# How do you evaluate #cos (arcsin (3/5))#?

##### 2 Answers

May 26, 2018

#### Explanation:

The ratio

Note that

So

Consider a

We have:

#sin A = "opposite"/"hypotenuse" = 3/5#

#cos A = "adjacent"/"hypotenuse" = 4/5#

So:

#cos(arcsin(3/5)) = cos(A) = 4/5#

Alternatively, we could note more generally that:

#cos^2 theta + sin^2 theta = 1#

Hence:

#cos theta = +-sqrt(1-sin^2 theta)#

If

#cos(arcsin(x)) = sqrt(1-x^2)#

May 27, 2018

The inverse sine is multivalued, so