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

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Answer 1 · 2016-09-24T08:52:14

#sqrt(1-x^2)#.

Let #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^2a)= sqrt(1-x^2)#, for #a in [0, pi]#..

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

be in #Q_3# wherein both sine and cosine are negative. And then,

#sin a = -sqrt(1-x^2)# ,

.