Where do the curves $y=\sin 2x$ and $y=\sin x$ cross?

Where do the curves $y=\sin 2x$ and $y=\sin x$ cross?

Question

The diagram shows part of the graphs of the curves $y=\sin 2x$ and $y=\sin x$.

$A$ is a maximum point of the curve $y=\sin 2x$. $B$ is a maximum point of the curve $y=\sin x$.

$C$ is the point of intersection of the curves shown. Find

1. the $x$ co-ordinate of $C$,

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1. the shaded area.

Solution

To find the $x$-coordinate of $C$, we need to solve the equation $\sin x = \sin 2x$ in the region $0 \le x \le \frac{1}{2}\pi$.

Now $\sin x = \sin 2x \iff \sin x = 2 \sin x \cos x$. We can see that $\sin x$ is not zero at $C$, so the solution we seek is $\cos x = \dfrac{1}{2}$, and so $x = \dfrac{\pi}{3}$.

Let’s call the shaded area $R$. To find $R$, we start with the area of the rectangle spanned by the line $AB$ and the $x$-axis, which is clearly $\dfrac{\pi}{4}$.

We then subtract the area below the curves $y=\sin x$ and $y=\sin 2x$ as shown by the red area and blue area respectively in the diagram. Thus

$$R=\frac{\pi}{4}-\int_{\pi/4}^{\pi/3} \sin 2x\:dx -\int_{\pi/3}^{\pi/2} \sin x\:dx.$$

Now $\int_{\pi/3}^{\pi/2} \sin x\:dx = \bigl[-\cos x\bigr]_{\pi/3}^{\pi/2} = \dfrac{1}{2}.$

And thinking about stretching horizontally $\int_{\pi/4}^{\pi/3}\sin 2x\:dx = \frac{1}{2}\int_{\pi/2}^{2\pi/3}\sin x\:dx = \frac{1}{2}\bigl[-\cos x\bigr]_{\pi/2}^{2\pi/3} = \dfrac{1}{4}.$

Thus $R = \dfrac{\pi}{4}-\dfrac{3}{4}.$