# How do you find the exact functional value sec (-11pi/12) using the cosine sum or difference identity?

Aug 9, 2015

Find $\sec \left(\frac{- 11 \pi}{12}\right)$

Ans: $- \frac{4}{\sqrt{2} + \sqrt{6}}$

#### Explanation:

sec = 1/cos
$\cos \left(\frac{- 11 \pi}{12}\right) = \cos \left(\frac{11 \pi}{12}\right) = \cos \left(\frac{3 \pi}{12} + \frac{8 \pi}{12}\right)$ =

$= \cos \left(\frac{\pi}{4} + \frac{2 \pi}{3}\right)$
Use the trig identity: cos (a + b) = cos a.cos b - sin a.sin b

$\cos a = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ ; $\cos b = \cos \left(\frac{2 \pi}{3}\right) = - \frac{1}{2}$
$\sin a = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ ; $\sin b = \sin \left(\frac{2 \pi}{3}\right) = \frac{\sqrt{3}}{2}$
Therefor:
$\cos \left(\frac{- 11 \pi}{12}\right) = \cos \left(\frac{\pi}{4} + \frac{2 \pi}{3}\right) =$

$\left(\frac{\sqrt{2}}{2}\right) \left(- \frac{1}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$ = $- \frac{\sqrt{2} + \sqrt{6}}{4}$
Check by calculator:
$\cos \left(\frac{- 11 \pi}{12}\right) = \cos \left(- 165\right) = - 0.966$
$- \frac{\sqrt{2} + \sqrt{6}}{4} = - 0.966 .$ OK

Finally:$\sec \left(\frac{- 11 \pi}{12}\right) = - \frac{4}{\sqrt{2} + \sqrt{6}}$ = $- \frac{1}{0.966} = - 1.035$