Question

show that cosx.cos2x.cos4x.cos8x=sin((2²)²x/(2²)²sinx

Answer 1

Remember : 2 sin(p) cos(p) = sin(2p)
Let,
A = cos(x) cos(2x) cos(4x) cos(8x)
2 sin(x) A = 2 sin(x) cos(x) cos(2x) cos(4x) cos(8x)
2 sin(x) A = sin(2x) cos(2x) cos(4x) cos(8x)
4 sin(x) A = 2 sin(2x) cos(2x) cos(4x) cos(8x)
4 sin(x) A = sin(4x) cos(4x) cos(8x)
8 sin(x) A = 2 sin(4x) cos(4x) cos(8x)
8 sin(x) A = sin(8x) cos(8x)
16 sin(x) A = 2 sin(8x) cos(8x)
16 sin(x) A = sin(16x)

$A = \frac{sin(16x)}{16 sin(x)}$

And, $16 = 4^{2} \\ 4 = 2^{2} \\ 16 = ( 2^{2} )^{2}$

Replacing 16 in terms of powers of 2

$A = \frac{sin(( 2^{2} )^{2}x)}{( 2^{2} )^{2}sin(x)} \\ \\ cos(x) cos(2x) cos(4x) cos(8x) = \frac{sin(( 2^{2} )^{2}x)}{( 2^{2} )^{2}sin(x)}$