why sin^-1(-x) is equal to - sin ^-1(x)
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Step-by-step explanation:
Let [tex] \sin^{-1}(x)=\theta[/tex]
Now,
[tex] \sin( \theta) = \sin( \sin^{ - 1} (x) ) [/tex]
[tex] \implies \sin( \theta) = x[/tex]
Now,
[tex] - x = - \sin( \theta) [/tex]
[tex] \implies - x = \sin( - \theta) [/tex]
[tex] \implies \sin^{ - 1} (- x) = \sin^{ - 1} (\sin( - \theta) ) \\ [/tex]
[tex] \implies \sin^{ - 1} (- x) = - \theta \\ [/tex]
[tex] \implies \sin^{ - 1} (- x) = - \sin^{ - 1} (x) \\ [/tex]