Chapter 7, Problem 80RE

To determine

To find:the exact value of the given expression.

Answer to Problem 80RE

  $\sin \frac{5\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$ .

Explanation of Solution

Given:

The given expression is $\sin \frac{5\pi }{8}$ .

Calculation:

As the given expressionis $\sin \frac{5\pi }{8}=\sin \frac{\raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{2}$ .

Now, using the identity $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ :

  $\begin{array}{l}\sin \frac{5\pi }{8}=\sin \frac{\raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{2}\\ =\sqrt{\frac{1-\cos \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{2}}\mathrm{......}(1)\end{array}$

Now to find the value of $\cos \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , then

  $\begin{array}{l}\cos \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=\cos \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=\cos \left(\pi +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)=-\cos \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\\ \{\text{As }\cos \left(\pi +\theta \right)=-\cos \pi \}\end{array}$

Therefore, substitute the value in (1)

  $\begin{array}{l}\sin \frac{5\pi }{8}=\sqrt{\frac{1-\cos \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{2}}\\ =\sqrt{\frac{1-\left(-\cos \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}{2}}\\ =\sqrt{\frac{1+\cos \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{2}}\\ =\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}\\ =\sqrt{\frac{\frac{2+\sqrt{2}}{2}}{2}}\\ =\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}\end{array}$

Hence, the value of $\sin \frac{5\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$ .