 cycles between 0 and 2π.

Replace x by x + c: y = sin(x + c). Each height that y = sin x reaches at x, the new graph reaches c earlier, at x − c. So the whole graph slides c units to the left when c is positive, and to the right when c is negative. This surprises many people, because the plus sign sends the graph left.

y = sin(x + π/3) is y = sin x moved π/3 to the left. At x = −π/3 it is 0, where y = sin x is 0 at x = 0.

Finally, add a number d outside: y = sin x + d. This lifts the whole graph up by d, so the midline moves from the x axis to the line y = d. Putting it all together gives the general wave y = a sin(bx + c) + d. Its amplitude is |a|, its period is 2π/b (with b > 0), it moves c/b to the left, and its midline is y = d. The greatest value is d + |a| and the least value is d − |a|.

Worked example

State the amplitude and period of y = 3 sin 2x, and its greatest and least values. Sketch it for 0 ≤ x ≤ 2π.

Compare with y = a sin bx. Here a = 3 and b = 2.
The amplitude is 3.
The period is 2π ÷ 2 = π. So the wave has 2 full cycles between 0 and 2π.
The greatest value is 3 and the least value is −3, because the midline is the x axis.
One cycle goes from 0 to π. It starts at 0 and reaches its peak at a quarter of the period, x = π/4.
At the peak 2x = π/2, so sin 2x = 1 and the height is 3 × 1 = 3.
It crosses the axis at x = π/2, reaches its trough −3 at x = 3π/4, and is back at 0 at x = π. Then the cycle repeats.

Answer: Amplitude 3, period π, greatest value 3, least value −3.

y = 3 sin 2x. The ticks between the numbers are π/4 apart. The first peak is at x = π/4 and the first trough is at x = 3π/4.

Here is the equation y = 2 sin 3x, which the first goal asks about. It has a = 2 and b = 3, so the amplitude is 2 and the period is 2π/3. It completes three cycles in 2π. It reaches its first peak at one quarter of a period, x = π/6, and its first trough at x = π/2.

One period of y = 2 sin 3x. The wave repeats every 2π/3.

Worked example

Sketch y = 1 + 2 cos x for 0 ≤ x ≤ 2π, and state its greatest value, its least value and its midline.

Start from y = cos x. It starts at 1, falls to −1 at π and returns to 1 at 2π.
The 2 in front stretches the wave up and down: the heights run from −2 to 2.
The 1 added lifts it up by 1. So the midline is y = 1.
The greatest value comes where cos x = 1, at x = 0 and x = 2π. It is 1 + 2 = 3.
The least value is 1 − 2 = −1, at x = π.

Answer: Greatest value 3, least value −1, midline y = 1.

y = 1 + 2 cos x. The dashed line y = 1 is the midline.

Worked example

The depth of water, h metres, in a harbour t hours after midnight is h = 8 + 3 sin(πt/6), for 0 ≤ t ≤ 12. Find the greatest and least depths, the period of the tide, and the times when the depth is 9.5 m.

The graph is a sine wave with midline h = 8 and amplitude 3.
The greatest depth is 8 + 3 = 11 m. The least depth is 8 − 3 = 5 m.
In sin(πt/6) the number next to t is π/6. So the period is 2π ÷ (π/6) = 2π × 6/π = 12 hours.
For a depth of 9.5 m we need 8 + 3 sin(πt/6) = 9.5. Then 3 sin(πt/6) = 1.5, so sin(πt/6) = 0.5.
For 0 ≤ t ≤ 12, πt/6 runs from 0 to 2π, which is one full cycle. In it, sin = 0.5 only at π/6 and 5π/6. So πt/6 = π/6, which gives t = 1.
The other time: πt/6 = 5π/6, which gives t = 5. Both are inside 0 ≤ t ≤ 12.

Answer: Greatest 11 m, least 5 m, period 12 hours. The depth is 9.5 m at t = 1 and t = 5.

The depth in the harbour over 12 hours. The dashed line h = 9.5 cuts the wave at t = 1 and t = 5.

The tan graph changes in the same way, with one difference. The period of y = tan bx is π ÷ b. For y = tan 2x the period is π/2, so the asymptotes are closer together. They stand at x = π/4, 3π/4, and so on.

Worked example

Sketch y = tan 2x for 0 ≤ x ≤ π. Give its period, and the equations of its asymptotes.

The number next to x is 2. The period of tan bx is π ÷ b, so the period is π/2.
tan 2x is not defined when 2x = π/2, 3π/2, and so on. So x = π/4 and x = 3π/4.
The graph passes through (0, 0), (π/2, 0) and (π, 0), where 2x = 0, π and 2π.
Between 0 and π/4 it rises from 0 towards the asymptote. Between π/4 and 3π/4 it rises from far below to far above, through (π/2, 0). Between 3π/4 and π it rises from far below to 0.

Answer: The period is π/2. The asymptotes are x = π/4 and x = 3π/4.

y = tan 2x for 0 ≤ x ≤ π. The dashed lines are the asymptotes x = π/4 and x = 3π/4.

Worked example

The graph of y = a + b sin cx has a greatest value of 4, a least value of −2 and a period of π. Find a, b and c, where b and c are positive.

The midline is halfway between the greatest and least values: a = (4 + (−2)) ÷ 2 = 1.
The amplitude is the distance from the midline to the top: b = 4 − 1 = 3.
The period is 2π ÷ c = π, so c = 2.
Check the least value: 1 − 3 = −2. This agrees. So y = 1 + 3 sin 2x.

Answer: a = 1, b = 3 and c = 2.

One cycle of y = 1 + 3 sin 2x: greatest value 4, least value −2, period π.

Take care when x has a number in front inside the bracket. In y = sin(bx + c), take out the b first: bx + c = b(x + c/b). So the graph of sin bx is moved c/b to the left, not c.

Worked example

Describe how the graph of y = sin(2x + π/3) comes from the graph of y = sin 2x. Give its period and its first peak for x > 0.

Take out the 2: 2x + π/3 = 2(x + π/6). So y = sin 2(x + π/6).
Replacing x by x + π/6 moves a graph π/6 to the left. So the graph of y = sin 2x moves π/6 to the left, not π/3.
The period does not change: 2π ÷ 2 = π.
The peak of sin 2x is at x = π/4. After the move it is at x = π/4 − π/6 = π/12.
Check: at x = π/12, 2x + π/3 = π/6 + π/3 = π/2, and sin(π/2) = 1. This agrees.

Answer: It moves π/6 to the left. The period is π. The first peak is at x = π/12.

y = sin(2x + π/3) (solid) is y = sin 2x (dashed) moved π/6 to the left.

See it three ways

With things

A swing moves back and forth. It goes as far to one side as to the other, so its middle is the line it hangs on. How far it swings is the amplitude. How long one full swing takes is the period.

As a picture

In words

The amplitude is half the distance between the greatest and least values. The period is the length of x for one full wave.

In symbols

For y = a sin bx: amplitude = |a| and period = 2π/b. For y = tan bx: period = π/b.

Try it

See the period change

You need: a graphing app or squared paper, a pencil.

Plot y = sin x for x from 0 to 2π, using 0, π/4, π/2, 3π/4, π and so on up to 2π.
Now plot y = sin 2x at the same values of x. Write 2x for each x first: 0, π/2, π, 3π/2, 2π and so on up to 4π.
Count the peaks of each wave between 0 and 2π.
Double each value of sin x. Those are the heights of y = 2 sin x. Compare them with sin x.

What you will see: The wave y = sin 2x has two peaks where y = sin x has one. So it repeats twice as fast. The values of 2 sin x are twice those of sin x, so its peaks are twice as tall.

If you cannot try it, this is what happens.

A wrong idea many people have

y = sin(x + π/3) is y = sin x moved to the right by π/3.

Why it feels right: A plus sign feels like "go forward", and forward is to the right on a graph.

The test that breaks it: Put x = −π/3 into y = sin(x + π/3). We get sin 0 = 0. So the point (−π/3, 0) is on the graph. The wave that starts at 0 for y = sin x now starts at −π/3, to the left.

What is true: y = f(x + c) moves the graph c units to the left, for positive c. To find the new place of a point, find the x that makes the bracket take its old value.

A wrong idea many people have

The period of y = sin 3x is 6π.

Why it feels right: A number multiplying something usually makes it bigger. Here 3 times 2π is 6π.

The test that breaks it: Put x = 2π/3 into sin 3x. We get sin 2π = 0, the same as at x = 0. The graph is already back at the start after 2π/3, not after 6π.

What is true: The b in sin bx makes the graph repeat b times as fast. The period is 2π ÷ b = 2π/3.

A wrong idea many people have

y = tan x has a greatest value, because the graph goes up.

Why it feels right: Every other graph in this chapter has a top. A graph that rises can seem to stop somewhere.

