Let us first recall the Defn. of sin^-1 fun. :
sin^-1x=theta, |x|le1 iff sintheta=x, theta in [-pi/2,pi/2].
Ans. (1)
We have, sin((5pi)/4)=sin(pi+pi/4)=-sin(pi/4)=-1/sqrt2,
so, sin^-1(sin((5pi)/4))=sin^-1(-1/sqrt2).
Now, knowing that sin(-pi/4)=-1/sqrt2, and, -pi/4 in [-pi/2,pi/2], we
conclude, from the Defn. of sin^-1, sin^-1(-1/sqrt2)=-pi/4, i.e.,
sin^-1(sin((5pi)/4))=sin^-1(-1/sqrt2)=-pi/4, which is in the
desired range, i.e., −π/2≤θ≤π.
Proceeding as above,
Ans.(2) : sin^-1(sin((5pi)/3))=-pi/3.
Ans.(3) : cos^-1(cos((-7pi)/4))=pi/4.
Ans.(4) : cos^-1(cos((2pi)/3))=(2pi)/3.
