How Many Local Maxima Does F(x) = (sin(Nx)^2)/(sin(x)^2) Have?

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The function F(x) = (sin(Nx)^2)/(sin(x)^2) has exactly N-2 local maxima within the interval 0 < x < π. The analysis reveals that the function possesses N-1 zeros in the same interval, leading to the conclusion that the number of local maxima is derived from the relationship between zeros and extrema. Specifically, with N-1 zeros, there are N-2 local maxima, confirming the initial assertion without the need for derivative calculations.

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1. Show that the function F(x)=(sin(Nx)^2)/(sin(x)^2) has N-2 local maxima in the interval 0<x<pi



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3. I am stuck after i have calculated the derivate, (2Nsin(Nx)cos(Nx)sin(x)^2-2sin(x)cos(x)sin(Nx)^2)/sin(x)^4 = 0

I am not sure how to simplify this equation, so far I have found 2N-1 local maximum and minimum, which is not correct. Please give me some hints.
 
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Try to plot the original function for small N-s. How many zeros has the original function in the interval (0, pi)?
What are the minimal values of the function? What are limits at x=0 and at x=pi?

ehild
 
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The original function has N-1 zeros on the interval (0,pi).
The minimal values for the function is 0. And F->N^2 as x->0 and x->pi
Wich means, if there are (N-1) zeros then there is (N-1)-1 = N-2 maximum.
So I don't have to calculate the derivative.

Thanks for the help ^^
 
You have found out the solution earlier than me:smile:

ehild
 

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