How do you evaluate #arcsin(sin(-8))#?

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Answer 1 · 2018-07-09T09:01:47

#8-3\pi\approx-81.633^\circ#

Let #\sin^{-1}(\sin(-8))=\theta\quad \quad \forall \ -\pi/2\le \theta\le\pi/2#

#\sin\theta=\sin(-8)#

#\theta=2n\pi+(-8)\ \ \text{Or}\ \ \theta=2n\pi+(\pi-(-8))#

#\theta=2n\pi-8\ \ \text{Or}\ \ \theta=(2n+1)\pi+8#

Where, #n# is any integer i.e. #n=0, \pm1, \pm2, \pm3, \ldots#

But, we know that #-\pi/2\le \theta\le\pi/2# , setting #n=-2# in second general solution we get

#\theta=(2\cdot (-2)+1)\pi+8#

#=-3\pi+8#

#=-81.633^\circ#