How do you find the exact value of #sin(arccos(-2/3))#?

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Answer 1 · 2017-02-09T15:56:33

#sin(arccos(-2/3)) = sqrt(5)/3#

First note that #theta = arccos(-2/3)# is in Q2 since #-2/3 < 0#.

In Q2 #sin# is positive.

From Pythagoras we have:

#cos^2 theta + sin^2 theta = 1#

and hence:

#sin theta = +-sqrt(1-cos^2 theta)#

In our case we want the positive square root and find:

#sin(arccos(-2/3)) = sqrt(1-(-2/3)^2) = sqrt(1-4/9) = sqrt(5/9) = sqrt(5)/3#