Lowering Powers

Question

Use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine.

cos⁴(x) sin²(x)

Answer 1

Hello,

Identities to keep in mind:

sin²x+cos²x=1 and cos²x=(1+cos(2x))/2

Using the first identity, we have sin²x=1-cos²x.

So we have that cos⁴x*sin²x=cos⁴x(1-cos²x)

Expanding, we have cos⁴x-cos⁶x.

Since cos²x=(1+cos(2x))/2, this implies cos⁴x=((1+cos(2x))/2)^2 and cos⁶x=((1+cos(2x))/2)^3.

Expanding each one, we have

cos⁴x-cos⁶x=1/4(1+2cos(2x)+cos²(2x))-1/8(1+3cos(2x)+3cos²(2x)+cos³(2x)).

Simplifying we get

cos⁴x-cos⁶x=1/8+1/8(cos(2x)-cos²(2x)-cos³(2x)).

cos²(2x)=1/2(1+cos(4x)) and cos³(2x)=cos²(2x)*cos(2x)=1/2(1+cos(4x))*cos(2x)=1/2(cos(2x)+cos(4x)*cos(2x))

Substituting, we get

cos⁴x-cos⁶x=1/8+1/8(cos(2x)-1/2(1+cos(4x))-1/2(cos(2x)+cos(4x)*cos(2x)))

Cleaning it up, we get

cos⁴x*sin²x=1/16+cos(2x)/16-cos(4x)/16-cos(4x)cos(2x)/16.