# Question e7339

Apr 17, 2017

It doesn't.

$\cos \left({\sin}^{-} 1 \left(\frac{2}{3}\right)\right) = \frac{\sqrt{5}}{3}$

#### Explanation:

Let's break this down:

color(red)(cos(color(blue)(sin^-1(2/3)))

The $\textcolor{red}{\cos}$ function is taking the cosine of the angle $\textcolor{b l u e}{{\sin}^{-} 1 \left(\frac{2}{3}\right)}$.

Let's say that $\textcolor{b l u e}{\theta = {\sin}^{-} 1 \left(\frac{2}{3}\right)}$. Remember that this is an angle.

In the triangle where this is true, we see that $\sin \left(\theta\right) = \frac{2}{3}$.

We can draw a triangle where $\sin \left(\theta\right) = \frac{2}{3}$: this means that the "opposite" side is $2$ and the hypotenuse is $3$.

Through the Pythagorean Theorem we see that the other leg is $\sqrt{5}$, since we have a right triangle.

What we want to find is color(red)(cos(color(blue)(sin^-1(2/3))), which is really color(red)(cos(color(blue)(theta))#.

That is, we just want cosine of theta in this triangle.

Cosine is "adjacent" over hypotenuse, so:

$\textcolor{red}{\cos \left(\textcolor{b l u e}{\theta}\right)} = \textcolor{red}{\cos \left(\textcolor{b l u e}{{\sin}^{-} 1 \left(\frac{2}{3}\right)}\right)} = \frac{\sqrt{5}}{3}$