Question

Prove that cos 4x = 1 - 8sin²xcos²x

Answer 1

Here is your answer.......

Answer 2

heya...

here is ua answer:

cos4x=cos2(2x)=cos^2(2x)-sin^2(2x)

=[cos^2(x)-sin^2(x)]^2-(2sinx cosx)^2

=[cos^4(x)+sin^4(x)-2cos^(x)sin^2(x)]-[4sin^2(x)cos^2(x)]

=cos^4(x)+sin^4(x)-6cos^2(x)sin^2(x)

=[cos^4(x)+sin^4(x)+2cos^2(x)cos^2(x)]-[8cos^2(x)sin^2(x)]

=[cos^2(x)+sin^2(x)]^2 - 8cos^2(x)sin^2(x)

= (1)^2 - 8sin^2(x)cos^2(x)

= 1- 8sin^2(x)cos^2(x)

HENCE PROOVED

USED FORMULAE

cos2x=cos^2(x)-sin^2(x)

Sin2x=2sinx cosx

(a+b)^2=a^2+b^2+2ab

hope it helps..!!