Derivation of a formula with trigonometric functions

  • Thread starter j1221
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  • #1
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Hi everyone,

Homework Statement



My problem is just to derive a simple formula, which is

http://www.texify.com/img/%5Cnormalsize%5C%21%28-1%29%5E%7Br%28r%2B1%29/2%7D%20%3D%5Csqrt%7B2%7D%20%5Cmbox%7Bcos%7D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%282r%2B1%29.gif [Broken]

Here r is a positive integer.

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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Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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http://www.wolframalpha.com/input/?i=%28-1%29^%28r%28r%2B1%29%2F2%29

go to derivate and click show steps.
 
  • #4
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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.

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.
 
  • #5
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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.


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

go to derivate and click show steps.
 
  • #6
HallsofIvy
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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.
 
  • #7
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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!!


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.
 
  • #8
SammyS
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This formula holds only for r being an integer. Right ?
 
  • #9
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Yes!

This formula holds only for r being an integer. Right ?
 
  • #10
SammyS
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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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