First slide

Multiple and sub- multiple Angles

Question

If $2 \sin^{2} ((\frac{π}{2}) \cos^{2} x) = 1 - \cos (πsin 2 x), x \neq (2 n + 1) \frac{π}{2}$ $, n \in I$ then cos 2x is equal to

Easy

A

$\frac{1}{5}$
B

$\frac{3}{5}$
C

$\frac{4}{5}$
D

1

Solution

The given equation is equivalent to
$\begin{array}{l} 2 \sin^{2} ((π / 2) \cos^{2} x) = 2 \sin^{2} ((π / 2) \sin 2 x) \\ or \cos^{2} x = \sin 2 x \\ or \cos x (\cos x - 2 \sin x) = 0 \\ \Rightarrow 1 - 2 \tan x = 0 as \cos x \neq 0, x \neq (2 n + 1) \frac{π}{2} \\ or \tan x = \frac{1}{2} \\ \Rightarrow \cos 2 x = \frac{1 - \tan^{2} x}{1 + \tan^{2} x} = \frac{3}{5} \end{array}$

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