How do you find the value for sec^-1( 2/ sqrt3)?

Jun 5, 2015

${\sec}^{-} 1 \left(\frac{2}{\sqrt{3}}\right) = {30}^{o}$

The value ${\sec}^{-} 1 \left(\frac{2}{\sqrt{3}}\right)$ returns will be an angle.

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

Take the secant of each side,

$\sec \left(\setminus \theta\right) = \frac{2}{\setminus} \sqrt{3}$

Substitute $\sec \left(\setminus \theta\right) = \frac{1}{\cos} \left(\theta\right)$,

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

$\cos \left(\theta\right) = \setminus \frac{\sqrt{3}}{2}$

These numbers are common. There is a specific triangle to be remembered that allows us to find these sorts of trigonometric values exactly.

Recall that the cosine of an angle is equal to the adjacent over the hypotenuse. From the diagram we see that $\cos \left({30}^{o}\right) = \setminus \frac{\sqrt{3}}{2}$
Therefore $\theta = {30}^{o}$ solves $\cos \left(\theta\right) = \setminus \frac{\sqrt{3}}{2}$