Trigonometry - 2

$\displaystyle (1+\sin{\frac{\pi}{8}})(1+\sin{\frac{5 \pi}{8}})(1+\sin{\frac{9 \pi}{8}})(1+\sin{\frac{13 \pi}{8}})$

$\displaystyle =(1+\sin{\frac{\pi}{8}})(1+\sin{\frac{5 \pi}{8}})(1-\sin{\frac{\pi}{8}})(1-\sin{\frac{5 \pi}{8}})$

$\displaystyle =(1-\sin^2\frac{\pi}{8})(1-\sin^2\frac{5 \pi}{8})$

$\displaystyle =(\cos^2\frac{\pi}{8})( \cos^2\frac{5 \pi}{8})$

$\displaystyle =\big[\frac{1}{2} (\cos \frac{3 \pi}{4} + \cos \frac{\pi}{2})\big]^2$

$=0.125$

We know that $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\approx0.70711$ ​. Hence using double angle formula gives $\cos\frac{\pi}{4}=2\cos^{2}\frac{\pi}{8}-1=\frac{1}{\sqrt{2}}\Rightarrow \cos^{2}\frac{\pi}{8}=\frac{2+\sqrt{2}}{4}\Rightarrow\cos\frac{\pi}{8}\approx0.92388$ Since $\sin^{2}\theta+\cos^{2}\theta=1$ , therefore we have $\sin^{2}\frac{\pi}{8}=\frac{2-\sqrt{2}}{4}\Rightarrow \sin\frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\approx0.38268$ Using the formula $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$ , we can obtain $\left\{\begin{matrix} \sin\frac{5\pi}{8}\approx0.92388\\ \sin\frac{9\pi}{8}\approx-0.38268 \\ \sin\frac{13\pi}{8}\approx-0.92388 \end{matrix}\right.$ This gives our desired answer to be $\boxed{0.125}$ .