Chapter 5 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

M5.3.5 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

  • 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

  • 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.

A circle with centre O and axes named x and y, with 1 and −1 marked on the axes. A ray from O to a point P in the first quadrant is marked with an arc labelled θ. Dashed lines from P run to the axes and are labelled cos θ and sin θ.1−11−1xycos θsin θθPO
P = (cos θ, sin θ) lies on the circle of radius 1. So (cos θ)² + (sin θ)² = 1².
A right angled triangle with hypotenuse labelled 1, a side labelled cos θ and a side labelled sin θ. The angle θ is marked between the hypotenuse and the side labelled cos θ.sin θcos θ1θ
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.

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −1 to 2 named y. A solid curve named sin²x rises from 0 to 1 at π/2 and returns to 0 at π, then repeats. A dashed curve named cos²x is the same shape starting at 1. A dashed horizontal line at height 1 is named sum.0π/2π3π/22π01x (radians)ysumsin²xcos²x
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.

A table with the head identity. The rows read sin²x + cos²x ≡ 1; sin²x ≡ 1 − cos²x; cos²x ≡ 1 − sin²x; tan x ≡ sin x ÷ cos x.identitysin²x + cos²x ≡ 1sin²x ≡ 1 − cos²xcos²x ≡ 1 − sin²xtan x ≡ sin x ÷ cos x
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.

  1. Start with the left side: (sin x + cos x)². Expand the bracket: sin²x + 2 sin x cos x + cos²x.
  2. Put the two squares together: (sin²x + cos²x) + 2 sin x cos x.
  3. Use sin²x + cos²x ≡ 1. The left side becomes 1 + 2 sin x cos x.
  4. 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).

  1. Start with the left side. Write tan x as sin x ÷ cos x. Then 1/tan x is cos x ÷ sin x.
  2. So the left side is sin x/cos x + cos x/sin x.
  3. Use the common denominator sin x cos x. The first fraction gets sin x on top and bottom, the second gets cos x.
  4. This gives (sin²x + cos²x)/(sin x cos x).
  5. Use sin²x + cos²x ≡ 1. The top becomes 1, so the left side is 1/(sin x cos x).
  6. 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.

  1. 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.
  2. The top becomes cos x (1 + sin x). The bottom becomes (1 − sin x)(1 + sin x) = 1 − sin²x.
  3. Use cos²x ≡ 1 − sin²x. The bottom is cos²x.
  4. So the left side is cos x (1 + sin x)/cos²x.
  5. Cancel one cos x from the top and the bottom. This leaves (1 + sin x)/cos x.
  6. 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.

  1. The equation has sin²x and cos x. Replace sin²x by 1 − cos²x.
  2. We get 2(1 − cos²x) + 3 cos x = 3, which is 2 − 2cos²x + 3 cos x = 3.
  3. Move everything to one side and tidy: 0 = 2cos²x − 3 cos x + 1.
  4. Factorise: (2 cos x − 1)(cos x − 1) = 0. Check: 2cos²x − 2 cos x − cos x + 1 = 2cos²x − 3 cos x + 1.
  5. So cos x = 1/2 or cos x = 1.
  6. cos x = 1/2 gives x = π/3 and x = 2π − π/3 = 5π/3. cos x = 1 gives x = 0 and x = 2π.
  7. 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

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −4 to 4 named y. A curve named f(x), which is 2 sin²x + 3 cos x, starts at height 3 and a dashed line at height 3 meets it at four dots, at x = 0, π/3, 5π/3 and 2π.0π2π−4−2024x (radians)yf(x)
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.

  1. Write tan x as sin x ÷ cos x. The equation is sin x/cos x = 2 sin x, where cos x is not 0.
  2. 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.
  3. Take out the common factor sin x: sin x (1/cos x − 2) = 0.
  4. So sin x = 0, or 1/cos x = 2. The second gives cos x = 0.5.
  5. sin x = 0 gives x = 0, π and 2π. cos x = 1/2 gives x = π/3 and x = 5π/3.
  6. None of these has cos x = 0, so all are allowed. In order: x = 0, π/3, π, 5π/3 or 2π.
  7. 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

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −4 to 4 named y. A curve named tan x has three branches with dashed asymptotes at π/2 and 3π/2. A curve named 2 sin x is a wave. They meet at five dots, at x = 0, π/3, π, 5π/3 and 2π.0π/2π3π/22π−4−2024x (radians)ytan x2 sin x
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

A right angled triangle with hypotenuse labelled 1, a side labelled cos θ and a side labelled sin θ. The angle θ is marked between the hypotenuse and the side labelled cos θ.sin θcos θ1θ

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.

  1. Choose any angle, such as 25°. Press sin 25 and square the answer. Write it down.
  2. Press cos 25 and square the answer. Write it down.
  3. Add the two squares.
  4. 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 table with the heads angle, sin², cos² and sum. The rows read 25°: 0.179, 0.821, 1. 100°: 0.970, 0.030, 1. 200°: 0.117, 0.883, 1.anglesin²cos²sum25°0.1790.8211100°0.9700.0301200°0.1170.8831

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.

