For any a, sin a = cos(pi/2-a) sina=cos(π2−a).
Use bijective 1-1 inversion f^(-1)(f(x))=xf−1(f(x))=x that returns x.
Here,
cos^(-1)(sin(7/2pi))cos−1(sin(72π))
=cos^(-1)(cos(pi/2-7/2pi))=cos−1(cos(π2−72π)).
=cos^(-1)(cos(-3pi))=cos−1(cos(−3π)).
=-3pi=−3π
Despite that both +-3pi±3π point to the same direction, the sense of
rotation is opposite, for the negative sign. So, I make this self-
correction to my previous answer from 3pi3π to -3pi−3π .
See how it works, in the reverse process.
sin^(-1)(cos(-3pi))sin−1(cos(−3π))
=sin^(-1)sin(pi/2-(-3pi)))=sin−1sin(π2−(−3π)))
-sin^(-1)(sin(7/2pi))−sin−1(sin(72π))
=7/2pi=72π, using f^(-1)f(x)=xf−1f(x)=x
Interestingly, owing to the principal value convention, your
calculator gives the answer as 180^o=pi180o=π. For the reversed
calculation, it gives # -90^o.=-pi/2#, and not #7/2pi#..
