# Q (15) Prove the following $\small \cot4x(\sin5x + \sin3x) = \cot x(\sin5x - \sin3x)$

S safeer

We know that
$\sin x + \sin y = 2\sin\left ( \frac{x+y}{2} \right )\cos\left (\frac{x-y}{2} \right )$
By using this , we get

sin5x + sin3x = 2sin4xcosx

$\frac{\cos4x}{\sin4x}\left ( 2\sin4x\cos x \right ) = 2\cos4x\cos x\\ \\$

now nultiply and divide by sin x

$\\\ \\ \frac{2\cos4x\cos x \sin x}{\sin x } \ \ \ \ \ \ \ \ \ \ \\ \\ =\cot x (2\cos4x\sin x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{\cos x}{\ sin x} = \cot x \right )\\ \\$

Now we know that

$\\ 2\cos x\sin y = \sin(x+y) - \sin(x-y)\\ \\$

By using this our equation becomes

$\\ \\=\cot x (\sin5x - sin3x)\\$
R.H.S.

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