Question

If $\frac{-\pi }{2}\le x\le \frac{\pi }{2}$, then the two curves $y=\cos x$ and $y=\sin 3x$ intersect at

• A. $\left(\frac{\pi }{4},\frac{1}{\sqrt{2}}\right)$ and $\left(\frac{\pi }{8},\cos \frac{\pi }{8}\right)$
• B. $\left(-\frac{\pi }{4},\frac{1}{\sqrt{2}}\right)$ and $\left(\frac{-\pi }{8},\cos \frac{\pi }{8}\right)$
• C. $\left(\frac{\pi }{4},\frac{-1}{\sqrt{2}}\right)$ and $\left(\frac{\pi }{8},-\cos \frac{\pi }{8}\right)$
• D. $\left(\frac{-\pi }{4},\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (A)

At the intersection point of $y=\cos x$ and $y=\sin 3x$,
we have $\cos x=\sin 3x$
$\Rightarrow \cos x=\cos \left(\frac{\pi }{2}-3x\right)$
$\Rightarrow x=2n\pi \pm \left(\frac{\pi }{2}-3x\right)$
$\Rightarrow x=\frac{\pi }{4},\frac{\pi }{8}[\because -\pi /2\le x\le \pi /2]$
So, $y=\cos \frac{\pi }{4}\text{ at }x=\frac{\pi }{4}$
and $y=\cos \frac{\pi }{8},\text{ at }x=\frac{\pi }{8}$
Thus, the points are $\left(\frac{\pi }{4},\frac{1}{\sqrt{2}}\right)$ and $\left(\frac{\pi }{8},\cos \frac{\pi }{8}\right)$