Question

Cos4a=1-8cos^2a+8cos^4a

Answer 1

cos (4A) = cos (2•(2A)) 
↔ cos (2•(2A)) = 2•cos² (2A) - 1 
↔ cos (4A) = 2•cos² (2A) - 1 
↔ cos (4A) = 2•(2•cos² A - 1)² - 1 
↔ cos (4A) = 2•(2•cos² A - 1)² - 1 
↔ cos (4A) = 2•{4•(cos² A)² - 4•(cos² A) + 1} - 1
↔ cos (4A) = 8•cos⁴ A - 8•cos² A + 2 - 1 
↔ cos (4A) = 8•cos⁴ A - 8•cos² A + 1 
↔ cos (4A) = 1 - 8•cos² A + 8•cos⁴ A .... qid 

Answer 2

$cos (4x) = cos(2x + 2x)$

Using, $cos (A+B) = cosAcosB - sinAsinB$

$cos(2x + 2x) = cos(2x) cos(2x) - sin(2x) sin(2x)$ = $(cos(2x))^2 - (sin(2x))^2$ = $(cos(x +x ))^2 - (sin(x + x))^2$

$(cos(x +x ))^2 - (sin(x + x))^2=( cos x cos x - sin x sin x )^2 - ( sin x cos x + sin x cos x )^2$

$= ( cos^2 x-sin^2 x )^2-( 2 sin x cos x)^2$

$= ( cos^2 x-sin^2 x )^2-4sin^2 x cos^2 x$

$= ( cos^2 x-(1-cos^2 x ) )^2-4(1 - cos^2 x ) cos^2 x$

$= ( 2 cos^2 x-1 )^2-4cos^2 x +4cos^4 x$

$= 8 cos^4 x-8 cos^2 x+1$

Hence proved.