Question

If $x<0$, then prove that ${\cos }^{-1}x=\pi -{\sin }^{-1}\sqrt{1-x^2}$.

Answer 1

We have for $x>0$, ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$.

The given problem is equivalent to the above i.e. for $x<0$, ${\cos }^{-1}x=\pi -{\cos }^{-1}x$.

Now the proof for $x>0$, ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$, is as follows

Let, ${\cos }^{-1}(-x)=z$

or, $-x=\cos z$

or, $x=-\cos z$

or, $x=\cos (\pi -z)$

or, $\pi -z={\cos }^{-1}x$

or, ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$

or, ${\cos }^{-1}(-x)=\pi -{\sin }^{-1}\sqrt{1-x^2}$ [Since ${\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}$ for $x>0$..]

So for $x>0$ we have ${\cos }^{-1}(-x)=\pi -{\sin }^{-1}\sqrt{1-x^2}$

or, for $x<0$ we have ${\cos }^{-1}\left(x\right)=\pi -{\sin }^{-1}\sqrt{1-x^2}$.