use half angle formulas or formula for reducing powers to fill in the blanks in the identity below…

Explanation:

Step1: Recall power - reducing formula

The power - reducing formula for $\sin^{2}\alpha$ is $\sin^{2}\alpha=\frac{1 - \cos(2\alpha)}{2}$. Let $\alpha = 4x$, so $\sin^{2}(4x)=\frac{1-\cos(8x)}{2}$.

Step2: Square $\sin^{2}(4x)$

$(\sin(4x))^{4}=(\sin^{2}(4x))^{2}$. Substitute $\sin^{2}(4x)=\frac{1 - \cos(8x)}{2}$ into it, we get $(\sin(4x))^{4}=\left(\frac{1-\cos(8x)}{2}\right)^{2}=\frac{1 - 2\cos(8x)+\cos^{2}(8x)}{4}$.

Step3: Use power - reducing formula for $\cos^{2}(8x)$

The power - reducing formula for $\cos^{2}\beta$ is $\cos^{2}\beta=\frac{1+\cos(2\beta)}{2}$. Let $\beta = 8x$, so $\cos^{2}(8x)=\frac{1+\cos(16x)}{2}$.

Step4: Substitute $\cos^{2}(8x)$ into the expression

$(\sin(4x))^{4}=\frac{1 - 2\cos(8x)+\frac{1+\cos(16x)}{2}}{4}=\frac{2-4\cos(8x)+1 + \cos(16x)}{8}=\frac{3}{8}-\frac{1}{2}\cos(8x)+\frac{1}{8}\cos(16x)$.

Answer:

$\frac{3}{8},8,16$