Question

Sin^2 P/8 * Cos^2 P/8

Answer 1

Sin^2 P/8 * Cos^2 P/8=(1-cosP/8)(1+сosP/8)=1-cos^2P/8=sin^2P/8

Answer 2

I can help you with math problems and other topics.

To answer your question, I will use the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ and the angle sum formula $\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$. Here are the steps:

\begin{align*} \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} &= \frac{1}{2} \left( \sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8} \right) - \frac{1}{2} \left( \sin^2 \frac{\pi}{8} - \cos^2 \frac{\pi}{8} \right) \\ &= \frac{1}{2} \cdot 1 - \frac{1}{2} \cos \frac{\pi}{4} \\ &= \frac{1}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{1 - \sqrt{2}}{4} \end{align*}

Therefore, $\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = \frac{1 - \sqrt{2}}{4}$.

You can check your answer using a trigonometry calculator or a trigonometric equation solver. I hope this helps.