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How do you evaluate #sin^-1(1/2)# without a calculator?

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Sep 3, 2016

By Defn. of #sin^-1# function, #sin^-1 x=theta, |x| le1 iff sin theta=x, theta in [-pi/2,pi/2]#.

As #sin (pi/6)=1/2, &, pi/6 in [-pi/2,pi/2], sin^-1 (1/2)=pi/6#.

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