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.#