How do you evaluate #arcsin((sqrt3)/2)# or #sin(Arccos(-15/17))#?

1 Answer
Apr 23, 2015

In this way.

The range of the function #y=arcsinx# is #[-pi/2,pi/2]#, so there is only this solution:

#x=pi/3#.

#sin(arccos(-15/17))=sinalpha#

where #alpha=arccos(-15/17)#, and the angle is in the second quadrant because the range of the function #y=arccosx# is #[0,pi]# and the value is negative.

So

#cosalpha=-15/17# and than

#sinalpha=+sqrt(1-cos^2alpha)=sqrt(1-225/289)=sqrt((289-225)/289)=#

#=sqrt(64/289)=8/17#

(with the #+# because in the second quadrant the sinus is positive).