Question

hola guys please solve it fast ⤵⤵
$\cos ^{2}2 x - \cos ^{2} 6x = \sin4x \sin8x$
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Answer 1

Step-by-step explanation:

Given Equation is cos²2x - cos²6x

$=(cos2x+cos6x)(cos2x-cos6x)$

$\boxed{cosA + cosB = 2 sin(\frac{A + B}{2})cos(\frac{A-B}{2}), cosA - cosB = -2sin(\frac{A+B}{2})sin(\frac{A-B}{2})}$

$=[2 cos(\frac{2x + 6x}{2})cos(\frac{2x - 6x}{2})][-2sin(\frac{2x + 6x}{2})sin(\frac{2x-6x}{2})]$

$=[2cos(4x)cos(-2x)][-2sin4xsin(-2x)]$

$=[2cos4xcos2x][-2sin4x(-sin2x)]$

$=(2sin4xcos4x)(2sin2xcos2x)$

$= sin8xsin4x$

Hope it helps!

Answer 2

Cos A + Cos B = 2Cos (A+B)/2 Cos (A-B)/2  

Cos A - Cos B = 2Sin (A+B)/2 Sin (A-B)/2  

Cos² 2x - Cos² 6x is in the form a² - b² which is (a+b)(a-b)  

Cos²2x - cos²6x = (Cos2x + Cos6x)(Cos2x - Cos6x)  

(2 Cos (2x + 6x)/2 * Cos(2x - 6x)/2 ) * -2Sin (2x + 6x)/2 * Sin (2x -6x)/2 )  

(2Cos4x * cos(-2x) ) (-2sin4x * sin (-2x))  

2Sin4xCos4x * 2Sin2xCos2x  

= Sin4x*Sin8x