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AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 21 of 160

Marks: +1, -0

\cos \frac{\pi}{4} \cos \frac{\pi}{8} \cos \frac{\pi}{16} \cos \frac{\pi}{32}

=2^{m} \csc \frac{\pi}{n}

m+n=

Solution:

\cos \frac{\pi}{4} \cos \frac{\pi}{8} \cos \frac{\pi}{16} \cos \frac{\pi}{32}

=\cos \frac{\pi}{2^2} \cos \frac{\pi}{2^3} \cos \frac{\pi}{2^4} \cos \frac{\pi}{2^5}

=\frac{1}{2\sin\left(\frac{\pi}{2^5}\right)}\left[2\sin\left(\frac{\pi}{2^5}\right)\cos\left(\frac{\pi}{2^5}\right)\cos\frac{\pi}{2^2}\cos\frac{\pi}{2^3}\cos\frac{\pi}{2^4}\right]

=\frac{1}{2}\csc\left(\frac{\pi}{2^5}\right)\left[\sin\frac{\pi}{2^4}\cdot\cos\frac{\pi}{2^4}\cdot\cos\frac{\pi}{2^2}\cdot\cos\frac{\pi}{2^3}\right]

=\frac{1}{4}\csc\left(\frac{\pi}{2^5}\right)\left[\sin\frac{\pi}{2^3}\cdot\cos\frac{\pi}{2^3}\cdot\cos\frac{\pi}{2^2}\right]

=\frac{1}{8}\csc\left(\frac{\pi}{2^5}\right)\left[\sin\frac{\pi}{2^2}\cdot\cos\frac{\pi}{2^2}\right]

=\frac{1}{16}\csc\frac{\pi}{2^5}\sin\frac{\pi}{2}

=2^{-4}\csc\frac{\pi}{2^5}

\left[\because \sin\frac{\pi}{2}=1\right]

\therefore m=-4

and

n=2^5

, then

m+n=-4+32=28

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