Question

Prove that: $cos^{2}2x-cos^{2} 6x =sin 4x sin 8x$

Answer 1

LHS = cos²(2x) - cos²(6x)

= (cos2x - cos6x)(cos2x + cos6x)

[as we know from algebraic identity a² - b² = (a - b)(a + b)]

now, use formula,
cosC - cosD = 2sin(C + D)/2. sin(D - C)/2
cosC + cosD = 2cos(C + D)/2.cos(C - D)/2.

cos2x - cos6x = 2sin(2x + 6x)/2.sin(6x - 2x)/2
= 2sin4x.sin2x

cos2x + cos6x = 2cos(2x + 6x)/2.cos(2x - 6x)/2.
= 2cos4x.cos(-2x) = 2cos4x.cos2x

now,
(cos2x - cos6x)(cos2x + cos6x) = (2sin4x.sin2x)(2cos4x.cos2x)

= (2sin4x.cos4x)(2sin2x.cos2x)

use formula, 2sinA.cosA = sin2A.

= sin8x.sin4x = RHS

hence proved.