Chapter 3 of 8

All 8 chapters
  1. 1Radians, readunderstood
  2. 2The unit circle and any angle, readunderstood
  3. 3Trig graphs and their changes, readunderstood
  4. 4Solving trig equations, readunderstood
  5. 5Identities, readunderstood
  6. 6Reciprocal and inverse trig functions, readunderstood
  7. 7Compound and double angles, readunderstood
  8. 8Combining sin and cos, readunderstood

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

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • 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.

A graph with the x axis named x (radians) from 0 to 2π, marked at π/2, π, 3π/2 and 2π, and a y axis from −2 to 2 named y. A solid wave named sin x starts at 0, rises to 1 at π/2, falls through 0 at π to −1 at 3π/2 and returns to 0 at 2π. A dashed wave named cos x starts at 1, falls through 0 at π/2 to −1 at π and returns to 1 at 2π.0π/2π3π/22π−2−1012x (radians)ysin xcos x
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.

A graph with the x axis named x (radians) from −π to π, marked at −π, −π/2, 0, π/2 and π, and a y axis from −4 to 4 named y. Dashed vertical lines stand at x = −π/2 and x = π/2. The graph named tan x has a branch rising from the left edge up towards the asymptote at −π/2, a central branch rising steeply through the origin between the two asymptotes, and a branch rising from the asymptote at π/2 towards the right edge.−π−π/20π/2π−4−2024x (radians)ytan x
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.

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −3 to 3 named y. A solid wave named sin x has height 1. A dashed wave named 2 sin x has the same zeros but height 2, with its peak at π/2 and its trough at 3π/2.0π/2π3π/22π−3−2−10123x (radians)ysin x2 sin x
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.

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −3 to 3 named y. A solid wave named sin x has one cycle. A dashed wave named sin 2x has two complete cycles in the same space, with peaks at π/4 and 5π/4 and troughs at 3π/4 and 7π/4.0π/2π3π/22π−3−2−10123x (radians)ysin xsin 2x
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.

A graph with the x axis named x (radians) from −π/2 to 3π/2, and a y axis from −3 to 3 named y. A solid wave named sin x starts at the origin. A dashed wave named sin(x + π/3) is the same wave moved to the left by π/3, with a peak at π/6.−π/20π/2π3π/2−3−2−10123x (radians)ysin xsin(x + π/3)
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.

  1. Compare with y = a sin bx. Here a = 3 and b = 2.
  2. The amplitude is 3.
  3. The period is 2π ÷ 2 = π. So the wave has 2 full cycles between 0 and 2π.
  4. The greatest value is 3 and the least value is −3, because the midline is the x axis.
  5. One cycle goes from 0 to π. It starts at 0 and reaches its peak at a quarter of the period, x = π/4.
  6. At the peak 2x = π/2, so sin 2x = 1 and the height is 3 × 1 = 3.
  7. 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.

A graph with the x axis named x (radians) from 0 to 2π, marked at π/2, π, 3π/2 and 2π, and a y axis from −4 to 4 named y. A wave named 3 sin 2x has two cycles. Dots mark its first peak at x = π/4 with height 3 and its first trough at x = 3π/4 with height −3, each with dashed guide lines.0π/2π3π/22π−4−2024x (radians)y3 sin 2x
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.

A graph with the x axis named x (radians) from 0 to 2π/3, marked at π/6, π/3, π/2 and 2π/3, and a y axis from −2 to 2 named y. One cycle of a wave named 2 sin 3x starts at 0, peaks at 2 at π/6, crosses at π/3, has a trough of −2 at π/2 and ends at 0 at 2π/3.0π/6π/3π/22π/3−2−1012x (radians)y2 sin 3x
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.

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

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

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −2 to 4 named y. A wave named 1 + 2 cos x has dots at its peaks (0, 3) and (2π, 3) and its trough (π, −1). A dashed horizontal line at height 1 is the midline.0π/2π3π/22π−2−101234x (radians)y1 + 2 cos x
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.

  1. The graph is a sine wave with midline h = 8 and amplitude 3.
  2. The greatest depth is 8 + 3 = 11 m. The least depth is 8 − 3 = 5 m.
  3. In sin(πt/6) the number next to t is π/6. So the period is 2π ÷ (π/6) = 2π × 6/π = 12 hours.
  4. 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.
  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.
  6. 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.

A graph with the horizontal axis named t (hours) from 0 to 12 and the vertical axis named h (m) from 0 to 12 for the depth. A wave rises from 8 at t = 0 to a peak of 11 at t = 3, falls to 8 at t = 6, to a trough of 5 at t = 9 and returns to 8 at t = 12. A dashed line at 9.5 meets the wave at t = 1 and t = 5.024681012024681012t (hours)h (m)
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.

