Question

How do you evaluate `sin(arccos (1/3))`?

Answer 1

`sin(arccos(1/3))=+-(2sqrt2)/3`

Explanation:

Let `arccos(1/3)=theta`. This means

`costheta=1/3` and hence

`sintheta=sqrt(1-(1/3)^2)=+-sqrt(1-1/9)=+-sqrt(8/9)=+-(2sqrt2)/3`

We are using both plus and minus as if `costheta` is in first quadrant, `sintheta` could be positive and if `costheta` is in fourth quadrant, `sintheta` could be negaitive.

As such `sin(arccos(1/3))=sintheta=+-(2sqrt2)/3`