Prove that cos 2x cos frac{x}{2} - cos 3x cos | vedclass

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(N/A) We have $L.H.S. = \cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2}$ Multiply and divide by $2$: $L.H.S. = \frac{1}{2} [2 \cos 2x \cos \frac{

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(N/A) We have $L.H.S. = \cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2}$
Multiply and divide by $2$:
$L.H.S. = \frac{1}{2} [2 \cos 2x \cos \frac{x}{2} - 2 \cos \frac{9x}{2} \cos 3x]$
Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$L.H.S. = \frac{1}{2} [(\cos(2x + \frac{x}{2}) + \cos(2x - \frac{x}{2})) - (\cos(\frac{9x}{2} + 3x) + \cos(\frac{9x}{2} - 3x))]$
$L.H.S. = \frac{1}{2} [\cos \frac{5x}{2} + \cos \frac{3x}{2} - \cos \frac{15x}{2} - \cos \frac{3x}{2}]$
$L.H.S. = \frac{1}{2} [\cos \frac{5x}{2} - \cos \frac{15x}{2}]$
Using the identity $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:
$L.H.S. = \frac{1}{2} [-2 \sin(\frac{\frac{5x}{2} + \frac{15x}{2}}{2}) \sin(\frac{\frac{5x}{2} - \frac{15x}{2}}{2})]$
$L.H.S. = -\sin(5x) \sin(-\frac{5x}{2})$
Since $\sin(-\theta) = -\sin \theta$:
$L.H.S. = \sin 5x \sin \frac{5x}{2} = R.H.S.$

Prove that $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$

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