How do you find $sin ({cos}^{- 1} x)$ $sin(cos^-1x)$ ?

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Answer 1 · 2016-09-24T08:52:14

$\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$ .

Let $a = {cos}^{- 1} x \in [0, π]$ $a =cos^(-1)x in [0, pi]$ . for the principal value. Then cos a =x, for

both positive and negative x,..

The given expression is

$sin a = \sqrt{1 - {cos}^{2} a} = \sqrt{1 - x^{2}}$ $sin a=sqrt(1-cos^2a)= sqrt(1-x^2)$ , for $a \in [0, π]$ $a in [0, pi]$ ..

If the principal value convention is relaxed and x is negative, a could

be in $Q_{3}$ $Q_3$ wherein both sine and cosine are negative. And then,

$sin a = - \sqrt{1 - x^{2}}$ $sin a = -sqrt(1-x^2)$ ,

.

How do you find sin(cos−1x)sin(cos^-1x)?