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 June 15th, 2010, 06:01 AM #1 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 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-x² ) b. 2 square root(1-x²) c. x d. 2x square root(1-x²) e. x square root(1-x²)
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:
 $\text{Simplify: }\: \sin\left(2\,\sin^{-1}x\right)$ [color=beige]. . [/color]$(a)\;\sqrt{1\,-\,x^2}\;\;\;\;\;(b)\;2\sqrt{1-x^2}\;\;\;\;\;(c)\; x \;\;\;\;\;(d)\; 2x\sqrt{1-x^2}\;\;\;\;\;(e)\;x\sqrt{1-x^2}$

$\text{Let }\,\theta \,=\,\sin^{-1}x\;\;\cdots\;\;\text{then: }\:\sin\theta \:=\:x$[color=beige] .[/color][color=blue][1][/color]

$\text{Since }\,\sin\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{hyp}$
[color=beige]. . [/color]$\theta\text{ is in a right triangle with: }\:opp\,=\,x,\;hyp \,=\,1$
$\text{Pythagoras tells us that: }\,adj= \sqrt{1-x^2}$
[color=beige]. . [/color]$\text{Hence: }\:\cos\theta \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}$[color=beige] .[/color][color=blue][2][/color]

$\text{We have: }\:\sin(2\theta) \;=\;2\,\sin\theta\,\cos\theta$

$\text{Substitute [1] and [2]: }\;\sin(2\theta) \;=\;2x\sqrt{1\,-\,x^2}\;\;\cdots\;\;\text{answer (d)}$

 June 15th, 2010, 06:22 AM #3 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Algebraic Expansion of Sin Excuse me, meant to put "Expression"* Does that change your answer?
 June 15th, 2010, 08:55 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 The term "equivalent expression" would be clearer, but the answer (except for the initial comment) would be the same. Since $\small{\sin^{-1}(x)}$ lies in the interval [-pi/2, pi/2], its cosine is non-negative, and so cos²? + sin²? = 1 (Pythagoras) implies cos? = ?(1 - sin²?) = ?(1 - x²), which leads to answer (d).
 June 15th, 2010, 11:25 AM #5 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Algebraic Expression of Sin Alright thanks

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