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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

If the period of the function

f(x)=\sin 5x \cdot \cos 3x

\alpha

\cos \alpha =

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

Solution:

Consider the function.

f(x)=\sin 5x \cdot \cos 3x

It can be simplified as,

f(x)=\frac{1}{2}(2 \sin 5x \cdot \cos 3x)

= \frac{1}{2}(\sin 8x + \sin 2x)

Now,

\sin 8x

is periodic with period

\frac{2\pi}{8}

And,

\sin 2x

is periodic with period

\frac{2\pi}{2}

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

So,

\alpha = \pi

Hence,

\cos \alpha = \cos \pi

= -1

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