The test that breaks it: Look at tan x near x = π/2. At 1.5 it is 14.1, at 1.57 it is 1256 and at 1.5707 it is 10 381. The values keep growing as x gets closer to π/2.

What is true: tan x has no greatest and no least value, so it has no amplitude. Its graph goes off to infinity at each asymptote, and the range is every real number.

Key words

period: the length of x after which a graph repeats exactly.
amplitude: the greatest height of a wave above its midline.
midline: the horizontal line halfway between the greatest and least values of a wave.
asymptote: a line that a graph gets closer and closer to without touching it.

Back to your guess

The answer is B: The graph of y = sin 2x repeats twice as often as y = sin x.

y = sin 2x repeats twice as often, so its period is π instead of 2π. It is still no taller than 1. It is y = 2 sin x that is twice as tall. A number inside, next to x, changes the width. A number outside changes the height.

Keep it

Copy this into your exercise book.

y = a sin(bx + c) + d: amplitude |a|, period 2π/b, left by c/b, midline y = d. tan: period π/b, asymptotes at odd multiples of π/2 ÷ b.

Draw: One wave of y = a sin bx with its amplitude, period and midline marked, and a tan graph with dashed asymptotes.

M5.3.4 Chapter 4

Solving trig equations

If sin x = 0.4, what is x, and why are there so many answers?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A There is exactly one value of x, and the calculator gives it.
You picked A. Read on. At the end, the chapter comes back to your guess.
B There are no values of x, because 0.4 is not a sine of a nice angle.
You picked B. Read on. At the end, the chapter comes back to your guess.
C There are two values of x between 0 and 2π, and the calculator gives only one.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Solve sin x = 0.4 for 0 ≤ x ≤ 2π
Solve 2cos²x − cos x − 1 = 0

An equation such as sin x = 0.4 asks: which angles have a sine of 0.4? On the unit circle, sin x is the height of the point. Draw the line at height 0.4, and it cuts the circle in two places. Turn full circles and you find more. So the equation has many solutions, and a question usually gives an interval to keep the list short.

The line y = 0.4 cuts y = sin x at two points between 0 and 2π. They are mirror images in the line x = π/2.

Draw it first

Draw the graph and the line

1 Draw the wave y = sin x from 0 to 2π.
2 Draw the horizontal line y = 0.4 across it.
3 Mark the two places where the line cuts the wave. Read down to the x axis to see the two solutions.

Here is the method. Step 1: get one trig function on its own, equal to a number. Step 2: use the calculator for the acute reference angle α, from the size of the number, ignoring any minus sign. Step 3: use the quadrants where the function has the right sign. Step 4: add or take away full turns until every solution in the interval is found.

Set the calculator to the right mode first: radians if the interval uses π, degrees if it uses degree signs. The key marked sin⁻¹ or arcsin gives the angle for a given sine. It gives the one in the first quadrant, or a negative angle when the number is negative. We use it only for α.

Solutions for 0 ≤ x ≤ 2π, with k above 0 (and at most 1 for sin and cos) and α the acute reference angle. The quadrants are those where the function has that sign. Use the same rows in degrees, with 180° for π and 360° for 2π.

Worked example

Solve sin x = 0.4 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

The calculator is in radian mode. sin⁻¹ 0.4 = 0.41152. This is the reference angle α.
sin is positive in the first and second quadrants. In the first, x = α = 0.41152.
In the second, x = π − α = 3.14159 − 0.41152 = 2.73007.
The next solutions would be 0.4115 + 2π and 2.7301 + 2π, which are more than 2π. So there are only two.
To 3 decimal places, x = 0.412 or x = 2.730.
Check: sin 0.412 = 0.40044 and sin 2.730 = 0.40007. Both are close to 0.4, and the small difference is from rounding.

Answer: x = 0.412 or x = 2.730

The same reference angle gives the answer in the unit circle picture. The two points at height 0.4 are mirror images in the y axis, so their angles are α and π − α.

P has angle α and Q has angle π − α. Both have height sin x = 0.4.

Worked example

Solve tan x = −2 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

The calculator gives tan⁻¹(−2) = −1.10715. This is a negative angle, so it is outside the interval.
The reference angle is α = 1.10715, from the size of the number 2.
tan is negative in the second and fourth quadrants. In the second, x = π − α = 3.14159 − 1.10715 = 2.03444.
In the fourth, x = 2π − α = 6.28319 − 1.10715 = 5.17604.
To 3 decimal places, x = 2.034 or x = 5.176.

Answer: x = 2.034 or x = 5.176

tan x = −2 has two solutions between 0 and 2π, one in each of the second and fourth quadrants, at x = 2.034 and x = 5.176.

When the angle is 2x, 3x or x − 20°, first change the interval. If 0 ≤ x ≤ 2π and the equation has 2x, then 2x runs from 0 to 4π. Solve for 2x in that whole longer interval, so that you find every solution, and only then divide by 2.

Worked example

Solve sin 2x = 0.5 for 0 ≤ x ≤ 2π. Give exact answers.

Let θ = 2x. Then 0 ≤ x ≤ 2π gives 0 ≤ θ ≤ 4π. We solve sin θ = 0.5 in that longer interval.
sin π/6 = 0.5, so α = π/6. In the first two quadrants, θ = π/6 and θ = π − π/6 = 5π/6.
Add one full turn, 2π, to each: θ = π/6 + 2π = 13π/6 and θ = 5π/6 + 2π = 17π/6.
The next ones, 25π/6 and 29π/6, are bigger than 4π, so we stop.
Divide each by 2: x = π/12, 5π/12, 13π/12 and 17π/12.
Check x = 5π/12: 2x = 5π/6 and sin 5π/6 = 0.5. This agrees.

Answer: x = π/12, 5π/12, 13π/12 and 17π/12

sin 2x = 0.5 has four solutions between 0 and 2π, because the graph has two cycles.

Worked example

Solve sin(x − 20°) = 0.6 for 0° ≤ x ≤ 360°. Give the answers to 1 decimal place.

Let θ = x − 20°. Then 0° ≤ x ≤ 360° gives −20° ≤ θ ≤ 340°. We solve sin θ = 0.6 in that interval.
The calculator is in degree mode. sin⁻¹ 0.6 = 36.87°. So in the first quadrant θ = 36.87°.
In the second quadrant θ = 180° − 36.87° = 143.13°.
Both lie between −20° and 340°. The next ones, 396.87° and 503.13°, are too big. The earlier ones, −323.13° and −216.87°, are too small.
Add 20° to each: x = 36.87° + 20° = 56.87° and x = 143.13° + 20° = 163.13°.
To 1 decimal place, x = 56.9° or x = 163.1°.
Check x = 163.1°: x − 20° = 143.1°, and sin 143.1° = 0.6004. This agrees.

Answer: x = 56.9° or x = 163.1°

The wave is y = sin x moved 20° to the right. The line y = 0.6 cuts it at x = 56.9° and x = 163.1°.

Many equations are quadratics in sin x or cos x. Treat sin x or cos x as one unknown. Move everything to one side, factorise, and solve each factor separately. Each factor gives an easy equation of the kind you already know how to solve. Always factorise: never divide away a trig function, because that can lose solutions.

Worked example

Solve 2cos²x − cos x − 1 = 0 for 0 ≤ x ≤ 2π. Give exact answers.

Let c = cos x. Then the equation is 2c² − c − 1 = 0.
Factorise: (2c + 1)(c − 1) = 0. Check by expanding: 2c² − 2c + c − 1 = 2c² − c − 1.
So 2c + 1 = 0 or c − 1 = 0. That gives cos x = −1/2 or cos x = 1.
For cos x = −1/2: the reference angle is π/3. cos is negative in the second and third quadrants, so x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3.
For cos x = 1: the graph touches the top at x = 0 and x = 2π. Both ends are inside the interval.
So x = 0, 2π/3, 4π/3 or 2π.

Answer: x = 0, 2π/3, 4π/3 or 2π

The curve y = 2cos²x − cos x − 1 (called f(x) here) meets the x axis at the four solutions: 0, 2π/3, 4π/3 and 2π.

General solutions give every solution at once, without an interval. Let n stand for any integer: positive, negative or 0. If sin x = sin α, then x = α + 2nπ or x = π − α + 2nπ. If cos x = cos α, then x = ±α + 2nπ. If tan x = tan α, then x = α + nπ. They are the quadrant rules with every full turn added: a full turn is 2π for sin and cos, but only π for tan.

Worked example

Find the general solution of 2cos²x + 3 sin x = 0, where n is any integer.

