use the power - reducing formulas to rewrite sin⁴x in terms of the first power of cosine. simplify your…

use the power - reducing formulas to rewrite sin⁴x 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. sin⁴x = - cosx + cosx sin⁴x = + cosx + cosx sin⁴x = - cosx sin⁴x = + cosx
Answer
Explanation:
Step1: Recall power - reducing formula for $\sin^{2}\alpha$
$\sin^{2}\alpha=\frac{1 - \cos(2\alpha)}{2}$
Step2: Rewrite $\sin^{4}x$ as $(\sin^{2}x)^{2}$
$\sin^{4}x = (\sin^{2}x)^{2}$
Step3: Substitute $\sin^{2}x=\frac{1 - \cos(2x)}{2}$ into $(\sin^{2}x)^{2}$
$\sin^{4}x=\left(\frac{1 - \cos(2x)}{2}\right)^{2}=\frac{1 - 2\cos(2x)+\cos^{2}(2x)}{4}$
Step4: Recall power - reducing formula for $\cos^{2}\beta$
$\cos^{2}\beta=\frac{1+\cos(2\beta)}{2}$, here $\beta = 2x$, so $\cos^{2}(2x)=\frac{1+\cos(4x)}{2}$
Step5: Substitute $\cos^{2}(2x)=\frac{1+\cos(4x)}{2}$ into $\frac{1 - 2\cos(2x)+\cos^{2}(2x)}{4}$
[ \begin{align*} \sin^{4}x&=\frac{1 - 2\cos(2x)+\frac{1+\cos(4x)}{2}}{4}\ &=\frac{\frac{2-4\cos(2x)+1+\cos(4x)}{2}}{4}\ &=\frac{3 - 4\cos(2x)+\cos(4x)}{8}\ &=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x) \end{align*} ]
Answer:
$\sin^{4}x=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$