My Math Forum Please clarify the final step of this proof (inflection point)

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 April 14th, 2017, 08:25 PM #1 Member   Joined: Nov 2016 From: USA Posts: 33 Thanks: 1 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 y'' 2cosx - xsinx = 0 xsinx=2cosx (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 (2cosx)^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?
 April 15th, 2017, 05:05 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,978 Thanks: 807 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). Thanks from Seventy7
April 15th, 2017, 06:57 AM   #3
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Joined: Jan 2016
From: Athens, OH

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First comment: I don't like the way you started the proof.
Quote:
 (2cosx)^2(x^2+4)=4x^2
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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