Replace cos²x by 1 − sin²x. The equation becomes 2(1 − sin²x) + 3 sin x = 0.
Expand: 2 − 2 sin²x + 3 sin x = 0. Move everything to one side: 2 sin²x − 3 sin x − 2 = 0.
Factorise: (2 sin x + 1)(sin x − 2) = 0. Check: 2 sin²x − 4 sin x + sin x − 2 = 2 sin²x − 3 sin x − 2.
So sin x = −1/2 or sin x = 2. The second has no solution, because sin x is never more than 1.
sin x = −1/2: the reference angle is π/6. sin is negative in the third and fourth quadrants.
So x = π + π/6 = 7π/6, or x = 2π − π/6, which is −π/6 + 2π.
Add every full turn: x = −π/6 + 2nπ or x = 7π/6 + 2nπ.

Answer: x = −π/6 + 2nπ or x = 7π/6 + 2nπ

See it three ways

With things

A swing passes the same height twice on each swing, once going up and once coming down. If you are asked at which times the swing is 40 cm high, there are two answers for each full swing.

As a picture

In words

The first solution comes from the calculator. The second is its mirror image in the line x = π/2. Then every full turn gives more.

In symbols

sin x = 0.5: x = π/6 + 2nπ or x = 5π/6 + 2nπ, for n = 0, 1, 2, … and for negative n too.

Try it

Find the second solution on a graph

You need: squared paper, a ruler, a pencil.

Draw y = sin x for x from 0° to 360°, plotting every 30°. Use 1 square for 0.1 up the side.
Draw a horizontal line at height 0.5. Mark the places where the line cuts the wave.
Read the two x values from the graph. They should be near 30° and 150°.
Check each with a calculator: sin 30° and sin 150°.

What you will see: The two solutions are the same distance either side of 90°, the peak of the wave. They are 30° from 0° and 30° from 180°.

If you cannot try it, this is what happens.

A wrong idea many people have

To solve 2 sin x cos x = cos x, divide both sides by cos x. Then 2 sin x = 1 gives x = π/6 and 5π/6.

Why it feels right: Dividing by the same thing on both sides is a normal step in algebra, and it removes a messy factor.

The test that breaks it: Try x = π/2. Then cos x = 0, so both sides are 0, and the equation holds. But the answers above do not include π/2. Dividing by cos x threw away every solution where cos x = 0.

What is true: Move everything to one side and factorise: cos x (2 sin x − 1) = 0. Then cos x = 0 gives x = π/2 and 3π/2, and sin x = 1/2 gives π/6 and 5π/6. There are four solutions.

A wrong idea many people have

sin x = 0.4 has the one solution x = 0.412 in 0 ≤ x ≤ 2π, because that is what the calculator says.

Why it feels right: The calculator gave an answer, and it looks complete.

The test that breaks it: Put x = 2.730 into the calculator: sin 2.730 = 0.4004. This is also a solution, and the calculator did not show it.

What is true: The key sin⁻¹ gives only an angle from −π/2 to π/2 (cos⁻¹ gives one from 0 to π). Use the graph or the unit circle to find the others. For sin and cos, expect two solutions in each full turn. For tan, expect one.

A wrong idea many people have

To solve sin 2x = 0.5 for 0 ≤ x ≤ 2π, find 2x = π/6 and then x = π/12 and 5π/12 only.

Why it feels right: The question is for x, so the interval seems to stay at 0 to 2π when we divide.

The test that breaks it: Check x = 13π/12. Then 2x = 13π/6, and sin 13π/6 = sin π/6 = 0.5. This x is in the interval, but the short list does not include it.

What is true: Change the interval for 2x first: 0 ≤ 2x ≤ 4π. Then find every solution for 2x in that interval and divide each by 2. This gives four solutions.

Key words

general solution: a formula, using an integer n, that gives every solution of a trig equation.
reference angle: the acute angle found from the size of the number, from which the solutions in every quadrant are built.

Back to your guess

The answer is C: There are two values of x between 0 and 2π, and the calculator gives only one.

Between 0 and 2π, the height 0.4 is reached twice, once going up and once coming down. The calculator gives only the first. We find the other from the symmetry of the graph, and the wave then repeats for ever.

Keep it

Copy this into your exercise book.

Reference angle α from |k|. Choose quadrants by CAST. Add full turns. For sin 2x, double the interval first. Factorise, never divide by a trig function that can be 0.

Draw: A sine wave with a horizontal line cutting it twice, and the two solutions marked α and π − α.

M5.3.5 Chapter 5

Identities

Which trig facts are true for every angle, and how do I prove one?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A sin x + cos x = 1 for every angle x.
You picked A. Read on. At the end, the chapter comes back to your guess.
B sin²x + cos²x = 1 for every angle x.
You picked B. Read on. At the end, the chapter comes back to your guess.
C sin x = cos x for every angle x.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use sin²x + cos²x = 1 to simplify an expression
Write tan x as sin x divided by cos x
Prove a simple trig identity

An equation is true for some values of x, such as sin x = 0.4. An identity is true for every value of x for which both sides make sense. We write an identity with the sign ≡, read "is identically equal to". For example, 2(x + 1) ≡ 2x + 2 is true for all x, but 2x + 1 = 7 is true only for x = 3.

Trig has a few key identities. The first is a form of Pythagoras. A point P on the unit circle has coordinates (cos x, sin x), and its distance from the origin is 1. So the square of the x coordinate plus the square of the y coordinate is 1.

P = (cos θ, sin θ) lies on the circle of radius 1. So (cos θ)² + (sin θ)² = 1².

The same fact for an acute angle, from Pythagoras: sin²θ + cos²θ = 1², with θ at the angle opposite the side sin θ.

We write sin²x for (sin x)². Then the first identity is sin²x + cos²x ≡ 1. It is true for every x, in radians or in degrees, since P is always on the circle. Rearranging gives sin²x ≡ 1 − cos²x and cos²x ≡ 1 − sin²x. These three are the same fact. Use whichever one fits.

sin²x and cos²x rise and fall, but when one is high the other is low. Together they always make 1: the level line named sum.

The second identity comes from the definition of tan. tan x ≡ sin x ÷ cos x, for every x where cos x is not 0. It is the gradient of OP: the rise sin x divided by the run cos x. At the values where cos x = 0, the left side is not defined.

The identities to learn in this chapter. The last one needs cos x not equal to 0.

Now we prove identities. To prove an identity, show that the left side (LHS) can be changed into the right side (RHS) by steps that are always valid. Start with one side only, usually the more complicated one, and change it step by step until it matches the other. Or work on each side separately until both reach the same expression. Do not start from the identity and rearrange it, because that assumes what you are trying to prove.

Worked example

Prove that (sin x + cos x)² ≡ 1 + 2 sin x cos x.

Start with the left side: (sin x + cos x)². Expand the bracket: sin²x + 2 sin x cos x + cos²x.
Put the two squares together: (sin²x + cos²x) + 2 sin x cos x.
Use sin²x + cos²x ≡ 1. The left side becomes 1 + 2 sin x cos x.
This is the right side. So the identity is proved.

Answer: The left side becomes the right side, so the identity is proved.

When an identity has tan, write it in sin and cos first. When it has fractions, put them over a common denominator. These two moves solve many proofs.

Worked example

Prove that tan x + 1/tan x ≡ 1/(sin x cos x).

Start with the left side. Write tan x as sin x ÷ cos x. Then 1/tan x is cos x ÷ sin x.
So the left side is sin x/cos x + cos x/sin x.
Use the common denominator sin x cos x. The first fraction gets sin x on top and bottom, the second gets cos x.
This gives (sin²x + cos²x)/(sin x cos x).
Use sin²x + cos²x ≡ 1. The top becomes 1, so the left side is 1/(sin x cos x).
This is the right side. So the identity is proved, for x where sin x and cos x are not 0.

Answer: The left side becomes the right side, so the identity is proved.

Some proofs need a trick. If a fraction has 1 − sin x on the bottom, multiply top and bottom by 1 + sin x. The bottom then becomes (1 − sin x)(1 + sin x) = 1 − sin²x, which is cos²x. This is a difference of two squares, and it is the key move in many exam questions.

Worked example

Prove that cos x/(1 − sin x) ≡ (1 + sin x)/cos x.

Start with the left side. Multiply top and bottom by 1 + sin x. This is allowed, because it is the same as multiplying by 1.
The top becomes cos x (1 + sin x). The bottom becomes (1 − sin x)(1 + sin x) = 1 − sin²x.
Use cos²x ≡ 1 − sin²x. The bottom is cos²x.
So the left side is cos x (1 + sin x)/cos²x.
Cancel one cos x from the top and the bottom. This leaves (1 + sin x)/cos x.
This is the right side. So the identity is proved, for x where cos x is not 0 and sin x is not 1.

Answer: The left side becomes the right side, so the identity is proved.

To show that something is not an identity, one value of x is enough. We call it a counterexample. Take sin x + cos x ≡ 1. At x = π/4 the left side is √2/2 + √2/2 = √2, which is not 1. So it is not an identity, even though it is true at x = 0.

