Question

Prove the identity: cos4x=1-8sin^2xcos^2x

Answer 1

I get that it uses double angle identities , but how you go from there?

Answer 2

Try
Cos(2x + 2x) = cos2xcos2x - sin2xsin2x

Answer 3

How do you proceed from there?

Answer 4

Now try using the double angle identities

Answer 5

cos4x=cos(2x+2x)
cos4x=cos^2 2x-sin^2 2x
cos4x=cos2xcos2x-sin2xsin2x
cos4x=(cos^2x-sin^2x)(cos^2x-sin^2x)-(2sinxcosx)(2sinxcosx)
Using sin^2x+cos^2x=1
cos4x=cos^4x-2cos^2xsin^2x+sin^4x-(4sin^2xcos^2x)

cos^4x+sin^4x=1

so cos^4x=1-2cos^2xsin^2x-(4sin^2xcos^2x)
finally cos^4x=1-8sin^2xcos^2x