# How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle (3pi)/8?

Jan 25, 2017

Use trig identities and trig table -->
$2 {\sin}^{2} t = 1 - \cos 2 t$
$2 {\cos}^{2} t = 1 + \cos 2 t$
Call $\frac{3 \pi}{8} = t$ --> $2 t = \frac{6 \pi}{8} = \frac{3 \pi}{4}$ --> $\cos 2 t = - \frac{\sqrt{2}}{2}$
In this case, we have:
$2 {\sin}^{2} t = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$
$\sin \left(\frac{3 \pi}{8}\right) = \sin t = \pm \frac{\sqrt{2 + \sqrt{2}}}{2}$
Since $\sin \left(\frac{3 \pi}{8}\right)$ is positive, take the positive value.
$2 {\cos}^{2} t = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$
$\cos \left(\frac{3 \pi}{8}\right) = \cos t = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}$
Since $\cos \left(\frac{3 \pi}{8}\right)$ is positive, take the positive value.
$\tan \left(\frac{3 \pi}{8}\right) = \frac{\sin t}{\cos t} = \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}}$
$\cot \left(\frac{3 \pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}}$
$\sec \left(\frac{3 \pi}{8}\right) = \frac{1}{\cos} = \frac{2}{\sqrt{2 + \sqrt{2}}}$
$\csc \left(\frac{3 \pi}{8}\right) = \frac{1}{\sin} = \frac{2}{\sqrt{2 - \sqrt{2}}}$