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If the period of the function $f(x)=\sin 5 x \cos 3 x$ is $\alpha$, then $\cos \alpha=$
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Verified Answer
The correct answer is:
-1
We have, $f(x)=\sin 5 x \cdot \cos 3 x$.
$$
=\frac{1}{2}(2 \sin 5 x \cdot \cos 3 x)=\frac{1}{2}(\sin 8 x+\sin 2 x)
$$
Clearly, $\sin 8 x$ is periodic with period $\frac{2 \pi}{8}=\frac{\pi}{4}$ and $\sin 2 x$ is periodic with period $\frac{2 \pi}{2}=\pi$.
$\therefore$ Period of $f(x)=$ L.C.M $\left\{\frac{\pi}{4}, \pi\right\}$
$$
\frac{\operatorname{L.C} M\{\pi, \pi\}}{\operatorname{H.CF}\{4,1\}}=\frac{\pi}{1}=\pi \Rightarrow \alpha=\pi
$$
Hence, $\cos \alpha=\cos \pi=-1$
$$
=\frac{1}{2}(2 \sin 5 x \cdot \cos 3 x)=\frac{1}{2}(\sin 8 x+\sin 2 x)
$$
Clearly, $\sin 8 x$ is periodic with period $\frac{2 \pi}{8}=\frac{\pi}{4}$ and $\sin 2 x$ is periodic with period $\frac{2 \pi}{2}=\pi$.
$\therefore$ Period of $f(x)=$ L.C.M $\left\{\frac{\pi}{4}, \pi\right\}$
$$
\frac{\operatorname{L.C} M\{\pi, \pi\}}{\operatorname{H.CF}\{4,1\}}=\frac{\pi}{1}=\pi \Rightarrow \alpha=\pi
$$
Hence, $\cos \alpha=\cos \pi=-1$
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