What Values Make f(23,x) Equal to f(33,x) for 100<x<200 Degrees?

AI Thread Summary
The discussion revolves around finding values of x for which the function f(23,x) equals f(33,x) within the range of 100 to 200 degrees. Participants express uncertainty about how to approach the problem, primarily considering plugging in values and testing cases. There is mention of using De Moivre's theorem but without success in finding a solution. The conversation highlights confusion over the relationship between the sums of sine and cosine functions and their tangents. Overall, the thread seeks assistance in solving the equation effectively.
newchie
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Homework Statement


consider the function
f(n,x) = (sinx+sin2x+sin3x +...+sin(n-1)x+sinnx)/)cosx+cos2x+cos3x+...+cos(n-1)x+cosnx)
Find the sum of all values for which f(23,x) = f(33,x)where x is measured in degrees from 100<x<200



Homework Equations


you get obviously tanx+tan2x...
not sure where to go other than pluggin in values


The Attempt at a Solution


Plugging in values, testing cases trying different methods including demoivres, but doesent help. Help is appreciated!
 
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newchie said:

Homework Statement


consider the function
f(n,x) = (sinx+sin2x+sin3x +...+sin(n-1)x+sinnx)/(cosx+cos2x+cos3x+...+cos(n-1)x+cosnx)
Find the sum of all values for which f(23,x) = f(33,x)where x is measured in degrees from 100<x<200

Homework Equations


you get obviously tanx+tan2x...
not sure where to go other than pluggin in values

The Attempt at a Solution


Plugging in values, testing cases trying different methods including demoivres, but doesent help. Help is appreciated!
You do not get tan(x)+tan(2x)+ ...

\displaystyle \frac{a+b+c}{e+f+g}\ne\frac{a}{e}+\frac{b}{f}+ \frac{c}{g}\ \ \
 
SammyS said:
You do not get tan(x)+tan(2x)+ ...

\displaystyle \frac{a+b+c}{e+f+g}\ne\frac{a}{e}+\frac{b}{f}+ \frac{c}{g}\ \ \

ah yes woops, i misread hehe, anyways do you have an approach on this problem
 
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