How do you evaluate #cos^-1(sin(7pi/2))#?

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Answer 1 · 2016-07-29T12:11:01

#-3pi#

For any a, #sin a = cos(pi/2-a) #.

Use bijective 1-1 inversion #f^(-1)(f(x))=x# that returns x.

Here,

#cos^(-1)(sin(7/2pi))#

#=cos^(-1)(cos(pi/2-7/2pi))#.

#=cos^(-1)(cos(-3pi))#.

#=-3pi#

Despite that both #+-3pi# 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 #3pi# to #-3pi# .

See how it works, in the reverse process.

#sin^(-1)(cos(-3pi))#

#=sin^(-1)sin(pi/2-(-3pi)))#

#-sin^(-1)(sin(7/2pi))#

#=7/2pi#, using #f^(-1)f(x)=x#

Interestingly, owing to the principal value convention, your

calculator gives the answer as #180^o=pi#. For the reversed

calculation, it gives # -90^o.=-pi/2#, and not #7/2pi#..