(a)[4]
By expanding $(\cos^2 x + \sin^2 x)^3$ first, or by another valid approach, prove that $\cos^6 x + \sin^6 x = 1 - \frac{3}{4}\sin^2 2x$.
(b)[4]
Hence solve the equation $\cos^6 x + \sin^6 x = \frac{2}{3}$, for $0^\circ < x < 180^\circ$.
Mathematics 9709 · AS & A Level · Trigonometry
Trigonometry — practice question
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Carry out a cubic expansion and then set the result equal to $1$” …
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