Question

Verify: sin^2(x)*cos^4(x) = 1/32 (2+cos(2 x)-2 cos(4 x)-cos(6 x)) Verify: sin^2(x)*cos^4(x) = 1/32 (2+cos(2 x)-2 cos(4 x)-cos(6 x)) @Mathematics

Answer 1

cos^4(x)-cos^6(x)
=sin^6(x)-2 sin^4(x)+sin^2(x)
=1/32 (cos(2 x)-2 cos(4 x)-cos(6 x)+2)


*computed by WolframAlpha

Answer 2

How did you get Wolfram Alpha to give you this answer?

Answer 3

Use syntax from mathematica.

Answer 4

Can you show me all of the steps used to arrive at the solution?

Answer 5

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

=1/16 + Cos[x]^2/32 - Cos[x]^4/16 - Cos[x]^6/32 - Sin[x]^2/32 + (3 Cos[x]^2 Sin[x]^2)/8 + (15 Cos[x]^4 Sin[x]^2)/32 - Sin[x]^4/16 - (15 Cos[x]^2 Sin[x]^4)/32 + Sin[x]^6/32

=1/32 (cos(2 x)-2 cos(4 x)-cos(6 x)+2)