If the period of the function f(x) = frac{tan | vedclass

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(C) The periods of $	an 5x$,$\cos 3x$,and $\sin 6x$ are $\frac{\pi}{5}$,$\frac{2\pi}{3}$,and $\frac{\pi}{3}$ respectively. To find the period $\alpha

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) The periods of $	an 5x$,$\cos 3x$,and $\sin 6x$ are $\frac{\pi}{5}$,$\frac{2\pi}{3}$,and $\frac{\pi}{3}$ respectively. To find the period $\alpha$ of $f(x) = \frac{	an 5x \cos 3x}{\sin 6x}$,we simplify the expression: $f(x) = \frac{	an 5x \cos 3x}{2 \sin 3x \cos 3x} = \frac{	an 5x}{2 \sin 3x}$. The period of $	an 5x$ is $\frac{\pi}{5}$ and the period of $\sin 3x$ is $\frac{2\pi}{3}$. The period $\alpha$ of $f(x)$ is the least common multiple of $\frac{\pi}{5}$ and $\frac{2\pi}{3}$,which is $\frac{2\pi}{\gcd(1, 1)} = 2\pi$. Thus,$\alpha = 2\pi$. We need to evaluate $f\left(\frac{\alpha}{8}ight) = f\left(\frac{2\pi}{8}ight) = f\left(\frac{\pi}{4}ight)$. $f\left(\frac{\pi}{4}ight) = \frac{	an(5\pi/4)}{2 \sin(3\pi/4)} = \frac{1}{2 	imes (1/\sqrt{2})} = \frac{1}{\sqrt{2}}$.

If the period of the function $f(x) = \frac{\tan 5x \cos 3x}{\sin 6x}$ is $\alpha$,then find the value of $f\left(\frac{\alpha}{8}\right)$.

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