Target Mathematics

1.89: Trigonometric identities and equations

4PM1/1/January/2015 — Question 8 · 12 marks

tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}
(a) Show that 1+tan⁡2θ=1cos⁡2θ1+\tan^2\theta=\dfrac1{\cos^2\theta}. (3)
(b) Show that
1+sin⁡θcos⁡θ+sin⁡2θcos⁡2θ=1+tan⁡θ+2tan⁡2θ\dfrac{1+\sin\theta\cos\theta+\sin^2\theta}{\cos^2\theta}=1+\tan\theta+2\tan^2\theta
(3)
(c) Solve, for 0∘≤θ≤180∘0^\circ\le\theta\le180^\circ,
1+sin⁡θcos⁡θ+sin⁡2θ=4cos⁡2θ1+\sin\theta\cos\theta+\sin^2\theta=4\cos^2\theta
giving your answers in degrees to 1 decimal place where appropriate. (6)