Pre-Calc

How do I write sin(arcsinx + arccosx) as an algebraic expression?

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  1. sin(arcsin(x)) = x
    cos(arcsin(x)) = √(1-x^2)

    so, if
    A = arcsinx
    B = arccosx

    ain(A+B) = sinAcosB + cosAsinB
    = x√(1-x^2) + √(1-x^2)*x
    = 2x√(1-x^2)

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  2. arcsinx is an angle Ø so that sinØ = x/1 = x
    arccosx is an angle Ø so that cosØ = x/1
    thus sinØ = cosØ
    and Ø = 45°

    sin(arcsinx + arccosx)
    = sin(45° + 45°)
    = sin 90°
    = 1

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  3. confirmation by Wolfram:

    http://www.wolframalpha.com/input/?i=simplify+sin%28arcsinx+%2B+arccosx%29

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  4. sinAcosB + cosAsinB
    = x*x + √(1-x^2)√(1-x^2)
    = x^2 + 1-x^2
    = 1

    of course- since if you draw the triangle, you are taking the sine of two angles which add to 90°

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  5. But doesn't sinØ also equal cosØ at 225 degrees?

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  6. yeah - so?

    the expression still evaluates to 1.

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