Question

How do you evaluate `cos(sin^-1(sqrt2/2))`?

Answer 1

`cos(sin^-1(sqrt2/2))=sqrt2/2`

Explanation:

`cos(sin^-1(sqrt2/2))`

The restriction for the range of arcsin x is `[-pi/2,pi/2]`. Since the argument is positive it means that our triangle is in quadrant I with opposite side 1 and hypotenuse `sqrt 2` and hence the adjcent is also 1. Remember that we don't need to know the angle let's just call it `theta` then we need to find the ratio for `cos theta` which is
`cos theta=a/h = 1/sqrt2 = sqrt2/2`