Check yourself

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

  1. Which expression is equal to 1 − cos²x for every x?

    Tap your answer.

    1. Not this one. sin²x + cos²x = 1 gives 1 − cos²x = sin²x, with the square. sin x alone is different: at x = π/6 it is 0.5, but sin²x is 0.25. Go to M5.3.5. Try another.

    2. Not this one. That has the opposite sign. cos²x − 1 is −sin²x, which is never positive. Go to M5.3.5. Try another.

    3. Not this one. That is 2 − cos²x. The identity is sin²x = 1 − cos²x, with no extra 1. Go to M5.3.5. Try another.

    4. Yes, that is right. The identity sin²x + cos²x ≡ 1 can be rearranged to sin²x ≡ 1 − cos²x. So 1 − cos²x is sin²x.

    See the answer

    sin²x

    The identity sin²x + cos²x ≡ 1 can be rearranged to sin²x ≡ 1 − cos²x. So 1 − cos²x is sin²x.

    • If you chose A: sin²x + cos²x = 1 gives 1 − cos²x = sin²x, with the square. sin x alone is different: at x = π/6 it is 0.5, but sin²x is 0.25. Go to M5.3.5.
    • If you chose B: That has the opposite sign. cos²x − 1 is −sin²x, which is never positive. Go to M5.3.5.
    • If you chose C: That is 2 − cos²x. The identity is sin²x = 1 − cos²x, with no extra 1. Go to M5.3.5.

  2. Which of these is true for every value of x?

    Tap your answer.

    1. Yes, that is right. Every point on the unit circle has (cos x)² + (sin x)² = 1. The other three fail at some value, such as x = π/4, x = π/6 and x = π/2.

    2. Not this one. tan x = sin x ÷ cos x. You turned the fraction upside down. At x = π/4 both are 1, but at π/6 they differ. Go to M5.3.5. Try another.

    3. Not this one. It is true at x = 0, but at x = π/2 the left side is 0 − 1 = −1. The identity has a plus sign. Go to M5.3.5. Try another.

    4. Not this one. It is true at x = 0, but at x = π/4 the left side is √2, which is not 1. Go to M5.3.5. Try another.

    See the answer

    sin²x + cos²x = 1

    Every point on the unit circle has (cos x)² + (sin x)² = 1. The other three fail at some value, such as x = π/4, x = π/6 and x = π/2.

    • If you chose B: tan x = sin x ÷ cos x. You turned the fraction upside down. At x = π/4 both are 1, but at π/6 they differ. Go to M5.3.5.
    • If you chose C: It is true at x = 0, but at x = π/2 the left side is 0 − 1 = −1. The identity has a plus sign. Go to M5.3.5.
    • If you chose D: It is true at x = 0, but at x = π/4 the left side is √2, which is not 1. Go to M5.3.5.

  3. Which expression is equal to tan x cos x for every x where it is defined?

    Tap your answer.

    1. Not this one. tan x = sin x ÷ cos x, and the cos x cancels with the one it multiplies. What is left is sin x, not cos x. Go to M5.3.5. Try another.

    2. Not this one. That would be tan x times tan x. Here tan x is multiplied by cos x, which cancels the bottom of tan x. Go to M5.3.5. Try another.

    3. Yes, that is right. Write tan x as sin x ÷ cos x. Then tan x cos x = (sin x ÷ cos x) × cos x. The cos x cancels, leaving sin x.

    4. Not this one. tan x cos x is (sin x ÷ cos x) × cos x. The cos x multiplies, so it cancels. It does not move to the bottom. Go to M5.3.5. Try another.

    See the answer

    sin x

    Write tan x as sin x ÷ cos x. Then tan x cos x = (sin x ÷ cos x) × cos x. The cos x cancels, leaving sin x.

    • If you chose A: tan x = sin x ÷ cos x, and the cos x cancels with the one it multiplies. What is left is sin x, not cos x. Go to M5.3.5.
    • If you chose B: That would be tan x times tan x. Here tan x is multiplied by cos x, which cancels the bottom of tan x. Go to M5.3.5.
    • If you chose D: tan x cos x is (sin x ÷ cos x) × cos x. The cos x multiplies, so it cancels. It does not move to the bottom. Go to M5.3.5.

  4. A learner says: "sin²x + cos²x = 1, so sin x + cos x = 1." What is the mistake?

    Tap your answer.

    1. Not this one. The identity is true for every angle. The mistake is in the step after it. Go to M5.3.5. Try another.

    2. Yes, that is right. The square root of sin²x + cos²x is the square root of 1, which is 1. But sin x + cos x is a different sum: at x = π/4 it is √2. We cannot take the square root of each term separately.

    3. Not this one. The unit circle has radius 1, so the squares add to 1, not 0. Go to M5.3.5. Try another.

    See the answer

    The square root of a sum is not the sum of the square roots.

    The square root of sin²x + cos²x is the square root of 1, which is 1. But sin x + cos x is a different sum: at x = π/4 it is √2. We cannot take the square root of each term separately.

