# How do you evaluate sin ((7pi)/8) using the half angle formula?

Jun 1, 2016

$- \frac{\sqrt{2 - \sqrt{2}}}{2}$

#### Explanation:

Trig table, unit circle -->
$\sin \left(\frac{7 \pi}{8}\right) = \sin \left(- \frac{\pi}{8} + 2 \pi\right) = - \sin \left(\frac{\pi}{8}\right)$
Find $\sin \left(\frac{\pi}{8}\right)$ by using trig identity:
$\cos 2 a = 1 - 2 {\sin}^{2} a .$
$\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = 1 - 2 {\sin}^{2} \left(\frac{\pi}{8}\right)$
$2 {\sin}^{2} \left(\frac{\pi}{8}\right) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$
${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{4}$
$\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$
The negative answer is rejected because sin (pi/8) is positive.
Finally,
$\sin \left(\frac{7 \pi}{8}\right) = - \frac{\sqrt{2 - \sqrt{2}}}{2}$

Jun 3, 2016

$\frac{\sqrt{2 - \sqrt{2}}}{2}$

#### Explanation:

This can also be shown through the sine half-angle formula:

$\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(x\right)}{2}}$

Here, since we want to find $\sin \left(\frac{7 \pi}{8}\right)$, we know that $\frac{x}{2} = \frac{7 \pi}{8}$ and $x = \frac{7 \pi}{4}$.

$\sin \left(\frac{7 \pi}{8}\right) = \sqrt{\frac{1 - \cos \left(\frac{7 \pi}{4}\right)}{2}}$

Note that the $\pm$ sign has just turned into a positive sign: the sine of $\frac{7 \pi}{8}$ will be positive since $\frac{7 \pi}{8}$ is in the second quadrant.

$\sin \left(\frac{7 \pi}{8}\right) = \sqrt{\frac{1 - \left(\frac{\sqrt{2}}{2}\right)}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

We can show that $\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(x\right)}{2}}$ using the cosine double-angle formula.

$\cos \left(2 x\right) = 1 - 2 {\sin}^{2} \left(x\right)$

This is the same as saying

$\cos \left(x\right) = 1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)$

The argument of the cosine function is double that of the sine function--just expressed differently.

Solving for $\sin \left(\frac{x}{2}\right)$, we see that

$2 {\sin}^{2} \left(\frac{x}{2}\right) = 1 - \cos \left(x\right)$

${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1 - \cos \left(x\right)}{2}$

$\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos \left(x\right)}{2}}$