Question #9adca

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Question #9adca

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Answer 1 · 2017-02-23T13:43:15

$- \frac{π}{6} .$ $-pi/6.$

Explanation:

Recall the following Defn. of $a r c sin$ $arc sin$ function :

$a r c sin x = θ, | x | \leq 1 \Leftrightarrow sin θ = x, θ \in [- \frac{π}{2}, \frac{π}{2}] .$ $arc sinx=theta, |x|le1 iff sin theta=x, theta in [-pi/2,pi/2].$

Now, $sin (- \frac{π}{6}) = - sin (\frac{π}{6}) = - 0.5, &, - \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ $sin (-pi/6)=-sin(pi/6)=-0.5, &, -pi/6 in [-pi/2,pi/2]$

$:. arc sin(-0.5)=-pi/6.$

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Question #9adca

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