Question

If the period of the function $f(x)=\sin 5x\cos 3x$ is $\alpha$, then $\cos \alpha =$

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

Answer: (D)

We have, $f(x)=\sin 5x\cdot \cos 3x$.
$=\frac{1}{2}(2\sin 5x\cdot \cos 3x)-\frac{1}{2}(\sin 8x+\sin 2x)$
Clearly, $\sin 8x$ is periodic with period $\frac{2\pi }{8}=\frac{\pi }{4}$ and
$\sin 2x$ is periodic with period $\frac{2\pi }{2}=\pi$
$\therefore$ Period of $f(x)=\text{ L.C.M }\left\{\frac{\pi }{4},\pi \right\}$
$\frac{\text{ L.C. }M\{\pi ,\pi \}}{\text{ H.C.F }\{4,1\}}=\frac{\pi }{1}=\pi \Rightarrow \alpha =\pi$
Hence, $\cos \alpha =\cos \pi =-1$