    • If you chose A: The identity is true for every angle. The mistake is in the step after it. Go to M5.3.5.
    • If you chose C: The unit circle has radius 1, so the squares add to 1, not 0. Go to M5.3.5.

  5. To prove that an identity is true, a learner writes LHS = RHS and rearranges both sides. Why is this not a proof?

    Tap your answer.

    1. Not this one. A calculator checks single values only. An identity covers every x, so it needs an algebra argument. Go to M5.3.5. Try another.

    2. Not this one. A chain of correct steps from the claim can still reach a true statement from a false claim, as 1 = −1 becomes 1 = 1 when we square. Go to M5.3.5. Try another.

    3. Yes, that is right. A proof starts from facts we know and ends at the claim. Starting from the claim and reaching something true does not show the claim. Work on one side, or on each side separately, until both give the same expression.

    See the answer

    It assumes the identity is true, which is what must be shown.

    A proof starts from facts we know and ends at the claim. Starting from the claim and reaching something true does not show the claim. Work on one side, or on each side separately, until both give the same expression.

    • If you chose A: A calculator checks single values only. An identity covers every x, so it needs an algebra argument. Go to M5.3.5.
    • If you chose B: A chain of correct steps from the claim can still reach a true statement from a false claim, as 1 = −1 becomes 1 = 1 when we square. Go to M5.3.5.

  6. From an earlier chapter

    The calculator gives sin⁻¹ 0.8 = 0.927 radians. What is the other solution of sin x = 0.8 for 0 ≤ x ≤ 2π?

    Tap your answer.

    1. Not this one. That is 2π − 0.927, the fourth quadrant, where sin is negative. The other solution of sin x = 0.8 is in the second quadrant. Go to M5.3.4. Try another.

    2. Not this one. That is π + 0.927, the third quadrant, where sin is negative. Go to M5.3.4. Try another.

    3. Not this one. sin is negative for a negative angle in this range, so −0.927 gives −0.8, not 0.8. Go to M5.3.4. Try another.

    4. Yes, that is right. sin is positive in the first and second quadrants. The second solution is π − 0.927 = 3.142 − 0.927 = 2.214.

    See the answer

    2.214

    sin is positive in the first and second quadrants. The second solution is π − 0.927 = 3.142 − 0.927 = 2.214.

    • If you chose A: That is 2π − 0.927, the fourth quadrant, where sin is negative. The other solution of sin x = 0.8 is in the second quadrant. Go to M5.3.4.
    • If you chose B: That is π + 0.927, the third quadrant, where sin is negative. Go to M5.3.4.
    • If you chose C: sin is negative for a negative angle in this range, so −0.927 gives −0.8, not 0.8. Go to M5.3.4.

  7. Given that sin x = 3/5 and x is acute, find the value of 1 + tan²x as a fraction.

    See the answer

    25/16

    cos²x = 1 − 9/25 = 16/25, so cos x = 4/5 and tan x = (3/5) ÷ (4/5) = 3/4. Then 1 + tan²x = 1 + 9/16 = 25/16. This is 1/cos²x, since 1/(16/25) = 25/16.

  8. Prove that sin⁴x − cos⁴x ≡ sin²x − cos²x.

    See the answer

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

    Start with the left side. It is a difference of two squares: sin⁴x − cos⁴x = (sin²x)² − (cos²x)² = (sin²x − cos²x)(sin²x + cos²x). Use sin²x + cos²x ≡ 1, so the second bracket is 1. The left side is then sin²x − cos²x, which is the right side.

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

    See the answer

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

    Use the common denominator cos x (1 + sin x). The left side is [cos²x + (1 + sin x)²] ÷ [cos x (1 + sin x)]. Expand the top: cos²x + 1 + 2 sin x + sin²x. Use sin²x + cos²x ≡ 1 to get 2 + 2 sin x = 2(1 + sin x). The factor (1 + sin x) cancels with the bottom, leaving 2/cos x. This is the right side. The identity holds for x where cos x is not 0 and sin x is not −1.

  10. Solve 2cos²x + 3 sin x = 3 for 0 ≤ x ≤ 2π. Give exact answers.

    See the answer

    x = π/6, π/2 or 5π/6, which are 0.524, 1.571 and 2.618

    Replace cos²x by 1 − sin²x: 2 − 2 sin²x + 3 sin x = 3. Then 2 sin²x − 3 sin x + 1 = 0, which factorises as (2 sin x − 1)(sin x − 1) = 0. So sin x = 1/2 or sin x = 1. sin x = 1/2 gives π/6 and π − π/6 = 5π/6. sin x = 1 gives π/2.

  11. Solve 5 sin x = 2 tan x for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

    See the answer

    x = 0, 1.159, 3.142, 5.124 or 6.283

    Write tan x as sin x ÷ cos x: 5 sin x = 2 sin x/cos x. Move everything to one side and factorise: sin x (5 − 2/cos x) = 0. So sin x = 0, which gives x = 0, π and 2π. Or 5 = 2/cos x, which gives cos x = 0.4, so x = 1.159 and x = 2π − 1.159 = 5.124. To 3 decimal places the solutions are 0, 1.159, 3.142, 5.124 and 6.283.

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.

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

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