Question

oneplus cos 2x + cos 4 X + Cos 6 X equal 4 cos x cos2x cos3x

Answer 1

$\textbf{Answer}$

We need to prove,
1 + cos 2x + cos 4x + cos 6x = 4.cosx.cos2x.cos3x

We will use following trigonometric identities-

$\textbf{cos 2x = 2.cos^2 x - 1}$ -----(1)

$\textbf{cos A + cos B = 2.cos (A+B)/2 cos (A-B)/2}$ -------(2)

$\textbf{cos (-A) cos A}$ -------(3)

Lets comeback to the question,

LHS = 1 + cos 2x + cos 4x + cos 6x

$\textbf{Using the identity (1),}$

=> LHS = 1 + 2cos^2 (2x) - 1 + cos 2x + cos 6x

=> LHS = 2cos^2 (2x) + cos 2x + cos 6x

$\textbf{Using the identity (2) now,}$

=> LHS = 2cos^2 (2x) + 2cos (2x+6x)/2 . cos (2x-6x)/2

=> LHS = 2cos^2 (2x) + 2cos 4x . cos (-2x)

$\textbf{Using the identity (3) now,}$

=> LHS = 2cos^2 (2x) + 2cos 2x . cos 4x

=> LHS = 2cos 2x (cos 2x + cos 4x)

$\textbf{Again using the identity (2) now,}$

=> LHS = 2cos 2x {2cos (2x + 4x)/2 . cos (2x - 4x)/2}

=> LHS = 4cos2x {cos 3x . cos (-x)}

$\textbf{Again using the identity (3),}$

=> LHS = 4.cosx.cos2x.cos3x

$\textbf{LHS = RHS}$


$\textbf{Hope My Answer Helped}$

$\textbf{Thanks}$

Answer 2

$= > 1 + cos2x + cos4x + cos6x \\ \\ = > 1 + cos4x + cos2x + cos6x \\ \\ = > 2 {cos}^{2} x + 2cos( \frac{6x + 2x}{2} ).cos( \frac{6x - 2x}{2} ) \\ \\ = > 2 {cos}^{2} 2x + 2cos4x.cos( 2x) \\ \\ = > 2 {cos}^{2} 2x + 2cos4x.cos2x \\ \\ = > 2cos2x(cos2x + cos4x) \\ \\ = > 2cos2x[2cos( \frac{4x + 2x}{2} ).cos( \frac{4x - 2x}{2} )] \\ \\ = > 2cos2x.2cos3x.cocx \\ \\ = > 4cosx.cos2x.cos3x$
_____________________[◢PROVED]

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