What is sin^-1(cos 5pi/6) ?

1 Answer
May 24, 2018

-pi/3π3.

Explanation:

Recall the definition of the sin^-1sin1 function :

sin^-1 x=theta, (-1 le x le 1) iff x=sintheta, -pi/2 le theta le pi/2sin1x=θ,(1x1)x=sinθ,π2θπ2.

Now, cos(5/6pi)=cos(pi-pi/6)=-cos(pi/6)=-sqrt3/2cos(56π)=cos(ππ6)=cos(π6)=32.

:. sin^-1(cos(5/6pi))=sin^-1(-sqrt3/2).

Since, sin(-pi/3)=-sin(pi/3)=-sqrt3/2, &, -pi/3 in [-pi/2,pi/2],

sin^-1(-sqrt3/2)=-pi/3.

:. sin^-1(cos(5/6pi))=sin^-1(-sqrt3/2)=-pi/3.