Identities also solve equations. If an equation has sin²x and cos x, replace sin²x by 1 − cos²x. Then every term is in terms of cos x, and the equation becomes a quadratic in cos x. This is what the second goal of chapter 4 needed.

Worked example

Solve 2 sin²x + 3 cos x = 3 for 0 ≤ x ≤ 2π. Give exact answers.

The equation has sin²x and cos x. Replace sin²x by 1 − cos²x.
We get 2(1 − cos²x) + 3 cos x = 3, which is 2 − 2cos²x + 3 cos x = 3.
Move everything to one side and tidy: 0 = 2cos²x − 3 cos x + 1.
Factorise: (2 cos x − 1)(cos x − 1) = 0. Check: 2cos²x − 2 cos x − cos x + 1 = 2cos²x − 3 cos x + 1.
So cos x = 1/2 or cos x = 1.
cos x = 1/2 gives x = π/3 and x = 2π − π/3 = 5π/3. cos x = 1 gives x = 0 and x = 2π.
So x = 0, π/3, 5π/3 or 2π.

Answer: x = 0, π/3, 5π/3 or 2π, which are 0, 1.047, 5.236 and 6.283

The curve y = 2 sin²x + 3 cos x (called f(x) here) meets the line y = 3 at the four solutions.

Worked example

Solve tan x = 2 sin x for 0 ≤ x ≤ 2π. Give exact answers.

Write tan x as sin x ÷ cos x. The equation is sin x/cos x = 2 sin x, where cos x is not 0.
Do not divide by sin x, because sin x may be 0. Move everything to one side: sin x/cos x − 2 sin x = 0.
Take out the common factor sin x: sin x (1/cos x − 2) = 0.
So sin x = 0, or 1/cos x = 2. The second gives cos x = 0.5.
sin x = 0 gives x = 0, π and 2π. cos x = 1/2 gives x = π/3 and x = 5π/3.
None of these has cos x = 0, so all are allowed. In order: x = 0, π/3, π, 5π/3 or 2π.
Check x = π/3: tan(π/3) = √3 and 2 sin(π/3) = 2 × √3/2 = √3. This agrees.

Answer: x = 0, π/3, π, 5π/3 or 2π, which are 0, 1.047, 3.142, 5.236 and 6.283

y = tan x and y = 2 sin x meet five times between 0 and 2π.

See it three ways

With things

Take a ladder of length 1 against a wall. However it leans, the height up the wall and the distance from the wall are the two shorter sides of a right angled triangle with hypotenuse 1. The squares of those two lengths always add to 1.

As a picture

In words

The squares of the sine and the cosine of an angle always add up to 1.

In symbols

sin²θ + cos²θ ≡ 1. Dividing by cos²θ for tan: tan²θ + 1 ≡ 1/cos²θ.

Try it

Check the identity with a calculator

You need: a calculator, a pencil.

Choose any angle, such as 25°. Press sin 25 and square the answer. Write it down.
Press cos 25 and square the answer. Write it down.
Add the two squares.
Try again with 100° and with 200°.

What you will see: The sum is 1 every time, up to the last digit shown. Now try sin 25 + cos 25: it is 1.329, not 1.

If you cannot try it, this is what happens.

A wrong idea many people have

Since sin²x + cos²x = 1, taking square roots gives sin x + cos x = 1.

Why it feels right: The square root cancels the squares, and it seems to give each term back.

The test that breaks it: At x = π/4: sin²x + cos²x = 1/2 + 1/2 = 1, but sin x + cos x = √2/2 + √2/2 = √2, about 1.414.

What is true: The square root of a sum is not the sum of the square roots. In general √(a² + b²) ≠ a + b. The identity stays with the squares.

A wrong idea many people have

To prove an identity, write LHS = RHS and rearrange until you reach something obviously true.

Why it feels right: That is how we solve an equation, and it usually works.

The test that breaks it: Take 1 = −1. Square both sides: 1 = 1. This is true, but 1 = −1 is false. Working from the result you want can reach a true statement even when the result is false.

What is true: Start with one side and change it into the other, with each step an equality. Or change each side separately to the same expression. Then the chain of equal expressions is a proof.

A wrong idea many people have

In the third proof I can cancel the sin x in 1 − sin x with the sin x in 1 + sin x.

Why it feels right: The same letters sit on the top and the bottom of the fraction, and cancelling looks natural.

The test that breaks it: Put sin x = 0.5. Then 1 − sin x = 0.5 and 1 + sin x = 1.5. The two are different. Only whole factors can be cancelled, and sin x is a term, not a factor.

What is true: Cancel only factors that multiply the whole top and the whole bottom. Here, multiply by (1 + sin x) first, to turn the bottom into cos²x. Then one cos x is a factor of the top and the bottom.

Key words

identity: an equation that is true for every value of x for which both sides are defined, written with ≡.
counterexample: one value that makes a statement false, which shows that it is not an identity.

Back to your guess

The answer is B: sin²x + cos²x = 1 for every angle x.

Only sin²x + cos²x = 1 is true for every angle. Put x = 0: sin 0 + cos 0 = 1 holds, but at x = π/4 the sum sin x + cos x is √2, not 1. The square version comes from Pythagoras, and it never fails.

Keep it

Copy this into your exercise book.

sin²x + cos²x ≡ 1. tan x ≡ sin x ÷ cos x. Prove: work on one side, tan into sin and cos, common denominators, multiply by 1 + sin x. One counterexample disproves.

Draw: A right angled triangle with hypotenuse 1 and sides sin θ and cos θ.

M5.3.6 Chapter 6

Reciprocal and inverse trig functions

What are sec, cosec and cot, and what does arcsin x mean?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A sin⁻¹ x means the angle whose sine is x.
You picked A. Read on. At the end, the chapter comes back to your guess.
B sin⁻¹ x means 1 ÷ sin x.
You picked B. Read on. At the end, the chapter comes back to your guess.
C sin⁻¹ x means the same as sin x, written another way.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Sketch the graphs of sec, cosec and cot
Use 1 + tan²x = sec²x
Give the domain and range of arcsin x

Three more trig functions are the reciprocals of the first three. Reciprocal means 1 divided by it. We define sec x = 1/cos x and cosec x = 1/sin x. We define cot x = cos x/sin x, which is 1/tan x wherever tan x is defined and not 0. Each is defined only where its bottom is not 0.

Where is each defined? The function sec x needs cos x ≠ 0, so x cannot be π/2, 3π/2 and so on. The function cosec x needs sin x ≠ 0, so x cannot be 0, π, 2π and so on. The same holds for cot x. At x = π/2, cot x is defined and equals 0, even though tan x is not defined there.

To sketch the graph of a reciprocal, start with its parent. Where the parent is 1, the reciprocal is 1. Where the parent is −1, so is the reciprocal. Where the parent is 0, the reciprocal has an asymptote. Where the parent is small, the reciprocal is large. The parent and its reciprocal always have the same sign.

y = sec x (solid) from y = cos x (dashed). The asymptotes are where cos x = 0. The curve never enters the strip between −1 and 1.

Two facts follow. Since cos x lies between −1 and 1, the size of sec x = 1/cos x is at least 1. So the range of sec x is y ≤ −1 or y ≥ 1. And the period is 2π, like cos x. The graph has U shaped branches that open up where cos is positive and open down where it is negative.

y = cosec x (solid) from y = sin x (dashed). The asymptotes are at x = 0, π and 2π, where sin x = 0.

y = cot x. It has period π and asymptotes at x = 0, π and 2π. It falls from left to right, and it is 0 at π/2 and 3π/2.

Now two new identities. Start with sin²x + cos²x ≡ 1 and divide every term by cos²x. This gives tan²x + 1 ≡ sec²x. Divide by sin²x instead, and we get 1 + cot²x ≡ cosec²x. The first needs cos x ≠ 0, and the second needs sin x ≠ 0. Learn both: they turn tan into sec, and cot into cosec, in equations.

Worked example

Find the exact values of sec 120°, cosec 150° and cot 135°.

Find the ordinary ratio first, then take the reciprocal.
cos 120° = −1/2, because 120° is in the second quadrant with reference angle 60°.
So sec 120° = 1 ÷ (−1/2) = −2.
sin 150° = 1/2, because 150° has reference angle 30° and sin is positive in the second quadrant. So cosec 150° = 1 ÷ (1/2) = 2.
tan 135° = −1, because the reference angle is 45° and tan is negative in the second quadrant. So cot 135° = 1 ÷ (−1) = −1.

Answer: sec 120° = −2, cosec 150° = 2, cot 135° = −1.

Worked example

Given that tan θ = −3/4 and θ is obtuse, find sec θ and cos θ.

