2cos x - cos 3x - cos 5x =

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3.Trigonometrical Ratios, Functions and Identities

easy

$2\cos x - \cos 3x - \cos 5x = $

A

$16{\cos ^3}x{\sin ^2}x$

B

$16{\sin ^3}x{\cos ^2}x$

C

$4{\cos ^3}x{\sin ^2}x$

D

$4{\sin ^3}x{\cos ^2}x$

Solution

(a) $2\,\cos x – \cos \,3x – \cos \,5x = 2\cos x(1 – \cos 4x)$

$ = 2\cos x\,2{\sin ^2}2x$$ = 4\,\cos x\,{\sin ^2}\,2x $

$= 16\,{\sin ^2}x\,{\cos ^3}x$.

Standard 11

Mathematics

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