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Trigonometric identities

Verify the Pythagorean identity; use it to find unknown trig ratios given one ratio and the quadrant; apply the identities 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta and 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta; use complementary angle identities; prove two-step identities by expressing everything in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta.

Worked examples

Finding a trig ratio using the Pythagorean identity

Straightforward

Problem

Given that cos⁡θ=45\cos\theta = \dfrac{4}{5} and θ\theta is in the first quadrant, find sin⁡θ\sin\theta.

Applying a Pythagorean identity and a complementary angle identity

Moderate

Problem

Given that tan⁡θ=512\tan\theta = \dfrac{5}{12} and θ\theta is acute, find sec⁡θ\sec\theta using sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta. Then evaluate sin⁡(90°−θ)\sin(90° - \theta).

Proving a two-step identity

Challenging

Problem

Prove the identity cos⁡θ1−sin⁡θ+cos⁡θ1+sin⁡θ=2sec⁡θ\dfrac{\cos\theta}{1 - \sin\theta} + \dfrac{\cos\theta}{1 + \sin\theta} = 2\sec\theta.

Practise

Q1·Straightforward
Using exact values, compute cos⁡260°+sin⁡260°\cos^2 60° + \sin^2 60°.
Q2·Straightforward
If cos⁡θ=45\cos\theta = \dfrac{4}{5} and θ\theta is in the first quadrant, find sin⁡θ\sin\theta.
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Q3·Straightforward
If sin⁡θ=1213\sin\theta = \dfrac{12}{13} and θ\theta is acute, find cos⁡θ\cos\theta.
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Q4·Straightforward
If cos⁡θ=−32\cos\theta = -\dfrac{\sqrt{3}}{2} and θ\theta is in the second quadrant, what is sin⁡θ\sin\theta?
Q5·Moderate
If tan⁡θ=3\tan\theta = 3, use the identity sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta to find sec⁡2θ\sec^2\theta.
Q6·Moderate
If cot⁡θ=23\cot\theta = \dfrac{2}{3}, use the identity csc⁡2θ=1+cot⁡2θ\csc^2\theta = 1 + \cot^2\theta to find csc⁡2θ\csc^2\theta.
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Q7·Moderate
If sec⁡θ=53\sec\theta = \dfrac{5}{3}, use the identity tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1 to find tan⁡2θ\tan^2\theta.
Q8·Moderate
Simplify sin⁡(90°−θ)⋅sec⁡θ\sin(90° - \theta) \cdot \sec\theta for all valid values of θ\theta.
Q9·Challenging
Simplify sin⁡2θ1−cos⁡θ\dfrac{\sin^2\theta}{1 - \cos\theta} by expressing sin⁡2θ\sin^2\theta in terms of cos⁡θ\cos\theta.
Q10·Challenging
What is the value of (1−sin⁡2θ)(1+tan⁡2θ)(1 - \sin^2\theta)(1 + \tan^2\theta) for all valid θ\theta?
Q11·Challenging
Simplify cos⁡θ1−sin⁡θ+cos⁡θ1+sin⁡θ\dfrac{\cos\theta}{1 - \sin\theta} + \dfrac{\cos\theta}{1 + \sin\theta}.
Q12·Challenging
What is the value of sin⁡θcsc⁡θ+cos⁡θsec⁡θ\dfrac{\sin\theta}{\csc\theta} + \dfrac{\cos\theta}{\sec\theta} for all valid θ\theta?