# derivation of a formula with trigonometric functions

by j1221
Tags: derivation, formula, functions, trigonometric
 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.
 PF Patron Sci Advisor Thanks Emeritus P: 38,416 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.
 P: 1 http://www.wolframalpha.com/input/?i=%28-1%29^%28r%28r%2B1%29%2F2%29 go to derivate and click show steps.
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 by 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.
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 by the_epi http://www.wolframalpha.com/input/?i=%28-1%29^%28r%28r%2B1%29%2F2%29 go to derivate and click show steps.
 PF Patron Sci Advisor Thanks Emeritus P: 38,416 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.
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 by 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.
 PF Patron HW Helper Sci Advisor Emeritus P: 7,077 This formula holds only for r being an integer. Right ?
P: 6
Yes!

 Quote by SammyS This formula holds only for r being an integer. Right ?
PF Patron
HW Helper
Then $\displaystyle\cos\left(\frac{\pi}{4}(2r+1)\right)=\cos\left(\frac{\pi}{ 2}r+\frac{\pi}{4}\right)\,.$