Derivation of a formula with trigonometric functions

[j1221]

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

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.

 

[HallsofIvy]

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]

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

go to derivate and click show steps.

 

[j1221]

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.

 

[j1221]

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.

 

[HallsofIvy]

For r a positive integer, 2r+ 1 is odd so, dropping multiples of 2\pi, cos(\pi/4(2r+1) is cos(\pi/4)= \sqrt{2}/2, cos(3\pi/4)= -\sqrt{2}/2, cos(5\pi/4)= \sqrt{2}/2, and cos(7\pi/4)= -\sqrt{2}/2. 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]

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 2\pi, cos(\pi/4(2r+1) is cos(\pi/4)= \sqrt{2}/2, cos(3\pi/4)= -\sqrt{2}/2, cos(5\pi/4)= \sqrt{2}/2, and cos(7\pi/4)= -\sqrt{2}/2. 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]

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

 

[j1221]

Yes!

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

 

[SammyS]

Yes!

Then \displaystyle\cos\left(\frac{\pi}{4}(2r+1)\right)=\cos\left(\frac{\pi}{2}r+\frac{\pi}{4}\right)\,.
Use the angle addition identity for the cosine.