Question

show that sin 8x. cos x - sin 6x. cos 3x / cos 2x. cos x - sin 3x. sin 4x = tan 2x

Answer 1

We have to prove that:

$\frac{\sin 8x \cos x-\sin 6x \cos 3x}{\cos 2x \cos x-\sin 3x \sin 4x}=\tan 2x$

Consider the Left hand side of the equation

$\frac{\sin 8x \cos x-\sin 6x \cos 3x}{\cos 2x \cos x-\sin 3x \sin 4x}$

Multiplying numerator and denominator by 2, we get

=$\frac{2\sin 8x \cos x-2\sin 6x \cos 3x}{2\cos 2x \cos x-2\sin 3x \sin 4x}$

Now using trigonometric identity in the above equation,

$2 \sin A \cos B = \sin(A+B)x + \sin(A-B)x$

$2 \cos A \cos B = \cos(A+B)x + \cos(A-B)x$

$2 \sin A \sin B = \cos(A-B)x - \cos(A+B)x$

We get,

=$\frac{\sin 9x+ \sin 7x-(\sin 9x + \sin 3x)}{\cos 3x+\cos x-(\cos(-x)-\cos 7x)}$

=$\frac{\sin 7x-\sin 3x}{\cos 3x+\cos 7x}$

Now using trigonometric identities as:

$\sin A-\sin B =2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$

$\cos A+\cos B =2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$

So, we get

=$\frac{2\cos(\frac{7x+3x}{2})\sin(\frac{7x-3x}{2})}{2\cos(\frac{3x+7x}{2})\cos(\frac{3x-7x}{2})}$

= $\frac{2\cos5x\sin 2x}{2\cos 5x\cos2x}$

= $\frac{\sin 2x}{\cos2x}$

= $\tan 2x$

= RHS

Hence, proved.