Question

Solve $1 + \cos 2x + \cos 4x + \cos 6x =$
A.$2\cos x\cos 2x\cos 3x$
B.$4\sin x\cos 2x\cos 3x$
C.$4\cos x\cos 2x\cos 3x$
D.None of these

Answer 1

Hint : In the given question the addition of trigonometric ratios are given , so we have to use the identity formula for compound angles for $\cos x$ . First solve for larger angles ($\cos 4x$and $\cos 6x$ ) than for smaller angles . Also , here direct $1 + \cos 2x$ is given which is equal to $1 + \cos 2x = 2{\cos ^2}x$

Complete step-by-step answer :
Given : $1 + \cos 2x + \cos 4x + \cos 6x$ ,
Applying the identity formula $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$ for $\cos 4x$and $\cos 6x$ we get ,
$= 1 + \cos 2x + 2\cos (\dfrac{{6x - 4x}}{2})\cos (\dfrac{{6x + 4x}}{2})$ , on solving we get ,
$= 1 + \cos 2x + 2\cos x\cos 5x$
Now applying formula $1 + \cos 2x = 2{\cos ^2}x$ we get ,
$= 2{\cos ^2}x + 2\cos x\cos 5x$ taking $2\cos x$common we get ,
$= 2\cos x(\cos x + \cos 5x)$
Now again applying identity $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$ we get ,
$= 2\cos x\left[ {2\cos (\dfrac{{5x + x}}{2})\cos (\dfrac{{5x - x}}{2})} \right]$ , on solving we get ,
$= 2\cos x\left[ {2\cos 3x\cos 2x} \right]$
On solving further we get ,
$= 4\cos x\cos 2x\cos 3x$ .
Therefore , option ( 3 ) is the correct answer .
So, the correct answer is “Option 3”.

Note: Alternative Method :
Given : $1 + \cos 2x + \cos 4x + \cos 6x$
Now , this time taking the terms $\cos 6x$ and $\cos 2x$ together and applying the formula $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$ , again we get
$= 1 + \cos 4x + 2\cos (\dfrac{{6x - 2x}}{2})\cos (\dfrac{{6x + 2x}}{2})$ , on solving we get ,
$= 1 + \cos 4x + 2\cos 2x\cos 4x$
Now using the formula $\cos 2x = 2{\cos ^2}x - 1$ we get ,
$= 2{\cos ^2}2x + 2\cos 2x\cos 4x$, now taking the $2\cos 2x$ common we get
$= 2\cos 2x(\cos 2x + \cos 4x)$
Now again applying the formula $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$ we get ,
$= 2\cos 2x\left[ {2\cos (\dfrac{{4x - 2x}}{2})\cos (\dfrac{{4x + 2x}}{2})} \right]$ on solving further we get ,
$= 2\cos 2x\left[ {2\cos x\cos 3x} \right]$, on simplifying we get
$= 4\cos x\cos 2x\cos 3x$ .