TRIGONOMETRY
posted by Matthew .
Write the following trigonometric expression as an algebraic expression in x free of trigonometric or inverse trigonometric functions
sin(cos^1 x) 1≤x≤1

draw a triangle with adjacent/hypotenuse = x/1
the other leg is √(1x^2)
so, sin(arccos(x)) = √(1x^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 = 1x^2
sinθ = √(1x^2)
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