Use 1 + tan²θ ≡ sec²θ. Here tan²θ = (−3/4)² = 9/16.
So sec²θ = 1 + 9/16 = 25/16.
Then sec θ = 5/4 or sec θ = −5/4, because both have a square of 25/16.
θ is obtuse, so it is in the second quadrant, where cos is negative. So sec θ is negative too: sec θ = −5/4.
Then cos θ = 1 ÷ sec θ = 1 ÷ (−5/4) = −4/5.

Answer: sec θ = −5/4 and cos θ = −4/5.

The acute angle φ has tan φ = 3/4 and sec φ = 5/4. The obtuse angle θ = 180° − φ has the same sizes with the signs of the second quadrant.

Exam questions combine the new identities with equations. Replace tan²x by sec²x − 1, or cot²x by cosec²x − 1, so that only one function is left. Then factorise. Remember that |sec x| ≥ 1 and |cosec x| ≥ 1: a factor that gives a value between −1 and 1 has no solution.

Worked example

Solve 3 tan²x + 5 sec x + 1 = 0 for 0 ≤ x ≤ 2π. Give exact answers.

Replace tan²x by sec²x − 1. The equation becomes 3(sec²x − 1) + 5 sec x + 1 = 0.
Expand and collect: 3 sec²x − 3 + 5 sec x + 1 = 0, so 3 sec²x + 5 sec x − 2 = 0.
This is a quadratic in sec x. Factorise: (3 sec x − 1)(sec x + 2) = 0.
So sec x = 1/3 or sec x = −2.
sec x = 1/3 has no solution, because |sec x| ≥ 1. We reject it.
sec x = −2 gives cos x = −1/2. The reference angle is π/3, and cos is negative in the second and third quadrants.
So x = π − π/3 = 2π/3 or x = π + π/3 = 4π/3.

Answer: x = 2π/3 or x = 4π/3

Now the inverse functions. The function sin x takes many angles to the same value. For example sin 30° = sin 150° = 1/2. A function has an inverse only if each output comes from exactly one input. So we cut sin x down to one stretch where it takes each value once: from −π/2 to π/2. On that stretch it rises from −1 to 1 and covers every value once.

On the solid part, from −π/2 to π/2, sin x takes each value from −1 to 1 exactly once. That part has an inverse.

The inverse of this restricted sine is arcsin, also written sin⁻¹. arcsin x is the angle between −π/2 and π/2 whose sine is x. Its domain is the set of x that go in: −1 ≤ x ≤ 1. Its range is the set of values that come out: −π/2 ≤ y ≤ π/2. Its graph is the restricted sine reflected in the line y = x.

y = arcsin x: domain −1 ≤ x ≤ 1, range −π/2 ≤ y ≤ π/2.

The same idea gives the others. cos x is cut to 0 ≤ x ≤ π, where it falls from 1 to −1. Its inverse arccos x has domain −1 ≤ x ≤ 1 and range 0 ≤ y ≤ π. tan x is cut to −π/2 < x < π/2, where it rises through every real value. Its inverse arctan x has domain all real numbers and range −π/2 < y < π/2.

y = arccos x: domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π. It falls from π to 0.

y = arctan x: domain all real numbers, range −π/2 < y < π/2. The lines y = π/2 and y = −π/2 are asymptotes.

The ranges matter. A calculator, and arcsin, arccos and arctan, give one answer each, taken from the range. So arcsin(sin x) = x only when x is between −π/2 and π/2. For other x, the answer is the angle in the range that has the same sine.

Worked example

Find the exact values of arcsin(−1/2), arccos(−1/2) and arctan 1 in radians.

arcsin(−1/2) is the angle between −π/2 and π/2 whose sine is −1/2. sin(π/6) = 1/2, and the sine is negative below the axis. So the angle is −π/6.
arccos(−1/2) is the angle between 0 and π whose cosine is −1/2. cos(π/3) = 1/2, and cos is negative in the second quadrant. So the angle is π − π/3 = 2π/3.
arctan 1 is the angle between −π/2 and π/2 whose tan is 1. tan(π/4) = 1, so the angle is π/4.

Answer: −π/6, 2π/3 and π/4

Worked example

Find the exact value of arcsin(sin(2π/3)).

First sin(2π/3). The angle is in the second quadrant, with reference angle π/3. So sin(2π/3) = √3/2.
Now arcsin(√3/2) is the angle between −π/2 and π/2 whose sine is √3/2. That angle is π/3.
So arcsin(sin(2π/3)) = π/3, not 2π/3. The angle 2π/3 is outside the range of arcsin.

Answer: π/3, which is 1.047

Worked example

Find the exact value of sin(arccos(3/5)).

Let θ = arccos(3/5). Then cos θ = 3/5, and θ is between 0 and π.
sin θ is not negative between 0 and π, so sin θ = +√(1 − cos²θ).
cos²θ = 9/25, so sin²θ = 1 − 9/25 = 16/25.
So sin θ = 4/5.

Answer: 4/5

Class 12 courses also use the other three inverses. The inverse sec⁻¹ x has domain |x| ≥ 1 and range 0 ≤ y ≤ π, with y ≠ π/2. The inverse cosec⁻¹ x has domain |x| ≥ 1 and range −π/2 ≤ y ≤ π/2, with y ≠ 0. The inverse cot⁻¹ x has domain all real numbers and range 0 < y < π. These are the usual principal ranges, so check the ranges your course uses.

Some results hold wherever both sides are defined. arcsin x + arccos x = π/2 for −1 ≤ x ≤ 1. To see this, let y = arcsin x, so sin y = x. Then cos(π/2 − y) = x, and π/2 − y lies between 0 and π, so arccos x = π/2 − y. Also arcsin(−x) = −arcsin x, and arccos(−x) = π − arccos x.

For the tangent, arctan x + arctan y = arctan((x + y)/(1 − xy)) when xy < 1. It comes from tan(A + B) = (tan A + tan B)/(1 − tan A tan B). When xy > 1 and x and y are positive, the sum is more than π/2, and it equals π + arctan((x + y)/(1 − xy)). When xy > 1 and both are negative, take away π instead. The next example shows this.

Worked example

Find the exact value of arctan 2 + arctan 3.

Let A = arctan 2 and B = arctan 3. Both are acute, so A + B lies between 0 and π.
tan(A + B) = (2 + 3)/(1 − 2 × 3) = 5/(−5) = −1.
Between 0 and π, the angle with tangent −1 is 3π/4. So A + B = 3π/4.
The formula arctan((x + y)/(1 − xy)) would give arctan(−1) = −π/4, which is wrong. It needs xy < 1, and here xy = 6.

Answer: 3π/4

See it three ways

With things

A lift takes you from a floor number to a height. To go back, from a height to a floor number, you reverse it. That reverse is the inverse. It works only if each height belongs to just one floor.

As a picture

In words

arcsin x is the angle between −π/2 and π/2 whose sine is x.

In symbols

y = arcsin x means sin y = x with −π/2 ≤ y ≤ π/2. Domain −1 ≤ x ≤ 1.

Try it

Find an angle with the inverse key

You need: a calculator, paper.

Set the calculator to radian mode. Press sin⁻¹ 0.5 and write the answer. It should be 0.5236, which is π/6.
Press sin⁻¹ of the sine of 2.5 radians. First work out sin 2.5, then press sin⁻¹ of that answer.
Compare the answer with 2.5.
Subtract: π − 2.5 = 0.6416. Compare with your answer.

What you will see: The calculator gives 0.6416, not 2.5. The angle 2.5 is outside the range of arcsin, so the inverse key gives the angle in the range with the same sine, which is π − 2.5.

If you cannot try it, this is what happens.

A wrong idea many people have

sin⁻¹ x is 1/sin x.

Why it feels right: In algebra, a number to the power −1 is its reciprocal. So x⁻¹ = 1/x, and it seems that sin⁻¹ should be the same.

The test that breaks it: Put x = 0.5. Then 1/sin 0.5 = 1/0.4794 = 2.086. But the calculator key sin⁻¹ gives sin⁻¹ 0.5 = 0.5236. These are different numbers, and the key is the inverse function.

What is true: sin⁻¹ x means arcsin x, the inverse function. The reciprocal is written (sin x)⁻¹ or cosec x or 1/sin x. In this book we write arcsin to keep the two apart.

A wrong idea many people have

arcsin(sin x) = x for every x.

Why it feels right: A function and its inverse undo each other, so they should cancel.

The test that breaks it: Take x = 2π/3. sin(2π/3) = √3/2, and arcsin(√3/2) = π/3. The answer is π/3, not 2π/3.

What is true: arcsin(sin x) = x only when −π/2 ≤ x ≤ π/2. For other x, arcsin(sin x) is the angle in that range with the same sine. On the other side, sin(arcsin x) = x for every x from −1 to 1.

A wrong idea many people have

The range of sec x is −1 ≤ y ≤ 1, like cos x.

Why it feels right: sec x is built from cos x, so the values may look alike.

