1. 0/1 points details my notes oscat1 9.3.146.tut. previous answers ask your teacher use the power reduction…

Explanation:

Step1: Recall power - reduction formula for $\tan^{2}\theta$

The power - reduction formula for $\tan^{2}\theta=\frac{1 - \cos(2\theta)}{1+\cos(2\theta)}$. Here $\theta = x$, so $\tan^{2}x=\frac{1 - \cos(2x)}{1+\cos(2x)}$.

Step2: Substitute into the original expression

The original expression is $\tan^{2}(x)\sin(x)$. Substituting $\tan^{2}x$ gives $\frac{1 - \cos(2x)}{1+\cos(2x)}\sin(x)$.

Step3: Multiply through

$\frac{\sin(x)-\sin(x)\cos(2x)}{\ 1+\cos(2x)}$. We also know the double - angle formula $\cos(2x)=1 - 2\sin^{2}x$. Another way: We know $\tan^{2}x=\frac{\sin^{2}x}{\cos^{2}x}$, and the power - reduction formula for $\sin^{2}x=\frac{1 - \cos(2x)}{2}$ and $\cos^{2}x=\frac{1+\cos(2x)}{2}$. So $\tan^{2}x\sin(x)=\frac{\sin^{2}x}{\cos^{2}x}\sin(x)=\frac{\frac{1 - \cos(2x)}{2}}{\frac{1+\cos(2x)}{2}}\sin(x)=\frac{1 - \cos(2x)}{1+\cos(2x)}\sin(x)=\frac{\sin(x)-\sin(x)\cos(2x)}{1+\cos(2x)}$

We can also use the identity $\cos(2x)=1 - 2\sin^{2}x$ to further simplify: [ \begin{align*} \tan^{2}x\sin(x)&=\frac{\sin^{2}x}{\cos^{2}x}\sin(x)\ &=\frac{1 - \cos(2x)}{1+\cos(2x)}\sin(x)\ &=\frac{\sin(x)-\sin(x)\cos(2x)}{1+\cos(2x)}\ &=\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*} ]

A more standard way using power - reduction: The power - reduction formula for $\tan^{2}x=\frac{1-\cos(2x)}{1 + \cos(2x)}$ [ \begin{align*} \tan^{2}(x)\sin(x)&=\frac{1-\cos(2x)}{1+\cos(2x)}\sin(x)\ &=\frac{\sin(x)-\sin(x)\cos(2x)}{1+\cos(2x)} \end{align*} ] We know that $\sin(A)\cos(B)=\frac{1}{2}[\sin(A + B)+\sin(A - B)]$, so $\sin(x)\cos(2x)=\frac{1}{2}[\sin(3x)-\sin(x)]$ [ \begin{align*} \tan^{2}(x)\sin(x)&=\frac{\sin(x)-\frac{1}{2}(\sin(3x)-\sin(x))}{1+\cos(2x)}\ &=\frac{\sin(x)-\frac{1}{2}\sin(3x)+\frac{1}{2}\sin(x)}{1+\cos(2x)}\ &=\frac{\frac{3}{2}\sin(x)-\frac{1}{2}\sin(3x)}{1+\cos(2x)} \end{align*} ]

The power - reduction formula for $\tan^{2}x$ is $\tan^{2}x=\frac{1-\cos(2x)}{1+\cos(2x)}$, so $\tan^{2}(x)\sin(x)=\frac{(1 - \cos(2x))\sin(x)}{1+\cos(2x)}$

We can also rewrite it as follows: Since $\tan^{2}x=\frac{\sin^{2}x}{\cos^{2}x}$ and $\sin^{2}x=\frac{1 - \cos(2x)}{2},\cos^{2}x=\frac{1+\cos(2x)}{2}$ [ \begin{align*} \tan^{2}(x)\sin(x)&=\frac{\sin^{2}x}{\cos^{2}x}\sin(x)\ &=\frac{\frac{1 - \cos(2x)}{2}}{\frac{1+\cos(2x)}{2}}\sin(x)\ &=\frac{1 - \cos(2x)}{1+\cos(2x)}\sin(x) \end{align*} ]

The final rewritten expression without exponents greater than 1: $\frac{(1 - \cos(2x))\sin(x)}{1+\cos(2x)}$

Answer:

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