Question

$1+\cos 2x+\cos 4x+\cos 6x=$

• 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. $Noneofthese$

Answer 1

The correct option is C

$4\cos x\cos 2x\cos 3x$

Explanation for the correct option:

Find the sum of the series using trigonometric identities

$1+\cos 2x+\cos 4x+\cos 6x=(1+\cos 2x)+(\cos 4x+\cos 6x)$

$=2{\cos }^{2}x+\left[2\cos \left(\frac{6x+4x}{2}\right)\cos \left(\frac{6x-4x}{2}\right)\right]$ $[\because \cos 2A=2{\cos }^{2}A-1\mathrm{and}\mathrm{cosA}+\mathrm{cosB}=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}]$

$=2{\cos }^{2}x+2\cos 5x\cos x$

$=2\cos x(\cos x+\cos 5x)$

$=2\cos x(2\cos 2x\cos 3x)$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathit{A}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathit{B}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{A}\mathbf{+}\mathbf{B}}{\mathbf{2}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{A}\mathbf{-}\mathbf{B}}{\mathbf{2}}\right]$

$\mathbf{=}\mathbf{4}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{x}$

Hence, Option ‘C’ is Correct.