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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(1x² ) b. 2 square root(1x²) c. x d. 2x square root(1x²) e. x square root(1x²) 
June 15th, 2010, 06:17 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Algebraic Expansion of Sin Hello, RMG46! I don't understand what you mean by "expansion" . . . Quote:
[color=beige] .[/color][color=blue][1][/color] [color=beige]. . [/color] [color=beige]. . [/color][color=beige] .[/color][color=blue][2][/color]  
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 lies in the interval [pi/2, pi/2], its cosine is nonnegative, 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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