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JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If A=(7)/(25) and (3 π)/(2) < A < 2 π , then (A)/(4)+ (A)/(2)- 2 A=

If

\cos A=\frac{7}{25}

and

\frac{3 \pi}{2} < A < 2 \pi

, then

\cos \frac{A}{4}+\cos \frac{A}{2}-\cos 2 A=

Options

A $\frac{1}{\sqrt{10}}+\frac{27}{625}$
B $\frac{3}{\sqrt{10}}-\frac{27}{625}$
C $\frac{3}{\sqrt{10}}+\frac{27}{625}$
D $\frac{1}{\sqrt{10}}-\frac{27}{625}$

Correct answer

A. $\frac{1}{\sqrt{10}}+\frac{27}{625}$

Step-by-step solution

Given,

\quad \cos A=\frac{7}{25}

and

\frac{3 \pi}{2} < A < 2 \pi

Then,

\quad \cos 2 A=2\left(\frac{49}{625}\right)-1=\frac{98-625}{625}=-\frac{527}{625}\cos \frac{A}{2}=-\frac{4}{5}

and

\cos \frac{A}{4}=\frac{1}{\sqrt{10}}

So,

\quad \cos \frac{A}{4}+\cos \frac{A}{2}-\cos 2 A=\frac{1}{\sqrt{10}}+\frac{27}{625}

.

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