How do you evalute ${sin}^{- 1} (- \frac{\sqrt{3}}{2})$ $sin^-1 (-sqrt(3)/2)$ ?

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How do you evalute ${sin}^{- 1} (- \frac{\sqrt{3}}{2})$ $sin^-1 (-sqrt(3)/2)$ ?

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Answer 1 · 2014-10-23T21:46:57

Let

$θ = {sin}^{- 1} (- \frac{\sqrt{3}}{2})$ $theta=sin^{-1}(-sqrt{3}/2)$ .

By rewriting it in terms of sine,

$\Rightarrow sin θ = - \frac{\sqrt{3}}{2}$ $Rightarrow sin theta=-sqrt{3}/2$

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

Since

$sin (- \frac{π}{3}) = - \frac{\sqrt{3}}{2}$ $sin(-pi/3)=-sqrt{3}/2$ ,

we have

$θ = {sin}^{- 1} (- \frac{\sqrt{3}}{2}) = - \frac{π}{3}$ $theta=sin^{-1}(-sqrt{3}/2)=-pi/3$ .

I hope that this was helpful.

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How do you evalute ${sin}^{- 1} (- \frac{\sqrt{3}}{2})$ $sin^-1 (-sqrt(3)/2)$ ?

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