The test that breaks it: Put x = π/3. Then cos x = 1/2, and sec x = 1 ÷ (1/2) = 2, which is outside −1 to 1. Dividing 1 by a number of size at most 1 gives a number of size at least 1.

What is true: The range of sec x is y ≤ −1 or y ≥ 1. The graph has no points in the strip between y = −1 and y = 1. The same holds for cosec x. cot x takes every real value.

Key words

reciprocal: 1 divided by a number. The reciprocal of 2 is 1/2.
inverse function: the function that undoes another. arcsin is the inverse of sin on −π/2 ≤ x ≤ π/2.
domain: the set of values that can go into a function.
range: the set of values that can come out of a function.

Back to your guess

The answer is A: sin⁻¹ x means the angle whose sine is x.

sin⁻¹ x means the angle whose sine is x. It is the inverse function of sin, also written arcsin x. The reciprocal 1 ÷ sin x has its own name: cosec x. The same small −1 is used for both ideas in different places, which is a common cause of mistakes.

Keep it

Copy this into your exercise book.

sec = 1/cos, cosec = 1/sin, cot = 1/tan. 1 + tan²x = sec²x, 1 + cot²x = cosec²x. arcsin: domain −1 to 1, range −π/2 to π/2. arccos range 0 to π. arctan range −π/2 to π/2.

Draw: The graph of sec x with its asymptotes and the cos x curve dashed beneath it.

M5.3.7 Chapter 7

Compound and double angles

If I know sin A and sin B, how do I find sin(A + B), and why is it not sin A + sin B?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A sin(A + B) = sin A + sin B.
You picked A. Read on. At the end, the chapter comes back to your guess.
B sin(A + B) = sin A × sin B.
You picked B. Read on. At the end, the chapter comes back to your guess.
C sin(A + B) = sin A cos B + cos A sin B.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use the formula for sin(A + B)
Use double angle formulas to solve an equation

We often know the trig ratios of two angles, A and B, and need those of A + B or A − B. A compound angle is a sum or difference of angles like these. The ratio of a sum is not the sum of the ratios, so we need new formulas. They all follow from one fact about the unit circle.

Take two points on the unit circle, P at angle A and Q at angle B. Then P = (cos A, sin A) and Q = (cos B, sin B). The angle POQ at the centre is A − B. We find the distance PQ in two ways.

OP makes angle A with the x axis and OQ makes angle B. The angle POQ between them is A − B.

First use the cosine rule in triangle POQ. Here OP = OQ = 1 and the angle at O is A − B. So PQ² = 1² + 1² − 2 × 1 × 1 × cos(A − B) = 2 − 2 cos(A − B). Second, use the distance formula. Then PQ² = (cos A − cos B)² + (sin A − sin B)². Expand this and use sin²x + cos²x = 1 twice. We get PQ² = (cos²A + sin²A) + (cos²B + sin²B) − 2 cos A cos B − 2 sin A sin B = 2 − 2(cos A cos B + sin A sin B).

The two results for PQ² are equal. So 2 − 2 cos(A − B) = 2 − 2(cos A cos B + sin A sin B). This gives the first formula: cos(A − B) ≡ cos A cos B + sin A sin B.

The others follow. Replace B by −B. Since cos(−B) = cos B and sin(−B) = −sin B, we get cos(A + B) ≡ cos A cos B − sin A sin B. For the sine, use the fact that the reflection of the point at angle θ in the line y = x is the point at angle 90° − θ. Reflection swaps the coordinates, so cos(90° − θ) = sin θ and sin(90° − θ) = cos θ.

Then sin(A + B) = cos(90° − (A + B)) = cos((90° − A) − B). Use the formula for cos(X − Y) with X = 90° − A and Y = B. This gives cos(90° − A) cos B + sin(90° − A) sin B, which is sin A cos B + cos A sin B. Replacing B by −B gives sin(A − B). Dividing sin(A ± B) by cos(A ± B) gives tan.

The compound angle formulas. The sign in sin matches the sign inside the bracket. In cos it is the opposite sign.

For the tangent, divide the top and the bottom of sin(A + B)/cos(A + B) by cos A cos B. This gives tan(A + B) ≡ (tan A + tan B)/(1 − tan A tan B), and tan(A − B) ≡ (tan A − tan B)/(1 + tan A tan B). They hold where every tangent in them is defined and the bottom is not 0.

The formulas are not the obvious guess. Here is why the guess sin(A + B) = sin A + sin B fails. The graph of y = sin(x + π/3) is not the graph of y = sin x raised by the number sin(π/3). The first has height 1 at its peak, and the second has height 1 + √3/2.

sin(x + π/3) (solid) and sin x + sin(π/3) (dashed) are different curves. The second is too high: at its peak it is above 1.

Worked example

Find the exact value of sin 75°.

Write 75° as 45° + 30°, two angles whose exact values we know.
Use sin(A + B) = sin A cos B + cos A sin B with A = 45° and B = 30°.
sin 75° = sin 45° cos 30° + cos 45° sin 30°.
Put in the exact values: (√2/2)(√3/2) + (√2/2)(1/2).
Multiply: √6/4 + √2/4.
Add the fractions: sin 75° = (√6 + √2)/4.
Check with a calculator: (2.4495 + 1.4142) ÷ 4 = 0.9659, and sin 75° = 0.9659.

Answer: sin 75° = (√6 + √2)/4

Often you know sin A and cos B, not both of each. Use sin²x + cos²x = 1 to find the missing ratios. And you must choose the correct sign for each, from the quadrant the problem names.

Worked example

A is acute with sin A = 3/5. B is obtuse with cos B = −12/13. Find sin(A + B).

We need sin A, cos A, sin B and cos B. We know sin A = 3/5 and cos B = −12/13.
A is acute, so cos A is positive: cos²A = 1 − 9/25 = 16/25, so cos A = 4/5.
B is obtuse, so sin B is positive: sin²B = 1 − 144/169 = 25/169, so sin B = 5/13.
sin(A + B) = sin A cos B + cos A sin B = (3/5)(−12/13) + (4/5)(5/13).
The top of the first term is 3 × (−12) = −36. The top of the second term is 4 × 5 = 20.
Both terms have the bottom 5 × 13 = 65. So sin(A + B) = (−36 + 20)/65 = −16/65.

Answer: sin(A + B) = −16/65

The triangles give sin A = 3/5 and cos A = 4/5 for the acute angle A, and cos X = 12/13 and sin X = 5/13. The obtuse angle B = 180° − X has cos B = −12/13 and sin B = 5/13: the same sizes, with the signs of the second quadrant.

Worked example

A and B are acute angles with tan A = 1/2 and tan B = 1/3. Show that A + B = π/4.

Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B).
The top is 1/2 + 1/3 = 5/6.
The bottom is 1 − (1/2)(1/3) = 1 − 1/6 = 5/6.
So tan(A + B) = (5/6) ÷ (5/6) = 1.
A and B are acute, so A + B is between 0 and π. The only angle in that range with tangent 1 is π/4.
So A + B = π/4.

Answer: tan(A + B) = 1, so A + B = π/4.

Now put B = A in the compound formulas. This gives the double angle formulas. sin 2A = sin A cos A + cos A sin A, so sin 2A ≡ 2 sin A cos A. cos 2A = cos²A − sin²A. Using sin²A + cos²A = 1, this also has two other forms: cos 2A ≡ 2cos²A − 1 and cos 2A ≡ 1 − 2 sin²A. And tan 2A ≡ 2 tan A/(1 − tan²A).

The double angle formulas. Use cos²A − sin²A when both appear, 2cos²A − 1 when only cos A appears, and 1 − 2 sin²A when only sin A appears.

Choose the form of cos 2A that leaves one trig function. If the equation has cos x and cos 2x, use cos 2x = 2cos²x − 1. If it has sin x and cos 2x, use 1 − 2 sin²x. Then solve the quadratic, exactly as in the last two chapters. The double angle formulas also make sin²A = (1 − cos 2A)/2 and cos²A = (1 + cos 2A)/2.

Worked example

Solve 3 cos 2x + cos x + 2 = 0 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

The equation has cos 2x and cos x. Use cos 2x = 2cos²x − 1, which has only cos x.
We get 3(2cos²x − 1) + cos x + 2 = 0, which is 6cos²x − 3 + cos x + 2 = 0.
This is a quadratic in cos x: 6cos²x + cos x − 1 = 0, because −3 + 2 = −1.
Factorise: (3 cos x − 1)(2 cos x + 1) = 0. Check: 6cos²x + 3 cos x − 2 cos x − 1 = 6cos²x + cos x − 1.
So cos x = 1/3 or cos x = −1/2.
cos x = 1/3: the calculator gives 1.2310. cos is positive in the first and fourth quadrants, so x = 1.231 and x = 2π − 1.2310 = 5.052.
cos x = −1/2: the reference angle is π/3. So x = π − π/3 = 2π/3 = 2.094 and x = π + π/3 = 4π/3 = 4.189.
In order: x = 1.231, 2.094, 4.189 or 5.052.

