How do you evalute #sin^-1 (-sqrt(3)/2)#?

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Answer 1 · 2014-10-23T21:46:57

Let

#theta=sin^{-1}(-sqrt{3}/2)#.

By rewriting it in terms of sine,

#Rightarrow sin theta=-sqrt{3}/2#

So, we need to find the angle #theta# between #-pi/2# and #pi/2# that makes sine equal to #-sqrt{3}/2#.

Since

#sin(-pi/3)=-sqrt{3}/2#,

we have

#theta=sin^{-1}(-sqrt{3}/2)=-pi/3#.

I hope that this was helpful.