  1. The number next to x is 2. The period of tan bx is π ÷ b, so the period is π/2.
  2. tan 2x is not defined when 2x = π/2, 3π/2, and so on. So x = π/4 and x = 3π/4.
  3. The graph passes through (0, 0), (π/2, 0) and (π, 0), where 2x = 0, π and 2π.
  4. 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.

A graph with the x axis named x (radians) from 0 to π, marked at π/4, π/2, 3π/4 and π, and a y axis from −4 to 4 named y. Dashed vertical lines stand at x = π/4 and x = 3π/4. The graph named tan 2x has three branches, passing through the points (0, 0), (π/2, 0) and (π, 0).0π/4π/23π/4π−4−2024x (radians)ytan 2x
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.

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

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

A graph with the x axis named x (radians) from 0 to π, marked at π/4, π/2, 3π/4 and π, and a y axis from −3 to 5 named y. One cycle of a wave named 1 + 3 sin 2x runs from (0, 1) up to a peak at (π/4, 4), down to a trough at (3π/4, −2) and back to (π, 1). A dashed line at height 1 is the midline.0π/4π/23π/4π−2024x (radians)y1 + 3 sin 2x
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.

  1. Take out the 2: 2x + π/3 = 2(x + π/6). So y = sin 2(x + π/6).
  2. 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.
  3. The period does not change: 2π ÷ 2 = π.
  4. The peak of sin 2x is at x = π/4. After the move it is at x = π/4 − π/6 = π/12.
  5. 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.

A graph with the x axis named x (radians) from −π/2 to π, and a y axis from −1.5 to 1.5 named y. A dashed wave named sin 2x starts at the origin. A solid wave named sin(2x + π/3) is the same wave moved π/6 to the left, with its peak at π/12.−π/20π/2π−101x (radians)ysin 2xsin(2x + π/3)
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

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −2 to 2 named y. A wave named 2 sin x with amplitude 2. The peak at π/2 and trough at 3π/2 have guide lines to the y axis.0π/2π3π/22π−2−1012x (radians)y2 sin x

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.

  1. 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.
  2. 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.
  3. Count the peaks of each wave between 0 and 2π.
  4. 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 table with the heads x, sin x and sin 2x. The rows read 0: 0, 0. π/4: 0.71, 1. π/2: 1, 0. 3π/4: 0.71, −1. π: 0, 0.xsin xsin 2x000π/40.711π/2103π/40.71−1π00

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.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which statement gives the amplitude and period of y = 4 sin(x/2)?

    Tap your answer.

    1. Not this one. You may have multiplied 2π by 1/2. The period is 2π divided by 1/2, which is 4π. Go to M5.3.3. Try another.

    2. Not this one. The amplitude is the number in front of the sine, which is 4. The 1/2 inside changes only the period. Go to M5.3.3. Try another.

    3. Not this one. The 1/2 next to x stretches the graph sideways, so the period is not 2π. Go to M5.3.3. Try another.

    4. Yes, that is right. In y = a sin bx the amplitude is |a| = 4. Here b = 1/2, so the period is 2π ÷ (1/2) = 4π.

    See the answer

    Amplitude 4, period 4π

    In y = a sin bx the amplitude is |a| = 4. Here b = 1/2, so the period is 2π ÷ (1/2) = 4π.

    • If you chose A: You may have multiplied 2π by 1/2. The period is 2π divided by 1/2, which is 4π. Go to M5.3.3.
    • If you chose B: The amplitude is the number in front of the sine, which is 4. The 1/2 inside changes only the period. Go to M5.3.3.
    • If you chose C: The 1/2 next to x stretches the graph sideways, so the period is not 2π. Go to M5.3.3.

  2. What are the greatest and least values of y = 3 − 2 cos x?

    Tap your answer.

    1. Yes, that is right. cos x runs from −1 to 1. Then −2 cos x runs from −2 to 2, and adding 3 gives 1 to 5. The greatest value, 5, is when cos x = −1.

    2. Not this one. That is the range of cos x itself. The 2 in front and the 3 added both change it. Go to M5.3.3. Try another.

    3. Not this one. You used 3 + 2 for the top and then the same size below the x axis. The wave is centred on 3, not on 0. Go to M5.3.3. Try another.

    4. Not this one. You used the amplitude only. The wave is lifted by 3, so its midline is y = 3. Go to M5.3.3. Try another.

    See the answer

    Greatest 5, least 1

    cos x runs from −1 to 1. Then −2 cos x runs from −2 to 2, and adding 3 gives 1 to 5. The greatest value, 5, is when cos x = −1.

