# How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle 75^circ?

Jul 21, 2017

See the explanation below

#### Explanation:

Conversion of ${75}^{\circ}$ into radians

$= \frac{75}{180} \pi = \frac{5}{12} \pi r a d$

and

$\frac{5}{12} \pi = \frac{3}{4} \pi - \frac{\pi}{3}$

Therefore,

$\sin \left({75}^{\circ}\right) = \sin \left(\frac{5}{12} \pi\right)$

$= \sin \left(\frac{3}{4} \pi - \frac{1}{3} \pi\right)$

$= \sin \left(\frac{3}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) - \cos \left(\frac{3}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

$= \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$

$= \frac{\sqrt{6} + \sqrt{2}}{4}$

$\cos \left({75}^{\circ}\right) = \cos \left(\frac{5}{12} \pi\right) = \cos \left(\frac{3}{4} \pi - \frac{1}{3} \pi\right)$

$= \cos \left(\frac{3}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) + \sin \left(\frac{3}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

$= - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$

$= \frac{\sqrt{6} - \sqrt{2}}{4}$

Therefore,

$\tan \left({75}^{\circ}\right) = \sin \frac{{75}^{\circ}}{\cos} \left({75}^{\circ}\right) = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}$