(i)[3]
By expanding $\cos(3x - x)$ and $\cos(3x + x)$, prove that $\frac{1}{2}(\cos 2x - \cos 4x) = \sin 3x \sin x$.
(ii)[3]
Hence show that $\int_{\pi/6}^{\pi/3} \sin 3x \sin x \, dx = \frac{1}{8}\sqrt{3}$.
Mathematics 9709 · AS & A Level · Trigonometry
Trigonometry — practice question
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State the correct expansion of $\cos(3x-x)$ or $\cos(3x+x)$” …
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