    • If you chose B: That is the range of cos x itself. The 2 in front and the 3 added both change it. Go to M5.3.3.
    • If you chose C: You used 3 + 2 for the top and then the same size below the x axis. The wave is centred on 3, not on 0. Go to M5.3.3.
    • If you chose D: You used the amplitude only. The wave is lifted by 3, so its midline is y = 3. Go to M5.3.3.

  3. The graph shows one cycle of a wave that starts at its peak, 1, at x = 0. Which equation does it show?

    A graph with the x axis named x (radians) from 0 to π, marked at π/4, π/2, 3π/4 and π, and a y axis from −2 to 2 named y. A wave starts at the point (0, 1), crosses the axis at π/4, has a trough of −1 at π/2, crosses at 3π/4 and returns to 1 at π.0π/4π/23π/4π−2−1012x (radians)y
    Tap your answer.

    1. Not this one. Then the period would be 4π, and the graph would not have finished a cycle by x = π. Go to M5.3.3. Try another.

    2. Not this one. A sine graph starts at 0, on the x axis. This graph starts at its peak, 1. Go to M5.3.3. Try another.

    3. Yes, that is right. The wave starts at 1, so it is a cosine, with amplitude 1. One cycle takes π, so b = 2π ÷ π = 2. That gives y = cos 2x.

    4. Not this one. That has amplitude 2 and period 2π. This graph has height 1 and period π. Go to M5.3.3. Try another.

    See the answer

    y = cos 2x

    The wave starts at 1, so it is a cosine, with amplitude 1. One cycle takes π, so b = 2π ÷ π = 2. That gives y = cos 2x.

    • If you chose A: Then the period would be 4π, and the graph would not have finished a cycle by x = π. Go to M5.3.3.
    • If you chose B: A sine graph starts at 0, on the x axis. This graph starts at its peak, 1. Go to M5.3.3.
    • If you chose D: That has amplitude 2 and period 2π. This graph has height 1 and period π. Go to M5.3.3.

  4. The graph of y = sin(x + π/3) is the graph of y = sin x moved...

    Tap your answer.

    1. Not this one. Adding to x moves the graph the other way. At x = −π/3 the new graph is sin 0 = 0, so the start has moved left. Go to M5.3.3. Try another.

    2. Not this one. A number added outside, as in sin x + π/3, moves the graph up. Here it is added inside, next to x. Go to M5.3.3. Try another.

    3. Not this one. Down would be sin x − π/3, with the number outside. Here the number is next to x. Go to M5.3.3. Try another.

    4. Yes, that is right. y = f(x + c) moves the graph c units to the left. The wave that crossed the axis at x = 0 now crosses it where x + π/3 = 0, which is x = −π/3.

    See the answer

    π/3 to the left

    y = f(x + c) moves the graph c units to the left. The wave that crossed the axis at x = 0 now crosses it where x + π/3 = 0, which is x = −π/3.

    • If you chose A: Adding to x moves the graph the other way. At x = −π/3 the new graph is sin 0 = 0, so the start has moved left. Go to M5.3.3.
    • If you chose B: A number added outside, as in sin x + π/3, moves the graph up. Here it is added inside, next to x. Go to M5.3.3.
    • If you chose C: Down would be sin x − π/3, with the number outside. Here the number is next to x. Go to M5.3.3.

  5. tan x is not defined at...

    Tap your answer.

    1. Not this one. At multiples of π, sin x = 0, so tan x = 0. That is where the graph crosses the x axis. Go to M5.3.3. Try another.

    2. Yes, that is right. tan x = sin x ÷ cos x is not defined when cos x = 0. That is at π/2, 3π/2, 5π/2, −π/2 and so on: π/2 times an odd number. The graph has an asymptote at each.

    3. Not this one. It is also undefined at −π/2, 3π/2 and every other angle where cos x = 0. Go to M5.3.3. Try another.

    4. Not this one. At multiples of 2π, tan x = 0. The graph crosses the x axis there. Go to M5.3.3. Try another.

    See the answer

    every odd multiple of π/2

    tan x = sin x ÷ cos x is not defined when cos x = 0. That is at π/2, 3π/2, 5π/2, −π/2 and so on: π/2 times an odd number. The graph has an asymptote at each.

    • If you chose A: At multiples of π, sin x = 0, so tan x = 0. That is where the graph crosses the x axis. Go to M5.3.3.
    • If you chose C: It is also undefined at −π/2, 3π/2 and every other angle where cos x = 0. Go to M5.3.3.
    • If you chose D: At multiples of 2π, tan x = 0. The graph crosses the x axis there. Go to M5.3.3.

