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June 15th, 2010, 06:01 AM   #1
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Algebraic Expression of Sin

I can't seem to understand which choice is an expression, because plugging in -1, 0, 1 equaling 0 all work except for just c.

Problem: sin(2sin^-1x)

Choice:
a. square root(1-x )
b. 2 square root(1-x)
c. x
d. 2x square root(1-x)
e. x square root(1-x)
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June 15th, 2010, 06:17 AM   #2
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Re: Algebraic Expansion of Sin

Hello, RMG46!

I don't understand what you mean by "expansion" . . .


Quote:


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[color=beige] .[/color][color=blue][1][/color]


[color=beige]. . [/color]

[color=beige]. . [/color][color=beige] .[/color][color=blue][2][/color]






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June 15th, 2010, 06:22 AM   #3
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Re: Algebraic Expansion of Sin

Excuse me, meant to put "Expression"*
Does that change your answer?
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June 15th, 2010, 08:55 AM   #4
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The term "equivalent expression" would be clearer, but the answer (except for the initial comment) would be the same.

Since lies in the interval [-pi/2, pi/2], its cosine is non-negative,
and so cos? + sin? = 1 (Pythagoras) implies cos? = ?(1 - sin?) = ?(1 - x), which leads to answer (d).
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June 15th, 2010, 11:25 AM   #5
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Re: Algebraic Expression of Sin

Alright thanks
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