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

Jul 26, 2018

$\sin {75}^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$
$\cos {75}^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$
$\tan {75}^{\circ} = 2 + \sqrt{3}$

#### Explanation:

We know that ,

color(red)((1)sin(x+y)=sinxcosy+cosxsiny

color(blue)((2)cos(x+y)=cosxcosy-sinxsiny

color(green)((3)tan(x+y)=(tanx+tany)/(1-tanxtany)

We have ,

color(brown)(75^circ=45^circ+30^circ

diamondsin75^circ=color(red)(sin(45^circ+30^circ) toApply(1)

=>sin75^circ=color(red)(sin45^circcos30^circ+cos45^circsin30^circ

$\implies \sin {75}^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}$
$\implies \sin {75}^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$
$\implies \sin {75}^{\circ} = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$\implies \sin {75}^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$

diamondcos75^circ=color(blue)(cos(45^circ+30^circ)toApply(2)

=>cos75^circ=color(blue)(cos45^circcos30^circ-sin45^circsin30^circ

$\implies \cos {75}^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2}$
$\implies \cos {75}^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
$\implies \cos {75}^{\circ} = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$\implies \cos {75}^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$

diamondtan75^circ=color(green)(tan(45^circ+30^circ)toApply(3)

=>tan75^circ=color(green)((tan45^circ+tan30^circ)/(1-tan45^circtan30^circ)
$\implies \tan {75}^{\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$
$\implies \tan {75}^{\circ} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$
$\implies \tan {75}^{\circ} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$
$\implies \tan {75}^{\circ} = {\left(\sqrt{3} + 1\right)}^{2} / \left({\left(\sqrt{3}\right)}^{2} - {1}^{2}\right)$
$\implies \tan {75}^{\circ} = \frac{3 + 2 \sqrt{3} + 1}{3 - 1}$
$\implies \tan {75}^{\circ} = \frac{4 + 2 \sqrt{3}}{2}$
$\implies \tan {75}^{\circ} = 2 + \sqrt{3}$

Note:
If it is $\left(- {75}^{\circ}\right)$ ,then take $- {75}^{\circ} = - {45}^{\circ} - {30}^{\circ} .$
So,
$\sin \left(- {75}^{\circ}\right) = - \sin {75}^{\circ} = - \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)$
$\cos \left(- {75}^{\circ}\right) = \cos {75}^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$\tan {\left(- 75\right)}^{\circ} = - \tan {75}^{\circ} = - \left(2 + \sqrt{3}\right)$