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

1 Answer
Jun 10, 2016

#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#