Question

Use the power reduction formulas to rewrite the expression. (Hint: Your answer should not contain any exponents greater than 1.)

tan^2(x) sin^3(x)

Answer 1

Some useful relations and identities:

`\tan x=(\sin x)/(\cos x)`

`\sin^2x=\frac{1-\cos2x}2`

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

By the first relation, we have

`\tan^2x\sin^3x=(\sin^5x)/(\cos^2x)=((\sin^2x)^2\sin x)/(\cos^2x)`

Applying the two latter identities, we get

`\frac{\left(\frac{1-\cos2x}2\right)^2\sin x}{\frac{1+\cos2x}2}=\frac{\frac{1-2\cos2x+\cos^22x}4\sin x}{\frac{1+\cos2x}2}=((1-2\cos2x+\cos^22x)\sin x)/(2(1+\cos2x))`

We can apply the third identity again:

`((1-2\cos2x+\cos^22x)\sin x)/(2(1+\cos2x))=\frac{\left(1-2\cos2x+\frac{1+\cos4x}2\right)\sin x}{2(1+\cos2x)}=((3-4\cos2x+\cos4x)\sin x)/(4(1+\cos2x))`

and this is probably as far as you have to go, but by no means is it the only possible solution.