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Derivation of a formula with trigonometric functions 
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#1
Mar212, 12:46 PM

P: 6

Hi everyone,
1. The problem statement, all variables and given/known data My problem is just to derive a simple formula, which is Here r is a positive integer. 3. The attempt at a solution I verified this formula by inserting r=4k ~ 4k+3 (k=0,1,2....), but I still have no idea how to derive it from the left hand side of the equation. Could anyone please help me out? Any help is appreciated. 


#2
Mar212, 01:53 PM

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P: 39,534

I don't know what you mean by "r=4k ~ 4k+3" but the equation is clearly NOT true for n= 0, 1, 2, etc.



#3
Mar212, 03:16 PM

P: 1

http://www.wolframalpha.com/input/?i=%281%29^%28r%28r%2B1%29%2F2%29
go to derivate and click show steps. 


#4
Mar212, 03:30 PM

P: 6

Derivation of a formula with trigonometric functions
Hello HallsofIvy,
Thank you very much for pointing out my mistake. I typed the wrong formula. I have corrected it. Would you please check it out again? Thank you again. 


#5
Mar212, 03:42 PM

P: 6

Hello the_epi,
Thanks for your help. But I checked the website and check the Derivative part, I still do not understand how the Derivative related to the formula above. Could you please explain? Thanks a lot. 


#6
Mar212, 05:09 PM

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PF Gold
P: 39,534

For r a positive integer, 2r+ 1 is odd so, dropping multiples of [itex]2\pi[/itex], [itex]cos(\pi/4(2r+1)[/itex] is [itex]cos(\pi/4)= \sqrt{2}/2[/itex], [itex]cos(3\pi/4)= \sqrt{2}/2[/itex], [itex]cos(5\pi/4)= \sqrt{2}/2[/itex], and [itex]cos(7\pi/4)= \sqrt{2}/2[/itex]. So what does the left side give? I would look at r= 4n, 4n+1, 4n+2, and 4n+ 3 and compare to those values.



#7
Mar212, 05:40 PM

P: 6

Thank you very much HallsofIvy. I did the same thing to check this equation.
But I do not know how to DERIVE it. Do you have any ideas? Thanks!! 


#8
Mar212, 06:59 PM

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This formula holds only for r being an integer. Right ?



#9
Mar212, 07:08 PM

P: 6

Yes!



#10
Mar312, 12:35 AM

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Use the angle addition identity for the cosine. 


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