# How do you evaluate sec ((5pi)/12)?

Apr 4, 2016

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

#### Explanation:

sec = 1/cos . Evaluate cos ((5pi)/12)
Trig unit circle, and property of complementary arcs give -->
$\cos \left(\frac{5 \pi}{12}\right) = \cos \left(\frac{6 \pi}{12} - \frac{\pi}{12}\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{12}\right) = \sin \left(\frac{\pi}{12}\right)$
Find sin (pi/12) by using trig identity:
$\cos 2 a = 1 - 2 {\sin}^{2} a$
$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} = 1 - 2 {\sin}^{2} \left(\frac{\pi}{12}\right)$
$2 {\sin}^{2} \left(\frac{\pi}{12}\right) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$
${\sin}^{2} \left(\frac{\pi}{12}\right) = \frac{2 - \sqrt{3}}{4}$
$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ --> $\sin \left(\frac{\pi}{12}\right)$ is positive.
Finally,
$\sec \left(\frac{5 \pi}{12}\right) = \frac{2}{\sqrt{2 - \sqrt{3}}}$

You can check the answer by using a calculator.

Apr 4, 2016

$\sec \left(\frac{5 \pi}{12}\right) = \sqrt{6} + \sqrt{2}$

#### Explanation:

$\sec x = \frac{1}{\cos} x$

$\sec \left(\frac{5 \pi}{12}\right) = \frac{1}{\cos} \left(\frac{5 \pi}{12}\right)$

$\frac{5 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$-> Break up into composite Argument

$= \frac{1}{\cos} \left(\frac{\pi}{4} + \frac{\pi}{6}\right)$

->use $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$

$= \frac{1}{\cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right)}$

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

$= \frac{1}{\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}} = \frac{1}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{4}{\sqrt{6} - \sqrt{2}}$

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

$= \frac{4 \left(\sqrt{6} + \sqrt{2}\right)}{6 - 2} = \frac{4 \left(\sqrt{6} + \sqrt{2}\right)}{4}$

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