Question

Prove that \(\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \).

Short answer

It is proved that \(\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \).

Step-by-step solution

Show that \(\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \)

Suppose that \(y = {\sin ^{ - 1}}x\) then \( - \frac{\pi }{2} \le y \le \frac{\pi }{2} \Rightarrow \cos y \ge 0\). Therefore,

\(\cos \left( {{{\sin }^{ - 1}}x} \right) = \cos y\,\,\,\,\,\,\,\,\,\,\,{\rm{ }}\left\{ {\cos x = \sqrt {1 - {{\sin }^2}x} } \right\}\)

\( = \sqrt {1 - {{\sin }^2}y} \)

Simplify further

Here, \(y = {\sin ^{ - 1}}x\).

\(\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \)

Thus, it is proved that \(\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \).