Posted by Matthew on Tuesday, December 4, 2012 at 4:46pm.
draw a triangle with adjacent/hypotenuse = x/1
the other leg is √(1-x^2)
so, sin(arccos(x)) = √(1-x^2)
or, if θ = arccos(x), then we have the ubiquitous identity
sin^2θ + cos^2θ = 1
but cosθ = x, so
sin^2θ + x^2 = 1
sin^2θ = 1-x^2
sinθ = √(1-x^2)
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