use the power reduction formulas to rewrite the expression. (hint: your answer should not contain any…

Explanation:

Step1: Recall power - reduction formula for $\tan^{2}(x)$

The power - reduction formula for $\tan^{2}(x)=\frac{1 - \cos(2x)}{1+\cos(2x)}$. So the original expression $\tan^{2}(x)\sin(x)$ becomes $\frac{1 - \cos(2x)}{1+\cos(2x)}\sin(x)$.

Step2: Multiply through

$\frac{\sin(x)-\sin(x)\cos(2x)}{\ 1+\cos(2x)}$. We know that $\cos(2x)=1 - 2\sin^{2}(x)$. Substitute $\cos(2x)$ into the expression: [ \begin{align*} \frac{\sin(x)-\sin(x)(1 - 2\sin^{2}(x))}{1+(1 - 2\sin^{2}(x))}&=\frac{\sin(x)-\sin(x)+2\sin^{3}(x)}{2 - 2\sin^{2}(x)}\ &=\frac{2\sin^{3}(x)}{2(1 - \sin^{2}(x))}\ &=\frac{\sin^{3}(x)}{\cos^{2}(x)} \end{align*} ] Another way: We know that $\tan^{2}(x)=\frac{\sin^{2}(x)}{\cos^{2}(x)}$, so $\tan^{2}(x)\sin(x)=\frac{\sin^{3}(x)}{\cos^{2}(x)}$. And we use the power - reduction formula $\sin^{2}(x)=\frac{1 - \cos(2x)}{2}$ and $\cos^{2}(x)=\frac{1+\cos(2x)}{2}$. [ \begin{align*} \frac{\sin^{3}(x)}{\cos^{2}(x)}&=\sin(x)\cdot\frac{\sin^{2}(x)}{\cos^{2}(x)}\ &=\sin(x)\cdot\frac{1 - \cos(2x)}{1+\cos(2x)} \end{align*} ] We can also rewrite it as follows: [ \begin{align*} \tan^{2}(x)\sin(x)&=\frac{\sin^{2}(x)}{\cos^{2}(x)}\sin(x)\ &=\frac{1 - \cos(2x)}{2}\cdot\frac{\sin(x)}{\frac{1+\cos(2x)}{2}}\ &=\frac{(1 - \cos(2x))\sin(x)}{1+\cos(2x)} \end{align*} ]

Answer:

$\frac{(1 - \cos(2x))\sin(x)}{1+\cos(2x)}$