Question

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

cos^4 x sin^2 x

Answer 1

The key identity is

`\cos^2x=\frac{1+\cos2x}2`

We have

`\cos^4x\sin^2x=\cos^4x(1-\cos^2x)`

`=\cos^4x-\cos^6x`

`=\left(\frac{1+\cos2x}2\right)^2-\left(\frac{1+\cos2x}2\right)^3`

`=\frac{1+\cos2x-\cos^22x-\cos^32x}8`

Then

`\cos^22x=\frac{1+\cos4x}2`

and

`\cos^32x=\cos2x\cos^22x=\frac{\cos2x(1+\cos4x)}2`

`\cos^32x=\frac{\cos2x+\frac{\cos6x+\cos2x}2}2=\frac{3\cos2x+\cos6x}4`

`\cos^4x\sin^2x=\frac{1+\cos2x-\frac{1+\cos4x}2-\frac{3\cos2x+\cos6x}4}8`

`=(2+\cos2x-2\cos4x-\cos6x)/(32)`