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TS EAMCET 04 May 2019 Shift 2 Solved Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)

\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)

=

$\frac{1+\sqrt{2}}{2\sqrt{2}}$
$\frac{\pi}{8}$
$\frac{1}{8}$
$\frac{1}{2}$

Solution:

Consider the expression.

\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)

\text{ Let } x=\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)

It is solved as,

x = \left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)

x = \left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1-\cos\frac{3\pi}{8}\right)\left(1-\cos\frac{\pi}{8}\right)

= \left(1-\cos^2\frac{\pi}{8}\right)\left(1-\cos^2\frac{3\pi}{8}\right)

= \left(\sin^2\frac{\pi}{8}\right)\left(\sin^2\frac{3\pi}{8}\right)

Solve further,

x = \sin^2\left(\frac{\pi}{8}\right) \sin^2\left(\frac{\pi}{2}-\frac{\pi}{8}\right)

= \sin^2\left(\frac{\pi}{8}\right) \cos^2\left(\frac{\pi}{8}\right)

= \frac{1}{4} \left(2 \sin\frac{\pi}{8} \cos\frac{\pi}{8}\right)^2

= \frac{1}{4} \sin^2\frac{\pi}{4}

Solve further

x = \frac{1}{4} \times \frac{1}{2}

= \frac{1}{8}

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