How do you simplify $sin^(-1)(- 1/2)$ ?

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Answer 1 · 2016-09-27T17:54:14

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

Let $sin^-1(-1/2)=theta.$ :. By defn. of $sin^-1$ fun., we have,

$sintheta=-1/2, and, theta in [-pi/2,pi/2]$ .

Recall that, $sin(pi/6)=1/2, so, sin(-pi/6)=-sin(pi/6)=-1/2$

Thus, $sin(-pi/6)=-1/2, &, -pi/6 in [-pi/2,pi/2]$ .

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

How do you simplify sin^(-1)(- 1/2)sin^(-1)(- 1/2)?