derivation of a formula with trigonometric functions


by j1221
Tags: derivation, formula, functions, trigonometric
j1221
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#1
Mar2-12, 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.
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HallsofIvy
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#2
Mar2-12, 01:53 PM
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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.
the_epi
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#3
Mar2-12, 03:16 PM
P: 1
http://www.wolframalpha.com/input/?i=%28-1%29^%28r%28r%2B1%29%2F2%29

go to derivate and click show steps.

j1221
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#4
Mar2-12, 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.

Quote Quote by HallsofIvy View Post
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.
j1221
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#5
Mar2-12, 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.


Quote Quote by the_epi View Post
http://www.wolframalpha.com/input/?i=%28-1%29^%28r%28r%2B1%29%2F2%29

go to derivate and click show steps.
HallsofIvy
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#6
Mar2-12, 05:09 PM
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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.
j1221
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#7
Mar2-12, 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!!


Quote Quote by HallsofIvy View Post
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.
SammyS
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#8
Mar2-12, 06:59 PM
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This formula holds only for r being an integer. Right ?
j1221
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#9
Mar2-12, 07:08 PM
P: 6
Yes!

Quote Quote by SammyS View Post
This formula holds only for r being an integer. Right ?
SammyS
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#10
Mar3-12, 12:35 AM
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Quote Quote by j1221 View Post
Yes!
Then [itex]\displaystyle\cos\left(\frac{\pi}{4}(2r+1)\right)=\cos\left(\frac{\pi}{ 2}r+\frac{\pi}{4}\right)\,.[/itex]
Use the angle addition identity for the cosine.


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