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

Question

Violet Anderson · Accepted Answer

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).

Charles Autterson · Answer

undefined To evaluate arcsin((√3)/2), you recognize that arcsin is the inverse function of sine. Since sine represents the ratio of the opposite side to the hypotenuse in a right triangle, and (√3)/2 corresponds to the y-coordinate of a point on the unit circle when the angle is π/3 radians (or 60 degrees), the arcsin((√3)/2) is π/3 radians (or 60 degrees).

To evaluate sin(Arccos(-15/17)), you first find the value of Arccos(-15/17), which represents the angle whose cosine is -15/17. Using the unit circle or trigonometric identities, you find that Arccos(-15/17) is approximately 131.81 degrees (or 2.30 radians). Then, you calculate sin(131.81 degrees) or sin(2.30 radians) to find the sine of that angle.