Mathematics 9709 · AS & A Level · Trigonometry

Trigonometry — practice question

The graph depicts the curve $y = \sin 3x \cos x$ for $0 \leq x \leq \frac{1}{2}\pi$, together with its minimum point $M$. The shaded region $R$ lies between the curve and the $x$-axis.
(i)[3]

Expand $\sin(3x + x)$ and $\sin(3x - x)$ to prove that $\sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)$.

(ii)[4]

Using the result from part (i) and showing all the required working, determine the exact area of the region $R$.

(iii)[5]

Using the result from part (i), express $\frac{dy}{dx}$ in terms of $\cos 2x$ and hence determine the $x$-coordinate of $M$, giving your answer correct to 2 decimal places.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: State the correct expansion of $\sin(3x+x)$ or $\sin(3x-x)$

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