How do you evaluate #sin^-1(sin((13pi)/10))#?

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Answer 1 · 2016-07-30T23:20:51

#-(3pi)/10#

The inverse sine function has domain #[-1,1]# which means it will have range #-pi/2<=y<=pi/2#

This means that any solutions we obtain must lie in this interval.

As a consequence of double angle formulae, #sin(x) = sin(pi-x)# so

#sin((13pi)/(10)) = sin(-(3pi)/10)#

Sine is #2pi# periodic so we can say that

#sin^(-1)(sin(x)) = x + 2npi, n in ZZ#

However any solutions must lie in the interval #-pi/2 <= y <= pi/2#.

There is no integer multiple of #2pi# we can add to #(13pi)/10# to get it within this interval so the only solution is #-(3pi)/10#.