Answer: x = 1.231, 2.094, 4.189 or 5.052

The curve y = 3 cos 2x + cos x + 2 (called f(x) here) crosses the x axis at the four solutions.

Worked example

Prove that (sin A + cos A)² ≡ 1 + sin 2A.

Start with the left side and expand: sin²A + 2 sin A cos A + cos²A.
Use sin²A + cos²A ≡ 1 for the squares: 1 + 2 sin A cos A.
Use sin 2A ≡ 2 sin A cos A for the last term: 1 + sin 2A.
This is the right side. So the identity is proved.

Answer: The left side becomes the right side, so the identity is proved.

Adding two compound formulas gives the factor formulas. Write C = P + Q and D = P − Q, so P = (C + D)/2 and Q = (C − D)/2. Then sin(P + Q) + sin(P − Q) = 2 sin P cos Q. In the new letters: sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2). The same steps give three more. Next, sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2). Then cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2). Last, cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2).

Worked example

Show that sin 75° + sin 15° = √6/2.

Use sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2) with C = 75° and D = 15°.
(C + D)/2 = 45° and (C − D)/2 = 30°.
So sin 75° + sin 15° = 2 sin 45° cos 30° = 2 × (√2/2) × (√3/2).
This is 2√6/4 = √6/2.
Check: sin 75° = 0.9659 and sin 15° = 0.2588. The sum is 1.2247, and √6/2 = 1.2247.

Answer: sin 75° + sin 15° = √6/2

See it three ways

With things

Two turns of a wheel, first by A and then by B, are the same as one turn by A + B. The height after both turns depends on the position after the first turn and the extra turn, so it needs both sines and both cosines.

As a picture

In words

The distance between two points on the unit circle depends on the difference of their angles. Working it out two ways gives the cosine of a difference.

In symbols

cos(A − B) = cos A cos B + sin A sin B. Then sin(A + B) = sin A cos B + cos A sin B. Put B = A: sin 2A = 2 sin A cos A.

Try it

Test the formula with a calculator

You need: a calculator, a pencil.

Choose A = 50° and B = 20°. Work out sin 50° and cos 50°, sin 20° and cos 20°.
Work out sin A cos B + cos A sin B using your numbers.
Work out sin 70° directly. Compare.
Work out sin A + sin B and compare it with sin 70°.

What you will see: The formula gives 0.9397, the same as sin 70°. The sum sin 50° + sin 20° = 1.108 is not even a possible sine.

If you cannot try it, this is what happens.

A wrong idea many people have

sin(A + B) = sin A + sin B, and sin 2A = 2 sin A.

Why it feels right: In algebra, 2(x + y) = 2x + 2y, so a function seems to pass through a bracket in the same way.

The test that breaks it: Put A = B = 45°. Then sin 90° = 1, but sin 45° + sin 45° = 1.414, which is more than 1. And 2 sin 45° = 1.414, while sin 90° = 1.

What is true: sin is a function, not a number, so it does not distribute over a sum. The right formulas are sin(A + B) = sin A cos B + cos A sin B and sin 2A = 2 sin A cos A.

A wrong idea many people have

cos(A + B) = cos A cos B + sin A sin B.

Why it feels right: It looks like the formula for sin(A + B), and a plus sign inside the bracket seems to give a plus sign.

The test that breaks it: Put A = B = 45°. Then cos 90° = 0. But cos 45° cos 45° + sin 45° sin 45° = 1/2 + 1/2 = 1. The right side is wrong.

What is true: In the cosine formulas the sign is the opposite of the sign in the bracket: cos(A + B) = cos A cos B − sin A sin B. Test every formula with A = B = 45° if you are unsure.

A wrong idea many people have

cos 2A = 2 cos A, or cos 2A = cos²A.

Why it feels right: A double angle seems to double the result, and squares are in the formula.

The test that breaks it: Put A = 60°. cos 120° = −1/2. But 2 cos 60° = 1, and cos²60° = 1/4. Neither matches.

What is true: cos 2A has three correct forms: cos²A − sin²A, 2cos²A − 1 and 1 − 2 sin²A. At A = 60°: 1/4 − 3/4 = −1/2, 2/4 − 1 = −1/2, and 1 − 6/4 = −1/2.

Key words

compound angle: a sum or difference of two angles, such as A + B or A − B.
double angle: the angle 2A, found by putting B = A in a compound angle formula.

Back to your guess

The answer is C: sin(A + B) = sin A cos B + cos A sin B.

The formula is sin(A + B) = sin A cos B + cos A sin B. Test option A: A = B = 90° gives sin 180° = 0, but sin 90° + sin 90° = 2. Option B gives 1 × 1 = 1. Only option C gives 1 × 0 + 0 × 1 = 0.

Keep it

Copy this into your exercise book.

sin(A ± B) = sin A cos B ± cos A sin B. cos(A + B) = cos A cos B − sin A sin B. For A − B, use plus. sin 2A = 2 sin A cos A. cos 2A = 2cos²A − 1 = 1 − 2 sin²A.

Draw: Two points on a unit circle at angles A and B joined by a chord.

M5.3.8 Chapter 8

Combining sin and cos

What do I get when I add a sine wave and a cosine wave?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The sum of 3 sin x and 4 cos x is a wave that never rises above 7.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The sum of 3 sin x and 4 cos x is a wave with a greatest value of 5.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The sum of 3 sin x and 4 cos x is not a wave at all.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Write 3 sin x + 4 cos x as 5 sin(x + α)
Find the largest value of 3 sin x + 4 cos x

Waves in nature often add together: two sounds, two tides, two light beams. If a sine wave and a cosine wave have the same period, their sum is again a single wave of the same period. We want to write it in harmonic form, a single sine R sin(x + α), because then its greatest value, its shape and its solutions are easy to read.

The two dashed waves, 3 sin x and 4 cos x, add up to the solid wave. It is a sine wave of the same period, with a peak of 5.

Suppose we want a sin x + b cos x ≡ R sin(x + α), where R is positive. Expand the right side with sin(A + B) = sin A cos B + cos A sin B. We get R sin x cos α + R cos x sin α. This is (R cos α) sin x + (R sin α) cos x.

Both sides are equal for every x, so the numbers in front of sin x agree, and so do those in front of cos x. So R cos α = a and R sin α = b. Square and add: R²cos²α + R²sin²α = a² + b², so R² = a² + b². Divide: (R sin α)/(R cos α) = b/a, so tan α = b/a.

R cos α = a and R sin α = b, so a and b are the two shorter sides of a right angled triangle with hypotenuse R. Here a = 3 and b = 4.

The method is: find R = √(a² + b²), then find α from R cos α = a and R sin α = b. When a and b are both positive, α is acute and tan α = b/a gives it directly. Always keep R positive. Always check the signs of cos α and sin α, because tan α = b/a alone cannot tell the quadrant.

Worked example

Write 3 sin x + 4 cos x in the form R sin(x + α), with R > 0 and α acute, and give α in radians to 3 decimal places.

Match 3 sin x + 4 cos x with R sin(x + α) = R cos α sin x + R sin α cos x.
So R cos α = 3 and R sin α = 4.
Square and add: R² = 3² + 4² = 25.
Since R is positive, R = 5.
Divide: tan α = 4/3. Both R cos α and R sin α are positive, so α is acute.
In radian mode, α = tan⁻¹(4/3) = 0.9273, which is 0.927 to 3 decimal places.
So 3 sin x + 4 cos x = 5 sin(x + 0.927).
Check at x = 0: the left side is 4. The right side is 5 sin 0.9273 = 5 × 0.8 = 4. They agree.

Answer: R = 5 and α = 0.927

The greatest and least values come straight from the form. sin(x + α) lies between −1 and 1. So R sin(x + α) lies between −R and R. The greatest value of a sin x + b cos x is R = √(a² + b²), and the least is −R. For 3 sin x + 4 cos x these are 5 and −5. The greatest value is reached when x + α = π/2.

So the greatest value is 5, reached when x + 0.9273 = π/2. That gives x = π/2 − 0.9273 = 0.6435, which agrees with the peak in the first graph. Note that the greatest value is not 3 + 4 = 7. Adding the amplitudes would need both waves to peak at once, and they do not.

When b is negative, or a is negative, take care. R cos α = a and R sin α = b still hold. The signs of a and b tell the quadrant. For −3 sin x + 4 cos x, R cos α = −3 is negative and R sin α = 4 is positive. So α is in the second quadrant, not the fourth where the calculator puts tan⁻¹(−4/3).

For −3 sin x + 4 cos x, cos α = −3/5 and sin α = 4/5. This point is in the second quadrant, so α = 2.214, not −0.927.

