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Trigonometric identitiesEdexcel International A Level Maths: Revision notes

Section 1

The identity tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}

For any angle θ\theta with cos⁡θ≠0\cos\theta\neq0: tan⁡θ=sin⁡θcos⁡θ.\tan\theta=\frac{\sin\theta}{\cos\theta}. This is true for all such angles, which is why it is an identity (written with ≡\equiv). It lets you replace tan⁡θ\tan\theta by sines and cosines, or recover tan⁡θ\tan\theta from the other two. When cos⁡θ=0\cos\theta=0 (for example at 90∘90^\circ), tan⁡θ\tan\theta is undefined. Example: if sin⁡θ=35\sin\theta=\frac35 and cos⁡θ=45\cos\theta=\frac45 then tan⁡θ=34\tan\theta=\frac34.

Key termsidentity
Exam tip

In a proof, convert every tan⁡\tan into sin⁡cos⁡\frac{\sin}{\cos} first; it usually makes the algebra much easier.

Section 2

The identity sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1

sin⁡2θ+cos⁡2θ≡1.\sin^2\theta+\cos^2\theta\equiv1. It follows from Pythagoras' theorem on a right-angled triangle with hypotenuse 1, and holds for every angle. 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. For instance 3−3cos⁡2θ=3sin⁡2θ3-3\cos^2\theta=3\sin^2\theta, and sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 at θ=30∘\theta=30^\circ: 14+34=1\frac14+\frac34=1.

Common mistake

Writing sin⁡2θ=sin⁡(θ2)\sin^2\theta=\sin\left(\theta^2\right) or cancelling the squares to leave sin⁡θ+cos⁡θ=1\sin\theta+\cos\theta=1. The square applies to the value of the sine.

Section 3

Finding the other ratios, with signs

Given one of sin⁡θ\sin\theta or cos⁡θ\cos\theta, use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 to find the other, then tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}. Taking a square root gives ±\pm, so decide the sign from the quadrant: sine is positive in the 1st and 2nd quadrants, cosine in the 1st and 4th, tangent in the 1st and 3rd. Example: sin⁡θ=817\sin\theta=\frac8{17} with θ\theta obtuse. cos⁡2θ=1−64289=225289\cos^2\theta=1-\frac{64}{289}=\frac{225}{289}. Cosine is negative in the 2nd quadrant, so cos⁡θ=−1517\cos\theta=-\frac{15}{17} and tan⁡θ=−815\tan\theta=-\frac8{15}. Example: cos⁡x=−23\cos x=-\frac23 with 180∘<x<270∘180^\circ<x<270^\circ: sin⁡2x=59\sin^2x=\frac59, and sine is negative in the 3rd quadrant, so sin⁡x=−53\sin x=-\frac{\sqrt5}3 and tan⁡x=52>0\tan x=\frac{\sqrt5}2>0.

Key termsquadrant
Common mistake

Giving both ±\pm values, or the wrong sign, when the angle's range is stated. Use the range of the angle to choose one sign.

Section 4

Simplifying expressions

Strategy: look for 1−sin⁡2θ1-\sin^2\theta or 1−cos⁡2θ1-\cos^2\theta (replace by a single square), for sin⁡2θ+cos⁡2θ\sin^2\theta+\cos^2\theta (replace by 1), and for tan⁡θ\tan\theta (replace by sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}).

  • sin⁡2θ1−sin⁡2θ=sin⁡2θcos⁡2θ=tan⁡2θ\frac{\sin^2\theta}{1-\sin^2\theta}=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta.
  • (sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=1+2sin⁡θcos⁡θ(\sin\theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta.
  • tan⁡θcos⁡θ=sin⁡θ\tan\theta\cos\theta=\sin\theta. Many problems reduce an expression to a single trigonometric function.
Exam tip

Expand brackets fully before looking for sin⁡2θ+cos⁡2θ\sin^2\theta+\cos^2\theta; the middle term 2sin⁡θcos⁡θ2\sin\theta\cos\theta stays.

Section 5

Proving identities

To prove an identity, start with one side (usually the more complicated) and transform it step by step until it equals the other side. Show every step and state when you use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 or tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}. Example: sin⁡θ1+cos⁡θ+1+cos⁡θsin⁡θ\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta}. Put over a common denominator: sin⁡2θ+(1+cos⁡θ)2sin⁡θ(1+cos⁡θ)\frac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}. Expand: sin⁡2θ+1+2cos⁡θ+cos⁡2θ=2+2cos⁡θ=2(1+cos⁡θ)\sin^2\theta+1+2\cos\theta+\cos^2\theta=2+2\cos\theta=2(1+\cos\theta). Cancelling gives 2sin⁡θ\frac2{\sin\theta}.

Key termsprove
Common mistake

Treating a proof as an equation: do not move terms across the equals sign or multiply both sides by something. Work on one side only.

Section 6

Using the identities to form equations

The identities can turn an equation with two functions into one in a single function. For 4sin⁡2θ−cos⁡2θ=24\sin^2\theta-\cos^2\theta=2, write 2=2(sin⁡2θ+cos⁡2θ)2=2\left(\sin^2\theta+\cos^2\theta\right) to get 2sin⁡2θ=3cos⁡2θ2\sin^2\theta=3\cos^2\theta. Dividing by cos⁡2θ\cos^2\theta gives 2tan⁡2θ=32\tan^2\theta=3, so tan⁡2θ=32\tan^2\theta=\frac32 and tan⁡θ=±62\tan\theta=\pm\frac{\sqrt6}2. Dividing by cos⁡2θ\cos^2\theta is allowed only because tan⁡θ\tan\theta must be defined, so cos⁡θ≠0\cos\theta\neq0.

Exam tip

Replace constants such as 1 or 2 by sin⁡2θ+cos⁡2θ\sin^2\theta+\cos^2\theta multiples to make both sides have squares only.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Trigonometric identities

  1. θ\theta is an obtuse angle with sin⁡θ=817\sin\theta=\frac{8}{17}.
    Find the exact value of sin⁡θcos⁡θ\sin\theta\cos\theta.2 marks
  2. cos⁡x=−23\cos x=-\frac23 and 180∘<x<270∘180^\circ<x<270^\circ.
    Find the exact value of tan⁡2x−sin⁡2x\tan^2x-\sin^2x.2 marks
  3. In this question, θ\theta is any angle for which the expressions are defined.
    Show that (sin⁡θ+cos⁡θ)2≡1+2sin⁡θcos⁡θ(\sin\theta+\cos\theta)^2\equiv1+2\sin\theta\cos\theta.3 marks
See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).