use the power reducing formulas to rewrite cot²4x sin⁴4x in terms of the first power of cosine. simplify…

use the power reducing formulas to rewrite cot²4x sin⁴4x in terms of the first power of cosine. simplify your answer as much as possible. to indicate your answer, first choose one of the four forms below. then fill in the blanks with the appropriate numbers. \n\no cot²4x sin⁴4x = - cosx + cosx\n\no cot²4x sin⁴4x = + cosx + cosx\n\no cot²4x sin⁴4x = - cosx\n\no cot²4x sin⁴4x = + cosx
Answer
Explanation:
Step1: Recall cotangent - sine relation
Since $\cot^{2}4x=\frac{\cos^{2}4x}{\sin^{2}4x}$, then $\cot^{2}4x\sin^{4}4x=\cos^{2}4x\sin^{2}4x$.
Step2: Apply power - reducing formulas
The power - reducing formulas are $\sin^{2}\alpha=\frac{1 - \cos2\alpha}{2}$ and $\cos^{2}\alpha=\frac{1+\cos2\alpha}{2}$. Here $\alpha = 4x$, so $\cos^{2}4x\sin^{2}4x=\left(\frac{1 + \cos8x}{2}\right)\left(\frac{1-\cos8x}{2}\right)$.
Step3: Expand the product
Using the difference of squares formula $(a + b)(a - b)=a^{2}-b^{2}$, we have $\left(\frac{1 + \cos8x}{2}\right)\left(\frac{1-\cos8x}{2}\right)=\frac{1-\cos^{2}8x}{4}$.
Step4: Apply power - reducing formula again
Since $\cos^{2}8x=\frac{1+\cos16x}{2}$, then $\frac{1-\cos^{2}8x}{4}=\frac{1-\frac{1 + \cos16x}{2}}{4}$.
Step5: Simplify the expression
[ \begin{align*} \frac{1-\frac{1+\cos16x}{2}}{4}&=\frac{\frac{2-(1 + \cos16x)}{2}}{4}\ &=\frac{2 - 1-\cos16x}{8}\ &=\frac{1}{8}-\frac{1}{8}\cos16x \end{align*} ]
Answer:
$\cot^{2}4x\sin^{4}4x=\frac{1}{8}-\frac{1}{8}\cos16x$ (corresponding to the form $\cot^{2}4x\sin^{4}4x=\square-\square\cos\square x$, where the first blank is $\frac{1}{8}$, the second blank is $\frac{1}{8}$, and the third blank is $16$)