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Learn7357Paper 1 · PureTrigonometryTrigonometric identities
Trigonometry · Subtopic 10 of 12

Trigonometric Identities

The two identities at the heart of AS trigonometry, where they come from, and how to use them to simplify expressions, find exact values and write clear proofs.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Know and use tan⁡θ≡sin⁡θcos⁡θ\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta} and sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta\equiv1.
  • Rearrange the Pythagorean identity: sin⁡2θ≡1−cos⁡2θ\sin^2\theta\equiv1-\cos^2\theta and cos⁡2θ≡1−sin⁡2θ\cos^2\theta\equiv1-\sin^2\theta.
  • Simplify trigonometric expressions and find exact values of other ratios.
  • Prove identities by working from one side to the other.
Pause and think

The point at angle θ\theta on the unit circle is (cos⁡θ, sin⁡θ)(\cos\theta,\ \sin\theta), and its distance from the origin is 11. What equation does Pythagoras give you?

Section 2 of 14

Core idea

Start hereEqual forever

2(x+3)2(x+3) and 2x+62x+6 look different but are always equal: an identity. Trigonometry has its own identities, and the most important one is Pythagoras in disguise. They let you swap one form for another, which is how hard trig equations become easy ones.

The big idea

tan⁡θ≡sin⁡θcos⁡θ\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta} and sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta\equiv1 are true for every angle. Use them to simplify, to prove other identities, and to rewrite equations in a single trig function.

1. The Pythagorean identity

Pythagoras on the unit circle
−2−1.5−1−0.50.511.52−1−0.50.510xycos²θ + sin²θ = 1 for every θ
The triangle has sides cos⁡θ\cos\theta, sin⁡θ\sin\theta and hypotenuse 11, wherever the point is. Squaring makes the signs irrelevant.
Proof of sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta\equiv1
  1. P=(cos⁡θ,sin⁡θ) on the unit circleP=(\cos\theta,\sin\theta)\text{ on the unit circle}
    Definition.
  2. (cos⁡θ)2+(sin⁡θ)2=12(\cos\theta)^2+(\sin\theta)^2=1^2
    The distance from the origin is 11: Pythagoras.
  3. sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta\equiv1
    Notation: sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2.
Rearranged: sin⁡2θ≡1−cos⁡2θ\sin^2\theta\equiv1-\cos^2\theta and cos⁡2θ≡1−sin⁡2θ\cos^2\theta\equiv1-\sin^2\theta.
Common misconceptionsin⁡2θ≠sin⁡(θ2)\sin^2\theta\neq\sin(\theta^2) sin⁡2θ\sin^2\theta squares the sine value. And sin⁡θ2\sin\theta^2 squares the angle first, which is different.

2. Proving identities

Start from one side (usually the more complicated) and transform it step by step into the other. Don't move terms across ≡\equiv as in an equation; you'd be assuming what you're proving.

Worked exampleProve 1−cos⁡2θcos⁡θ≡sin⁡θtan⁡θ\dfrac{1-\cos^2\theta}{\cos\theta}\equiv\sin\theta\tan\theta

Prove the identity.

Goal
Transform LHS into RHS.
Strategy
Use the Pythagorean identity, then split into sin⁡\sin and tan⁡\tan.
  1. LHS=1−cos⁡2θcos⁡θ\text{LHS}=\frac{1-\cos^2\theta}{\cos\theta}
    Start.
  2. =sin⁡2θcos⁡θ=\frac{\sin^2\theta}{\cos\theta}
    1−cos⁡2θ≡sin⁡2θ1-\cos^2\theta\equiv\sin^2\theta.
  3. =sin⁡θ⋅sin⁡θcos⁡θ=sin⁡θtan⁡θ=RHS=\sin\theta\cdot\frac{\sin\theta}{\cos\theta}=\sin\theta\tan\theta=\text{RHS}
    Split; sin⁡cos⁡=tan⁡\frac{\sin}{\cos}=\tan.
AnswerProved ∎.
✓ Checkθ=60∘\theta=60^\circ: LHS =3/41/2=32=\frac{3/4}{1/2}=\frac32; RHS =32⋅3=32=\frac{\sqrt3}2\cdot\sqrt3=\frac32 ✓.
Worked exampleProve 1sin⁡θ−sin⁡θ1+cos⁡θ≡1tan⁡θ\dfrac1{\sin\theta}-\dfrac{\sin\theta}{1+\cos\theta}\equiv\dfrac1{\tan\theta}

Prove the identity.

