Hello, I was wondering if someone could help me with the proof of. I'm a little confused, I was using cos=sqrt(1-x^2) and arcsin=1/(sqrt(1-x^2)) and getting no where. A trigonometric proof isn't needed but if it's the only way then that's fine. Thanks for all help!
letOriginally Posted by Dudealadude
Hello, I was wondering if someone could help me with the proof of
this can also be written as ...
![\displaystyle \cos[\arcsin(x)] = \cos(y) = \frac{adjacent \, side}{hypotenuse} = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}](http://latex.codecogs.com/png.latex?\displaystyle \cos[\arcsin(x)] = \cos(y) = \frac{adjacent \, side}{hypotenuse} = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2})
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