Question

sin4xcos4x=1-2sin2xcos2x​

Answer 1

LHS- sin4x + cos4x

= (sin2x)2 + (cos2x)2

= (sin2x + cos2x)2 - 2sin2xcos2x {since a2 +b2 = (a+b)2 - 2ab}

= 12 - 2sin2xcos2x {since sin2x +cos2x = 1}

= 1 - 2sin2xcos2x

= RHS

Answer 2

ANSWER

Here we have to prove=>

sin^4x . cos^4x = 1-2sin^2x . cos^2x

LHS

$sin^4x + cos^4x$

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

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

{since $a^2 +b^2$=$(a+b)^2$- 2ab}

=> $1^2 -2sin^2x.cos^2x$

{since$sin^2x +cos^2x$ = 1}

=> 1 - $2sin^2xcos^2x$

= RHS

Hence Proved.

REMEMBER

$sin^2x +cos^2x$ = 1

$(A + B)^2$ =$A^2 + B^2$ + 2× A×B