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

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Answer 1 · 2018-05-24T17:55:09

# -pi/3#.

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

#sin^-1 x=theta, (-1 le x le 1) iff x=sintheta, -pi/2 le theta le pi/2#.

Now, #cos(5/6pi)=cos(pi-pi/6)=-cos(pi/6)=-sqrt3/2#.

#:. 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#.