How do you find the value for #sin^-1 (-1/sqrt2)#?

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Answer 1 · 2018-07-27T14:55:55

#-\pi/4#

Notice, #-\pi/2\le \sin x\le \pi/2\ \ \forall \ \ x\in R#

#\therefore \sin^{-1}(-1/\sqrt2)#

#=-\sin^{-1}(1/\sqrt2)#

#=-\pi/4#

Answer 2 · 2018-07-27T15:13:29

As below

#theta = sin ^-1 (-1/sqrt2)#

#sin theta = - 1/sqrt2 #

#theta = 135^@ # or #-45^@# # = ((5pi)/4)^c # or # ((7pi)/4)^c#

Generalizing, #theta = (n pi + (pi/4))^c#, where n is odd integer & #theta = (n pi - (pi/4))^c# where n is even integer.