Target Mathematics

1.9: Cosec identity and compound-angle equation

0606/11/O/N/20 — Question 8 · 12 marks

(a)(i)
Show that 1(1+csc⁡θ)(sin⁡θ−sin⁡2θ)=sec⁡2θ\tfrac{1}{(1 + \csc\theta)(\sin\theta - \sin^{2}\theta)} = \sec^{2}\theta.
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(a)(ii)
Hence solve (1+csc⁡θ)(sin⁡θ−sin⁡2θ)=34(1 + \csc\theta)(\sin\theta - \sin^{2}\theta) = \tfrac{3}{4} for −180°≤θ≤180°-180^{\degree} \le \theta \le 180^{\degree}.
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(b)
Solve sin⁡(3ϕ+2π3)=cos⁡(3ϕ+2π3)\sin\left(3\phi + \tfrac{2\pi}{3}\right) = \cos\left(3\phi + \tfrac{2\pi}{3}\right) for 0≤ϕ≤2π30 \le \phi \le \tfrac{2\pi}{3} radians, giving your answers in terms of π\pi.
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