Prove that

Solution:

To Prove: $\cos \mathrm{x} \cos 2 \mathrm{x} \cos 4 \mathrm{x} \cos 8 \mathrm{x}=\frac{\sin 16 \mathrm{x}}{16 \sin \mathrm{x}}$

Taking LHS,

= cosx cos2x cos4x cos8x

Multiply and divide by 2sinx, we get

$=\frac{1}{2 \sin x}[2 \sin x \cos x \cos 2 x \cos 4 x \cos 8 x]$

$=\frac{1}{2 \sin x}[(2 \sin x \cos x) \cos 2 x \cos 4 x \cos 8 x]$

$=\frac{1}{2 \sin x}[\sin 2 x \cos 2 x \cos 4 x \cos 8 x][\because \sin 2 x=2 \sin x \cos x]$

Multiply and divide by 2, we get

$=\frac{1}{2 \times 2 \sin x}[(2 \sin 2 x \cos 2 x) \cos 4 x \cos 8 x]$

We know that,

sin 2x = 2 sinx cosx

Replacing x by 2x, we get

sin 2(2x) = 2 sin(2x) cos(2x)

or sin 4x = 2 sin 2x cos 2x

$=\frac{1}{4 \sin x}[\sin 4 x \cos 4 x \cos 8 x]$

Multiply and divide by 2, we get

$=\frac{1}{2 \times 4 \sin x}[2 \sin 4 x \cos 4 x \cos 8 x]$

We know that,

sin 2x = 2 sinx cosx

Replacing x by 4x, we get

sin 2(4x) = 2 sin(4x) cos(4x)

or sin 8x = 2 sin 4x cos 4x

$=\frac{1}{8 \sin x}[\sin 8 x \cos 8 x]$

Multiply and divide by 2, we get

$=\frac{1}{2 \times 8 \sin x}[2 \sin 8 x \cos 8 x]$

We know that

sin 2x = 2 sinx cosx

Replacing x by 8x, we get

sin 2(8x) = 2 sin(8x) cos(8x)

or sin 16x = 2 sin 8x cos 8x

$=\frac{1}{16 \sin x}[\sin 16 x]$

= RHS

∴ LHS = RHS

Hence Proved