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April 23rd, 2009, 02:56 PM   #1
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Rewrite the expression as an algebraic expression in x

sin(2sin^-1x)
*Note sine^-1 is sin inverse

sin2x=2sinxcosx
Drawing a triangle I found side a=x b=sq root (1-x^2) and c=1
Based on the triangle, I found cos to be sq root (1-x^2) so sin2x=2*x/1*sq root (1-x^2)=2xsqroot(1-x^2)
Is the answer 2xsqroot(1-x^2)
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April 24th, 2009, 05:06 AM   #2
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Yes, that's correct.
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