If cos^2sin^4x= A+Bcos2x+Ccos4x+Dcos2xcos4x, then A= B= C= D=

Question

Please tell me how you got it by showing steps!!

what does A, B, C, and D equal in the second equation?

Answer 1

Use the "half-angle formulas"

sin²x = (1 + cos2x)/2

cos²x = (1 - cos2x)/2

From the first one: sin⁴x = ( (1 + cos2x)/2 )² = 1/4 (1 + 2cos2x + cos²2x)

cos²2x = 1/2 (1 - cos4x), so:

sin⁴x = 1/4 (1 + 2cos2x +(1/2 (1 - cos4x)))

= 1/4 (1 + 2cos2x +1/2 - 1/2(cos4x))

= 1/4 (3/2 + 2cos2x - 1/2(cos4x))

= 3/8 + 1/2 (cos2x) - 1/8 (cos4x)

Then multiply by the equality for cos²x, so:

cos²x * sin⁴x = 1/2 (1 - cos2x) * (3/8 + 1/2 (cos2x) - 1/8 (cos4x))

Expand the right hand side, which will once again give a term including cos²2x, which will need to be converted as before.

When you finish expanding and simplifying, I think this will result in an equation of the form given.