<s>[INST] Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$ Question:
Let $x<0$ then $\cos^{-1}\sqrt{1-x^2}=$
(1) $\pi-\cos^{-1}x$
(2) $-\sin^{-1}x$
(3) $\pi-\sin^{-1}x$
(4) $\sin^{-1}x$
My approach is as follow
Given  $x < 0,{\cos ^{ - 1}}\sqrt {1 - {x^2}} $
Let  $x = \sin \theta $, therefore ${\cos ^{ - 1}}\sqrt {1 - {{\sin }^2}\theta }  = {\cos ^{ - 1}}\sqrt {{{\cos }^2}\theta }  = {\cos ^{ - 1}}\left| {\cos \theta } \right| = \theta $
Hence  ${\cos ^{ - 1}}\sqrt {1 - {{\sin }^2}\theta }  = {\cos ^{ - 1}}\sqrt {{{\cos }^2}\theta }  = {\cos ^{ - 1}}\left| {\cos \theta } \right| = {\sin ^{ - 1}}x$
But as per the graph answer is $-\sin^{-1}x$ for $x<0$.
 [/INST] Because $x=\sin \theta <0$ and $\cos \theta = \sqrt{1-x^2} >0$, angle $\theta$ is in the fourth quadrant. Thus, $\cos^{-1} |\cos \theta|=-\theta$
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