How do you prove cos4x-sin4x=cos2x?
Proof of given relation:
cos4x-sin4x=cos2x
Here we have LHS=cos4x-sin4x
Therefore,
cos4x-sin4x=cos2x2-sin2x2
=cos2x+sin2xcos2x-sin2x [∵a2-b2=a+ba-b]
=1.cos2x-sin2x [∵sin2x+cos2x=1]
=cos2x [∵cos2x-sin2x=cos2x]
=RHS
Hence, cos4x-sin4x=cos2x is proved.