Question #e7339

1 Answer
Apr 17, 2017

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.

google drawings

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#