Question

Question #e7339

Answer 1

It doesn't.

`cos(sin^-1(2/3))=sqrt5/3`

Explanation:

Let's break this down:

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

The `color(red)(cos)` function is taking the cosine of the angle `color(blue)(sin^-1(2/3))`.

Let's say that `color(blue)(theta=sin^-1(2/3))`. Remember that this is an angle.

In the triangle where this is true, we see that `sin(theta)=2/3`.

We can draw a triangle where `sin(theta)=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 `sqrt5`, 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:

`color(red)(cos(color(blue)(theta)))=color(red)(cos(color(blue)(sin^-1(2/3))))=sqrt5/3`