# How do you find the exact value of cos^-1 0?

Jun 1, 2018

see below

#### Explanation:

${\cos}^{- 1} 0 = \arccos 0$

What is the arc which cosine is zero?: two posibilities $90 = \frac{\pi}{2}$ and $270 = 3 \frac{\pi}{2}$

This is assuming that ${\cos}^{- 1}$ is the inverse of cosine. There is no missunderstanding if use $\arccos$ instead of ${\cos}^{- 1}$ Because ${\cos}^{- 1}$ is also understud as $\frac{1}{\cos} = \sec$ which is different

Jun 1, 2018

Think of right-angled triangles

#### Explanation:

The trig ratios are defined as ratios of side lengths of a right-angled triangle. $C o s$ is the adjacent side divided by the hypotenuse, i.e. the length of the short side next to the angle being considered divided by the length of the long side next to it that isn't adjacent to the right angle.

The question here is: What angle in the right-angled triangle results in the ratio of these two sides being zero? Imagine opening up the angle towards another right angle. The closer it gets to 90 degrees, the longer the hypotenuse becomes in order to get across to the far side of the triangle to meet the remaining side. When it gets to 90 degrees, the side is infinitely long - it never meets the other side of the triangle because it's parallel to it. So $\cos$ has become $\frac{a}{\infty} = 0$, where $a$ is the length of the adjacent side.

Thus ${\cos}^{- 1} 0 = 90$ degrees, or, in radians, $\frac{\pi}{2}$.