# 2 cos x − c o s 3 x − cos 5 x − 16 cos 3 x sin 2 x - Mathematics

MCQ

$2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x$

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#### Solution

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$\text{ We have,}$

$2\text{ cos } x - \cos3x - \cos5x - 16 \cos^3 x \sin^2 x$

$= 2\text{ cos } x - \cos3x - \cos5x - 16\left[ \frac{\cos3x + 3\text{ cos } x}{4} \times \frac{\left( 1 - \cos2x \right)}{2} \right]$

$= 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \left( \cos3x + 3\text{ cos } x \right)\left( 1 - \cos2x \right) \right]$

$= 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \cos3x - \cos3x \cos2x + 3\text{ cos } x - 3\text{ cos } x \cos2x \right]$

$= 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \cos3x + 3\text{ cos } x \right] + 2\cos3x \cos2x + 3\left[ 2\text{ cos } x \cos2x \right]$

$= 2\text{ cos } x - \cos3x - \cos5x - 2\left[ \cos3x + 3\text{ cos } x \right] + \cos5x + \text{ cos } x + 3\cos3x + 3\text{ cos } x$

$\left[ 2cosAcosB = \cos\left( A + B \right) + \cos\left( A - B \right) \right]$

$= 2\text{ cos } x - \cos3x - \cos5x - 2\cos3x - 6\text{ cos } x + \cos5x + \text{ cos } x + 3\cos3x + 3\text{ cos } x = 0$

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Q 16 | Page 44