Question

How do you write an algebraic expression that is equivalent to `sin(arccosx)`?

Answer 1

Use the identity `sin(u) = +-sqrt(1 - cos^2(u))`
where `u = cos^-1(x)`
Then use the property `cos(cos^-1(x)) = x`

Explanation:

Use the identity `sin(u) = +-sqrt(1 - cos^2(u))`
where `u = cos^-1(x)`:

`sin(cos^-1(x)) = +-sqrt(1-cos^2(cos^-1(x)))`

Use the property `cos(cos^-1(x)) = x`

`sin(cos^-1(x)) = +-sqrt(1-x^2)`