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 September 2nd, 2011, 08:13 AM #1 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Simplify the Expression Simplify the expression y = sin (tan^?1 x/?7)  by writing it in algebraic form Can someone explain this to me?
 September 2nd, 2011, 08:42 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Simplify the Expression [color=#000000]If you mean $\sin\left(\arctan\left(\frac{x}{\sqrt{7}}\right)\r ight)=\frac{\frac{x}{\sqrt{7}}}{\sqrt{\left(\frac{ x}{\sqrt{7}}\right)^{2}+1}}=\frac{x}{\sqrt{7}\sqrt {\left(\frac{x^2}{7}\right)+1}}$.[/color]
September 2nd, 2011, 10:00 AM   #3
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Re: Simplify the Expression

Hello, jaredbeach!

Quote:
 Simplify:[color=beige] .[/color]$y \:=\: \sin\left[\tan^{-1}\left(\frac{x}{\sqrt{7}}\right)\right]$

$\text{Let }\,\theta \:=\:\tan^{-1}\left(\frac{x}{\sqrt{7}}\right) \;\;\;\Rightarrow\;\;\;\tan\theta \:=\:\frac{x}{\sqrt{7}} \:=\:\frac{opp}{adj}$

$\text{Hence, }\theta\text{ is in a right triangle with: }\:opp\,=\,x,\;\;adj\,=\,\sqrt{7}$

$\text{Pythagoras says: }\:hyp \;=\;\sqrt{x^2\,+\,7}$

$\text{Therefore: }\:\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{x}{\sqrt{x^2\,+\,7}}$

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