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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 21 of 160

Marks: +1, -0

If

\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15}

\cos \frac{7\pi}{15} \cos \frac{30\pi}{15}=x,

\frac{1}{8x}=

4
$\frac{1}{4}$
8
$\frac{4}{3}$

Solution:

Consider the function.

\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15}

\cos \frac{7\pi}{15} \cos \frac{30\pi}{15}=x

It is solved as,

\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}

=2x

\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \left(\pi - \frac{8\pi}{15}\right)

=2x

-\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos 2^{2} \frac{\pi}{15} \cos 2^{3} \frac{\pi}{15}

=2x

-\frac{\sin 2^{4} \frac{\pi}{15}}{2^{4} \sin \frac{\pi}{15}}=2x

Thus,

2x=-\frac{\sin \frac{16\pi}{15}}{16 \sin \frac{\pi}{15}}

2x=-\frac{\sin \left(\pi + \frac{\pi}{15}\right)}{16 \sin \frac{\pi}{15}}

2x=-\frac{\sin \left(\frac{\pi}{15}\right)}{16 \sin \frac{\pi}{15}}

2x=\frac{1}{16}

Hence,

x=\frac{1}{32}

\frac{1}{x}=32

\frac{1}{8x}=4

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