# What is the sec(inverse cos (5^0.5/2))?

Dec 4, 2015

$\sec \left(\arccos \left(\frac{\sqrt{5}}{2}\right)\right) = \frac{2}{\sqrt{5}}$

#### Explanation:

For this problem, we just need two facts:

• $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ (by definition of the secant function)

• $\cos \left(\arccos \left(x\right)\right) = \arccos \left(\cos \left(x\right)\right) = x$ (by definition of the inverse of a function)

Applying those gives us

$\sec \left(\arccos \left(\frac{\sqrt{5}}{2}\right)\right) = \frac{1}{\cos} \left(\arccos \left(\frac{\sqrt{5}}{2}\right)\right) = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$