Question

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

Answer 1

The key identity is

$\cos^2x=\dfrac{1+\cos2x}2$

We have

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

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

$=\left(\dfrac{1+\cos2x}2\right)^2-\left(\dfrac{1+\cos2x}2\right)^3$

$=\dfrac{1+\cos2x-\cos^22x-\cos^32x}8$

Then

$\cos^22x=\dfrac{1+\cos4x}2$

and

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

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

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

$=\dfrac{2+\cos2x-2\cos4x-\cos6x}{32}$