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.

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