How do you prove $Sin^-1(-1/2)= -pi/6$ ?

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Answer 1 · 2016-10-02T10:23:07

See the Explanation.

The Defn. of $sin^-1$ fun. is :

$sin^-1 x=theta, |x| le 1 iff sintheta=x, theta in [-pi/2,pi/2].$

So, we have to find a $theta,$ which satisfies the conds., namely,

$(1) : -pi/2 le theta le pi/2," & "(2) : sintheta=-1/2.$

Knowing that, $sin (-pi/6)=-sin(pi/6)=-1/2,$ and, noting

that, $-pi/6 in [-pi/2,pi/2]$ , we have,

$sin^-1 (-1/2)=-pi/6.$

How do you prove Sin^-1(-1/2)= -pi/6Sin^-1(-1/2)= -pi/6?