Goal
LHS → RHS.
Strategy
Single fraction, replace sin⁡2\sin^2, factorise, cancel.
  1. (1+cos⁡θ)−sin⁡2θsin⁡θ(1+cos⁡θ)\frac{(1+\cos\theta)-\sin^2\theta}{\sin\theta(1+\cos\theta)}
    Common denominator.
  2. =1+cos⁡θ−1+cos⁡2θsin⁡θ(1+cos⁡θ)=\frac{1+\cos\theta-1+\cos^2\theta}{\sin\theta(1+\cos\theta)}
    sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta.
  3. =cos⁡θ(1+cos⁡θ)sin⁡θ(1+cos⁡θ)=\frac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)}
    Factorise the top.
  4. =cos⁡θsin⁡θ=1tan⁡θ=\frac{\cos\theta}{\sin\theta}=\frac1{\tan\theta}
    Cancel (1+cos⁡θ)(1+\cos\theta).
AnswerProved ∎.
✓ Checkθ=45∘\theta=45^\circ: 2−2/21+2/2=2−(2−1)=1=1tan⁡45∘\sqrt2-\frac{\sqrt2/2}{1+\sqrt2/2}=\sqrt2-(\sqrt2-1)=1=\frac1{\tan45^\circ} ✓.

3. Solving equations with identities

Worked example2sin⁡2x=3cos⁡x2\sin^2x=3\cos x, 0≤x≤360∘0\le x\le360^\circ

Solve.

Goal
All xx in range.
Strategy
Replace sin⁡2x\sin^2x so only cos⁡x\cos x appears; solve the quadratic.
  1. 2(1−cos⁡2x)=3cos⁡x2(1-\cos^2x)=3\cos x
    Identity.
  2. 2cos⁡2x+3cos⁡x−2=02\cos^2x+3\cos x-2=0
    Rearrange.
  3. (2cos⁡x−1)(cos⁡x+2)=0(2\cos x-1)(\cos x+2)=0
    Factorise (like 2c2+3c−22c^2+3c-2).
  4. cos⁡x=12 or cos⁡x=−2\cos x=\tfrac12\ \text{or}\ \cos x=-2
    −2-2 is impossible (∣cos⁡x∣≤1|\cos x|\le1).
  5. x=60∘, 300∘x=60^\circ,\ 300^\circ
    Cos positive in Q1 and Q4.
Answerx=60∘x=60^\circ or 300∘300^\circ.
✓ Checkx=60∘x=60^\circ: 2⋅34=1.52\cdot\frac34=1.5 and 3⋅12=1.53\cdot\frac12=1.5 ✓.
Pause and think

Show (sin⁡θ+cos⁡θ)2≡1+2sin⁡θcos⁡θ(\sin\theta+\cos\theta)^2\equiv1+2\sin\theta\cos\theta.

Summary

  • tan⁡=sin⁡/cos⁡\tan=\sin/\cos; sin⁡2+cos⁡2=1\sin^2+\cos^2=1.
  • Prove by transforming one side.
  • Reduce equations to one trig function.
Key words
Identity
True for all values; written with ≡\equiv.
HistoryOlder than trigonometry. sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 is just Pythagoras' theorem in a circle of radius 11. Pythagoras' theorem itself was known to Babylonian scribes over 3,500 years ago: the clay tablet Plimpton 322 lists Pythagorean triples, about a thousand years before Pythagoras was born.
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