How do you prove #cos(sin^-1x)=sqrt(1-x^2)#?

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Answer 1 · 2016-11-15T05:45:05

For the conventional inverse sine, the answer is #sqrt(1-x^2)#. See explanation.

Let #a = sin^(-1)x in [-pi/2, pi/2]# ( for the conventional inverse ),

wherein cosine is positive. Then, sin a = x.

The given expression is cos a = sqrt(1-sin^2a)=sqrt(1-x^2)#, for the

conventional #sin^(-1)x#.

The general value of #a in (-oo, oo)#, and so, might be in any

quadrant, for an arbitrary #x in [-1, 1]#. The cosine might be >=< 0.

So, here, the answer is #+-sqrt(1-x^2)#