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\({sin(2x)-cos(x) \over 4sin^2x - 1} = {sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\)

 

It looks like the denominator for the right expression is a perfect square... but after that, I'm not sure what else to do!

 Feb 21, 2019
 #1
avatar+105634 
+1

Lets see..

 

\({sin(2x)-cos(x) \over 4sin^2x - 1} = {sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\\ LHS\\ ={sin(2x)-cos(x) \over 4sin^2x - 1} \\ =\frac{2sinxcosx-cos(x) } {(2sinx - 1)(2sinx+1)} \\ =\frac{cosx(2sinx-1) } {(2sinx - 1)(2sinx+1)} \\ =\frac{cosx } {2sinx+1} \\ \)

 

 

\(RHS\\ ={sin^2 x (cos x) + cos^2x \over 2 sin(x)+1}\\ \ne LHS\)

 

 

I suspect this is an error on our part on on the part of your teacher or text.

 Feb 21, 2019
edited by Melody  Feb 21, 2019
 #2
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+1

Ah, I see, I believe it's an error since the textbook's answer says it's equal. You're probably right.

Guest Feb 21, 2019
 #3
avatar+105634 
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Here are the graphs.

They are clearly different.

 

 Feb 21, 2019

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