A simpler way for a negative b is to use R sin(x − α) with α acute. Then R sin(x − α) = R cos α sin x − R sin α cos x. So R cos α = a and R sin α = −b, and both are positive when a > 0 and b < 0. The next example does this.

Worked example

Write 8 sin x − 15 cos x in the form R sin(x − α), with R > 0 and α acute. Give α in radians to 3 decimal places.

Expand R sin(x − α) = R cos α sin x − R sin α cos x. Match with 8 sin x − 15 cos x.
So R cos α = 8 and R sin α = 15.
Square and add: R² = 8² + 15² = 64 + 225 = 289.
So R = 17.
tan α = 15/8, and both R cos α and R sin α are positive, so α is acute.
In radian mode, α = tan⁻¹(15/8) = 1.0808, which is 1.081.
So 8 sin x − 15 cos x = 17 sin(x − 1.081).

Answer: R = 17 and α = 1.081

For 8 sin x − 15 cos x, the sides are 8 and 15, and R = 17. The angle α has tan α = 15/8.

Now equations. To solve a sin x + b cos x = c, write the left side as R sin(x + α). Then solve sin(x + α) = c/R. This needs |c| ≤ R, because a sine is never more than 1. If |c| is more than R, there are no solutions. Then change the interval: if 0 ≤ x ≤ 2π, then x + α lies between α and 2π + α.

Worked example

Solve 3 sin x + 4 cos x = 2 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

We know 3 sin x + 4 cos x = 5 sin(x + 0.9273). So the equation is 5 sin(x + 0.9273) = 2.
Divide by 5: sin(x + 0.9273) = 0.4.
This is possible, since 0.4 is between −1 and 1.
Let θ = x + 0.9273. Then 0 ≤ x ≤ 2π gives 0.9273 ≤ θ ≤ 2π + 0.9273 = 7.2105.
sin⁻¹ 0.4 = 0.4115. In the second quadrant θ = π − 0.4115 = 2.7301. Adding 2π to 0.4115 gives 6.6947.
The others are outside: 0.4115 is less than 0.9273, and 2.7301 + 2π = 9.0133 is more than 7.2105. So θ = 2.7301 or θ = 6.6947.
Subtract 0.9273 from each. First, x = 2.7301 − 0.9273 = 1.8028.
Second, x = 6.6947 − 0.9273 = 5.7674.
To 3 decimal places, x = 1.803 or x = 5.767.
Check x = 1.803: 3 sin 1.803 + 4 cos 1.803 = 3 × 0.9732 + 4 × (−0.2301) = 2.919 − 0.920 = 1.999, close to 2.

Answer: x = 1.803 or x = 5.767

The wave y = 3 sin x + 4 cos x (called f(x) here) cuts the line y = 2 twice between 0 and 2π.

Worked example

Write √3 sin x + cos x in the form R sin(x + α) with α in radians as a multiple of π. Hence solve √3 sin x + cos x = 1 for 0 ≤ x ≤ 2π. Give exact answers.

Match with R cos α sin x + R sin α cos x. So R cos α = √3 and R sin α = 1.
R² = 3 + 1 = 4, so R = 2.
tan α = 1/√3, and both are positive, so α is acute. tan 30° = 1/√3, so α = π/6.
So √3 sin x + cos x = 2 sin(x + π/6).
The equation becomes 2 sin(x + π/6) = 1, so sin(x + π/6) = 1/2.
Let θ = x + π/6. Then 0 ≤ x ≤ 2π gives π/6 ≤ θ ≤ 2π + π/6 = 13π/6.
sin θ = 1/2 gives θ = π/6, 5π/6 and 13π/6 in this interval. All three are inside, including both ends.
Subtract π/6: x = 0, x = 5π/6 − π/6 = 4π/6 = 2π/3, and x = 13π/6 − π/6 = 2π.

Answer: x = 0, 2π/3 or 2π

The wave y = √3 sin x + cos x has greatest value 2. The line y = 1 meets it at x = 0, 2π/3 and 2π, including both ends of the interval.

The same idea works for other forms. a sin x + b cos x can also be written R cos(x − β), because cos(x − β) = cos x cos β + sin x sin β. Then R cos β = b and R sin β = a. The cosine form is neat when the question asks for R cos(x ± α). Whichever form is asked for, R is the same and equals √(a² + b²).

Worked example

Write 12 cos x + 35 sin x in the form R cos(x − α), with R > 0 and α acute. Give α in radians to 3 decimal places.

Expand R cos(x − α) = R cos x cos α + R sin x sin α. Match with 12 cos x + 35 sin x.
So R cos α = 12 and R sin α = 35.
Square and add: R² = 12² + 35² = 144 + 1225 = 1369.
So R = 37.
tan α = 35/12, and both R cos α and R sin α are positive, so α is acute.
In radian mode, α = tan⁻¹(35/12) = 1.240499..., which is 1.240 to 3 decimal places.
So 12 cos x + 35 sin x = 37 cos(x − 1.240).

Answer: R = 37 and α = 1.240

See it three ways

With things

Two people push a swing, one at the back and one at the side, at the same rhythm. The swing moves as if one person were pushing in the in-between direction with a combined strength. That combined strength is R, and the direction is α.

As a picture

In words

The two coefficients are the sides of a right angled triangle. R is its hypotenuse and α is its angle.

In symbols

a sin x + b cos x = R sin(x + α), R = √(a² + b²), tan α = b/a, R cos α = a, R sin α = b.

Try it

Check the greatest value

You need: a calculator, a pencil.

Work out 3 sin x + 4 cos x for x = 0, 0.2, 0.4, 0.6, 0.8 and 1.0 radians. Use radian mode.
Look for the greatest value in your list and note which x gives it.
Is any value bigger than 5?
Compare with 5 sin(x + 0.927) at the same x values.

What you will see: The values rise to 5.00 near x = 0.6 and then fall. None is bigger than 5. The values of 5 sin(x + 0.927) match the first list.

If you cannot try it, this is what happens.

A wrong idea many people have

The greatest value of 3 sin x + 4 cos x is 3 + 4 = 7, because each part is at most as big as its number.

Why it feels right: Each wave reaches its own greatest value, and adding greatest values seems to give the greatest total.

The test that breaks it: At x = π/2, sin x = 1 but cos x = 0, so the total is 3. At x = 0, sin x = 0 and cos x = 1, so the total is 4. The two waves never peak together, so 7 is never reached.

What is true: The greatest value is R = √(3² + 4²) = 5. In general it is √(a² + b²), which is less than a + b whenever both a and b are positive.

A wrong idea many people have

For −3 sin x + 4 cos x, tan α = −4/3, so α = tan⁻¹(−4/3) = −0.927.

Why it feels right: The calculator gives an answer for tan⁻¹ of any number, and the algebra for tan α = b/a is correct.

The test that breaks it: Put α = −0.927 back: R cos α = 5 × 0.6 = 3 and R sin α = 5 × (−0.8) = −4. We needed −3 and 4. Both signs are wrong.

What is true: tan α = b/a gives the angle but loses the quadrant. Use R cos α = a and R sin α = b. Here cos α = −3/5 and sin α = 4/5, which puts α in the second quadrant: α = π − 0.927 = 2.214.

A wrong idea many people have

To solve 3 sin x + 4 cos x = 2 for 0 ≤ x ≤ 2π, solve sin(x + 0.927) = 0.4 and use 0 ≤ x + 0.927 ≤ 2π.

Why it feels right: The interval for x is 0 to 2π, and the bracket looks like a plain x.

The test that breaks it: Take x = 5.767, which is a solution. Then x + 0.927 = 6.694, which is more than 2π = 6.283. A search only up to 2π would miss it.

What is true: The bracket x + α runs from α to 2π + α. Solve for x + α in that whole interval, then subtract α from each answer.

Key words

harmonic form: the form R sin(x + α), which is a single wave equal to a sin x + b cos x.

Back to your guess

The answer is B: The sum of 3 sin x and 4 cos x is a wave with a greatest value of 5.

The sum is a wave with greatest value 5. The two parts do not reach their tops at the same time: 3 sin x peaks at π/2 and 4 cos x peaks at 0. So the total never reaches 3 + 4 = 7. It is a single sine wave with amplitude √(3² + 4²) = 5, shifted sideways.

Keep it

Copy this into your exercise book.

a sin x + b cos x = R sin(x + α). R = √(a² + b²). R cos α = a, R sin α = b. Greatest value R. Solve: sin(x + α) = c/R in the interval shifted by α. Needs |c| ≤ R.

Draw: A right angled triangle with sides a and b, hypotenuse R and angle α.

Measure a radian with string

Measure sin and cos on a big circle

See the period change

Draw the graph and the line

Find the second solution on a graph

Check the identity with a calculator

Find an angle with the inverse key

Test the formula with a calculator

Check the greatest value

Key words, A to Z