A-level Mathematics
    All books

    5.7 Trigonometric identities — Practise with help

    Start every proof on one side. Write the chain of equals signs downwards, and change only the side you started with.

    Warm-up: remember

    1. Expand (a+b)2(a + b)^2.
    2. Write tan⁡θ\tan\theta as a fraction in terms of sine and cosine.
    3. Simplify x2x\dfrac{x^2}{x} for x≠0x \neq 0.
    Answers
    1. (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.
    2. tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}.
    3. x2x=x\dfrac{x^2}{x} = x.

    If question 2 was hard, revisit the definition of tangent before you continue.

    Practise with help

    1. Prove that sin⁡2θsin⁡θ≡sin⁡θ\dfrac{\sin^2\theta}{\sin\theta} \equiv \sin\theta for sin⁡θ≠0\sin\theta \neq 0.

    Hint

    Cancel one factor of sine from the top and the bottom.

    Next step

    Write sin⁡2θsin⁡θ=sin⁡θ×sin⁡θsin⁡θ\dfrac{\sin^2\theta}{\sin\theta} = \dfrac{\sin\theta \times \sin\theta}{\sin\theta}.

    Full working sin⁡2θsin⁡θ=sin⁡θ×sin⁡θsin⁡θ=sin⁡θ.\frac{\sin^2\theta}{\sin\theta} = \frac{\sin\theta \times \sin\theta}{\sin\theta} = \sin\theta.

    The left-hand side becomes the right-hand side, so the identity holds wherever sin⁡θ≠0\sin\theta \neq 0.

    2. Prove that cos⁡θtan⁡θ≡sin⁡θ\cos\theta\tan\theta \equiv \sin\theta for cos⁡θ≠0\cos\theta \neq 0.

    Hint

    Replace tangent with sin⁡θcos⁡θ\dfrac{\sin\theta}{\cos\theta}, then cancel cosine.

    Next step

    Write cos⁡θ×sin⁡θcos⁡θ\cos\theta \times \dfrac{\sin\theta}{\cos\theta}.

    Full working cos⁡θtan⁡θ=cos⁡θ×sin⁡θcos⁡θ=sin⁡θ.\cos\theta\tan\theta = \cos\theta \times \frac{\sin\theta}{\cos\theta} = \sin\theta.

    3. Prove that 1+tan⁡2θ≡sec⁡2θ1 + \tan^2\theta \equiv \sec^2\theta.

    Hint

    Write the left-hand side in sines and cosines over a common denominator.

    Next step

    1+sin⁡2θcos⁡2θ=cos⁡2θ+sin⁡2θcos⁡2θ1 + \dfrac{\sin^2\theta}{\cos^2\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\cos^2\theta}. Now use the first identity on the numerator.

    Full working 1+tan⁡2θ=1+sin⁡2θcos⁡2θ=cos⁡2θ+sin⁡2θcos⁡2θ=1cos⁡2θ=sec⁡2θ.1 + \tan^2\theta = 1 + \frac{\sin^2\theta}{\cos^2\theta} = \frac{\cos^2\theta + \sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} = \sec^2\theta.

    4. Prove that (sin⁡θ−cos⁡θ)2≡1−2sin⁡θcos⁡θ(\sin\theta - \cos\theta)^2 \equiv 1 - 2\sin\theta\cos\theta.

    Hint

    Expand the square, then look for sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta.

    Next step

    (sin⁡θ−cos⁡θ)2=sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ(\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta. Group the two squares.

    Full working (sin⁡θ−cos⁡θ)2=sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=(sin⁡2θ+cos⁡2θ)−2sin⁡θcos⁡θ=1−2sin⁡θcos⁡θ.(\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = (\sin^2\theta + \cos^2\theta) - 2\sin\theta\cos\theta = 1 - 2\sin\theta\cos\theta.

    5. Prove that cos⁡2θ1−sin⁡θ≡1+sin⁡θ\dfrac{\cos^2\theta}{1 - \sin\theta} \equiv 1 + \sin\theta for sin⁡θ≠1\sin\theta \neq 1.

    Hint

    Replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta, then factorise the top as a difference of two squares.

    Next step

    1−sin⁡2θ=(1−sin⁡θ)(1+sin⁡θ)1 - \sin^2\theta = (1 - \sin\theta)(1 + \sin\theta). Cancel one factor of 1−sin⁡θ1 - \sin\theta.

    Full working cos⁡2θ1−sin⁡θ=1−sin⁡2θ1−sin⁡θ=(1−sin⁡θ)(1+sin⁡θ)1−sin⁡θ=1+sin⁡θ.\frac{\cos^2\theta}{1 - \sin\theta} = \frac{1 - \sin^2\theta}{1 - \sin\theta} = \frac{(1 - \sin\theta)(1 + \sin\theta)}{1 - \sin\theta} = 1 + \sin\theta.

    6. Prove that sec⁡2θ+cosec⁡2θ≡sec⁡2θ cosec⁡2θ\sec^2\theta + \operatorname{cosec}^2\theta \equiv \sec^2\theta\,\operatorname{cosec}^2\theta for sin⁡θcos⁡θ≠0\sin\theta\cos\theta \neq 0.

    Hint

    Write both sides in sines and cosines. The right-hand side is simpler than it looks.

    Next step

    The right-hand side is 1cos⁡2θsin⁡2θ\dfrac{1}{\cos^2\theta\sin^2\theta}. On the left, put the two fractions over the common denominator cos⁡2θsin⁡2θ\cos^2\theta\sin^2\theta.

    Full working sec⁡2θ+cosec⁡2θ=1cos⁡2θ+1sin⁡2θ=sin⁡2θ+cos⁡2θcos⁡2θsin⁡2θ=1cos⁡2θsin⁡2θ=sec⁡2θ cosec⁡2θ.\sec^2\theta + \operatorname{cosec}^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta\sin^2\theta} = \frac{1}{\cos^2\theta\sin^2\theta} = \sec^2\theta\,\operatorname{cosec}^2\theta.

    Settings

    Colour scheme

    Applies to every book you read here.

    Reading theme

    Applies to every book you read here.