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Trigonometry: Question 8

Syllabus 1.5

Structured AS 6 marks

(a) Prove that sin⁡θtan⁡θ+cos⁡θ≡1cos⁡θ\sin\theta\tan\theta + \cos\theta \equiv \dfrac{1}{\cos\theta} for all θ\theta where tan⁡θ\tan\theta is defined. [4]

(b) Hence, or otherwise, find the exact value of sin⁡60°tan⁡60°+cos⁡60°\sin 60°\tan 60° + \cos 60°. [2]

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Worked solution

Part (a): Proving the identity

Start from the left-hand side and use tan⁡θ≡sin⁡θcos⁡θ\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}: sin⁡θtan⁡θ+cos⁡θ=sin⁡θ⋅sin⁡θcos⁡θ+cos⁡θ=sin⁡2θcos⁡θ+cos⁡θ\sin\theta\tan\theta + \cos\theta = \sin\theta\cdot\frac{\sin\theta}{\cos\theta} + \cos\theta = \frac{\sin^2\theta}{\cos\theta} + \cos\theta

Write cos⁡θ\cos\theta as a fraction with the same denominator, cos⁡θ=cos⁡2θcos⁡θ\cos\theta = \dfrac{\cos^2\theta}{\cos\theta}: sin⁡2θcos⁡θ+cos⁡2θcos⁡θ=sin⁡2θ+cos⁡2θcos⁡θ\frac{\sin^2\theta}{\cos\theta} + \frac{\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta+\cos^2\theta}{\cos\theta}

Apply the identity sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta \equiv 1 to the numerator: sin⁡2θ+cos⁡2θcos⁡θ=1cos⁡θ\frac{\sin^2\theta+\cos^2\theta}{\cos\theta} = \frac{1}{\cos\theta}

So sin⁡θtan⁡θ+cos⁡θ≡1cos⁡θ\sin\theta\tan\theta+\cos\theta \equiv \dfrac{1}{\cos\theta}, as required.

Part (b): Applying the identity

Taking θ=60°\theta = 60° in the identity from part (a): sin⁡60°tan⁡60°+cos⁡60°=1cos⁡60°\sin 60°\tan 60° + \cos 60° = \frac{1}{\cos 60°}

Using the exact value cos⁡60°=12\cos 60° = \dfrac12: 1cos⁡60°=112=2\frac{1}{\cos 60°} = \frac{1}{\tfrac12} = 2

Check by direct calculation: sin⁡60°=32\sin 60° = \dfrac{\sqrt3}{2}, tan⁡60°=3\tan 60° = \sqrt3, so sin⁡60°tan⁡60°+cos⁡60°=32×3+12=32+12=2✓\sin 60°\tan 60° + \cos 60° = \frac{\sqrt3}{2}\times\sqrt3 + \frac12 = \frac32 + \frac12 = 2 \checkmark

Final answers

  • (a) sin⁡θtan⁡θ+cos⁡θ≡1cos⁡θ\sin\theta\tan\theta+\cos\theta \equiv \dfrac{1}{\cos\theta}, as shown above.
  • (b) sin⁡60°tan⁡60°+cos⁡60°=2\sin 60°\tan 60° + \cos 60° = \boxed{2}