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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 √(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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