How do you prove ${cos}^{- 1} (x) = \frac{π}{2} - {sin}^{- 1} (x)$ $cos^-1(x)=pi/2-sin^-1(x)$ ?

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Answer 1 · 2016-02-27T05:24:38

Let, ${sin}^{- 1} x = θ$ $sin^(-1)x=theta$ ; then $sin θ = x$ $sintheta=x$

now, $sinθ = cos( π/2 - θ)=x$ and $cos^(-1)x= π/2 - θ$

therefore, $sin^(-1)x+cos^(-1)x= θ+π/2 - θ= π/2$

How do you prove cos^-1(x)=pi/2-sin^-1(x)cos−1(x)=π2−sin−1(x)cos^-1(x)=pi/2-sin^-1(x)?