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
\[ = 4\cos x\cos 2x\cos 3x\] .