# How do you find the exact value of sec75 using the half angle formula?

Sep 25, 2016

$\sec 75 = \sqrt{6} - \sqrt{2}$.

#### Explanation:

As $\sec 75 = \frac{1}{\cos} 75 \text{, we have to first find the value of } \cos 75$.

We have to use the following Half-Angle Formula :

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

where, the sign (+ or -) is to be determined in accordance with

$\cos \left(\frac{\theta}{2}\right)$

Taking, $\frac{\theta}{2} = 75 , i . e . , \theta = 150$, and, noting that $75$ lies in the

First Quadrant , where, $\cos \text{ is } + v e$, we get,

cos75=+sqrt((1+cos150)/2)=sqrt((1+cos(180-30)/2)

$= \sqrt{\frac{1 - \cos 30}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}}$

$= \frac{1}{2} \sqrt{2 - \sqrt{3}} = \frac{1}{2} \sqrt{2 - 2 \sqrt{\frac{3}{4}}}$

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

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

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

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

Therefore,

$\sec 75 = \frac{1}{\cos} 75 = \frac{2 \sqrt{2}}{\sqrt{3} - 1}$

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

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

$= \sqrt{6} - \sqrt{2}$.

Enjoy Maths!