# How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle 165^circ?

Jan 5, 2018

$\sin \left({165}^{\circ}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$,

$\cos \left({165}^{\circ}\right) = - \frac{\sqrt{2 + \sqrt{3}}}{2}$,

$\tan \left({165}^{\circ}\right) = - \sqrt{7 - 4 \sqrt{3}}$

#### Explanation:

${165}^{\circ}$ is in QII, so we know that $\sin \left({165}^{\circ}\right)$ is positive and $\cos \left({165}^{\circ}\right)$ is negative.

${165}^{\circ} = \frac{{330}^{\circ}}{2}$, so we'll use ${330}^{\circ}$ in the half angle formulas. $\cos \left({330}^{\circ}\right) = - \frac{\sqrt{3}}{2}$.

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

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

Combining what we know so far we can find sine:

$\sin \left({165}^{\circ}\right) = \sin \left({330}^{\circ} / 2\right) = \sqrt{\frac{1 - \cos \left({330}^{\circ}\right)}{2}}$

$= \sqrt{\frac{1 - \left(- \frac{\sqrt{3}}{2}\right)}{2}} = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}$

$= \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$

Now let's find cosine in much the same way:

$\cos \left({165}^{\circ}\right) = \cos \left({330}^{\circ} / 2\right) = - \sqrt{\frac{1 + \cos \left({330}^{\circ}\right)}{2}}$

$= - \sqrt{\frac{1 + \left(- \frac{\sqrt{3}}{2}\right)}{2}} = - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}}$

$= - \sqrt{\frac{2 + \sqrt{3}}{4}} = - \frac{\sqrt{2 + \sqrt{3}}}{2}$

So we've found:

$\sin \left({165}^{\circ}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$

and

$\cos \left({165}^{\circ}\right) = - \frac{\sqrt{2 + \sqrt{3}}}{2}$

To find $\tan \left({165}^{\circ}\right)$ we'll use the identity:

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$.

So,

$\tan \left({165}^{\circ}\right) = \sin \frac{{165}^{\circ}}{\cos} \left({165}^{\circ}\right)$

=(sqrt(2-sqrt(3))/2)/(- sqrt(2+sqrt(3))/2) = -sqrt(2-sqrt(3))/(sqrt(2+sqrt(3))

we can do more work on this, if we want, although that's finished answer for many people:

$- \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} = - \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}}$

$= - \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}}}$

$= - \sqrt{\frac{7 - 4 \sqrt{3}}{1}} = - \sqrt{7 - 4 \sqrt{3}}$