# How do you evaluate cos [Sec ^-1 (-5)]?

Aug 28, 2016

$- \frac{1}{5}$

#### Explanation:

As cosine is the reciprocal of secant,

$a = {\sec}^{- 1} \left(- 5\right) = {\cos}^{- 1} \left(- \frac{1}{5}\right)$,

the given expression is

$\cos a = - \frac{1}{5}$.

Aug 29, 2016

$\cos \left({\sec}^{-} 1 \left(- 5\right)\right) = - \frac{1}{5}$

#### Explanation:

Let:

$x = \cos \left({\sec}^{-} 1 \left(- 5\right)\right)$

We can then say that:

${\cos}^{-} 1 \left(x\right) = {\sec}^{-} 1 \left(- 5\right)$

Using the same principle to now isolate the $- 5$, we say that:

$\sec \left({\cos}^{-} 1 \left(x\right)\right) = - 5$

Since $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$, rewrite the left-hand side:

$\frac{1}{\cos} \left({\cos}^{-} 1 \left(x\right)\right) = - 5$

$\cos \left(x\right)$ and ${\cos}^{-} 1 \left(x\right)$ undo one another, being inverse functions:

$\frac{1}{x} = - 5$

Taking the reciprocal of both sides:

$x = - \frac{1}{5}$

Thus:

$\cos \left({\sec}^{-} 1 \left(- 5\right)\right) = - \frac{1}{5}$