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April 14th, 2017, 08:25 PM   #1
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Please clarify the final step of this proof (inflection point)

Here is the solution to proving the inflection point of y=xsinx lies on the curve y^2(x^2+4)=4x^2

2cosx - xsinx = 0

(hereafter all are x's are x sub zero)
inflection point of the curve (x, 2cos x)
Prove (x,2cosx) lies on the curve y^2(x^2+4)=4x^2

4cos^2x[(4 cos^2x/sin^2x) +4] = 4[(2cosx)/sinx]^2
16 (cos^2x/sin^2x) = 16 (cos^2x/sin^2x)

How does the first bolded equation equal the second bolded equation?
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April 15th, 2017, 05:05 AM   #2
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4cos^2(x)/sin^2(x)+ 4= 4((cos^2(x)/sin^2(x))+ 1)= 4(cos^2(x)sin^2(x)+ sin^2(x)/sin^2(x))= 4((cos^2(x)+ sin^2(x))/sin^2(x))= 4/sin^2(x).

Multiplying that by 4 cos^2(x) gives 16 cos^2(x)/sin^2(x).
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Country Boy is offline  
April 15th, 2017, 06:57 AM   #3
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First comment: I don't like the way you started the proof.
This is really starting with what you want to prove. You should start with the left side of the equation and derive the right side as follows:

The following graphs show the result and also the fact that there are infinitely many inflection points. But also notice not every intersection of the two graphs is an inflection point.

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