  6. From an earlier chapter

    What is sin 240°?

    Tap your answer.

    1. Yes, that is right. 240° is in the third quadrant, where sin is negative. The reference angle is 240° − 180° = 60°, and sin 60° = √3/2. So sin 240° = −√3/2.

    2. Not this one. The reference angle of 240° is 240° − 180° = 60°, and sin 60° = √3/2, not 1/2. Go to M5.3.2. Try another.

    3. Not this one. The size is right but the sign is wrong. 240° is in the third quadrant, where sin is negative. Go to M5.3.2. Try another.

    4. Not this one. That is sin 30° with the wrong sign. The reference angle of 240° is 60°. Go to M5.3.2. Try another.

    See the answer

    −√3/2

    240° is in the third quadrant, where sin is negative. The reference angle is 240° − 180° = 60°, and sin 60° = √3/2. So sin 240° = −√3/2.

    • If you chose B: The reference angle of 240° is 240° − 180° = 60°, and sin 60° = √3/2, not 1/2. Go to M5.3.2.
    • If you chose C: The size is right but the sign is wrong. 240° is in the third quadrant, where sin is negative. Go to M5.3.2.
    • If you chose D: That is sin 30° with the wrong sign. The reference angle of 240° is 60°. Go to M5.3.2.

  7. Find the period of y = 5 sin 3x.

    See the answer

    2π/3

    For y = a sin bx the period is 2π ÷ b. Here b = 3, so the period is 2π/3. The 5 changes only the height.

  8. The temperature T °C in a greenhouse, t hours after midnight, is T = 18 + 6 sin(πt/12), for 0 ≤ t ≤ 24. Find the greatest temperature, and the first time it is reached.

    See the answer

    24 °C at t = 6

    The greatest value of sin is 1, so the greatest temperature is 18 + 6 = 24 °C. It is reached when πt/12 = π/2. Then t = 12 ÷ 2 = 6, which is 6 hours after midnight.

  9. The graph shows one cycle of y = a sin bx. Find a and b, where b is positive.

    A graph with the x axis named x (radians) from 0 to 2π/3, marked at π/6, π/3, π/2 and 2π/3, and a y axis from −4 to 4 named y. One cycle of a wave starts at 0, peaks at 4, crosses the axis at π/3, has a trough of −4 and ends at 0 at 2π/3.0π/6π/3π/22π/3−4−2024x (radians)y
    See the answer

    a = 4, b = 3

    The peak is 4, so a = 4. One cycle takes from 0 to 2π/3, so the period is 2π/3 and b = 2π ÷ (2π/3) = 3.

  10. Find the period of y = tan(x/2). Where is its one asymptote between x = 0 and x = 2π?

    See the answer

    The period is 2π. The asymptote is x = π.

    For y = tan bx the period is π ÷ b. Here b = 1/2, so the period is π ÷ (1/2) = 2π. The graph is not defined when x/2 = π/2, which gives x = π. The next asymptote, at x = 3π, is outside the interval.

  11. The graph of y = sin(4x + π/2) is the graph of y = sin 4x moved...

    Tap your answer.

    1. Not this one. That would be the move for y = sin(x + π/2). Here 4x + π/2 = 4(x + π/8), so the move is π/8. Go to M5.3.3. Try another.

    2. Not this one. Adding inside the bracket moves a graph to the left. The move is to the left. Go to M5.3.3. Try another.

    3. Yes, that is right. Take out the 4: sin(4x + π/2) = sin 4(x + π/8). Replacing x by x + π/8 moves the graph π/8 to the left.

    4. Not this one. Adding inside moves a graph to the left, and the size is c/b = (π/2) ÷ 4, not π/2. Go to M5.3.3. Try another.

    See the answer

    π/8 to the left

    Take out the 4: sin(4x + π/2) = sin 4(x + π/8). Replacing x by x + π/8 moves the graph π/8 to the left.

    • If you chose A: That would be the move for y = sin(x + π/2). Here 4x + π/2 = 4(x + π/8), so the move is π/8. Go to M5.3.3.
    • If you chose B: Adding inside the bracket moves a graph to the left. The move is to the left. Go to M5.3.3.
    • If you chose D: Adding inside moves a graph to the left, and the size is c/b = (π/2) ÷ 4, not π/2. Go to M5.3.3.

  12. The graph of y = −2 cos x is drawn for 0 ≤ x ≤ 2π. Find the coordinates of its greatest value, giving x to 3 decimal places.

    See the answer

    (3.142, 2)

    The minus sign turns the cosine wave upside down, and the 2 doubles its height. The greatest value is 2, where cos x = −1. That is at x = π, which is 3.142 to 3 decimal places.

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.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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