# Derivation of a formula with trigonometric functions

• j1221
In summary, the conversation discusses a formula involving r being a positive integer and the attempt to derive it. The formula is verified for certain values of r and the left side is compared to cosines of different angles. The formula only holds for r being an integer and the angle addition identity for cosine is suggested for deriving it.
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.

Last edited by a moderator:
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.

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

go to derivate and click show steps.

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.

HallsofIvy said:
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.

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.

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

go to derivate and click show steps.

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.

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!

HallsofIvy said:
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.

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

Yes!

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

j1221 said:
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.

## 1. What is the purpose of deriving a formula with trigonometric functions?

The purpose of deriving a formula with trigonometric functions is to find a mathematical relationship between different trigonometric functions. This can help in simplifying complex expressions and solving problems in various fields such as physics, engineering, and mathematics.

## 2. What are the basic trigonometric functions used in deriving a formula?

The basic trigonometric functions used in deriving a formula are sine, cosine, and tangent. These functions are ratios of sides of a right triangle and are defined as opposite/hypotenuse, adjacent/hypotenuse, and opposite/adjacent respectively.

## 3. How is the process of deriving a formula with trigonometric functions carried out?

The process of deriving a formula with trigonometric functions involves using mathematical principles such as the Pythagorean theorem, trigonometric identities, and algebraic manipulation to simplify the expression. This is done step by step until the final formula is obtained.

## 4. Can derived formulas with trigonometric functions be used in real-life applications?

Yes, derived formulas with trigonometric functions have many real-life applications. They can be used to solve problems in fields such as navigation, surveying, and astronomy. They are also used in creating mathematical models for various physical phenomena.

## 5. Are there any limitations to deriving formulas with trigonometric functions?

Yes, there are limitations to deriving formulas with trigonometric functions. The process can become very complex for certain expressions, and it may not be possible to find a simplified formula. It is also important to note that the derived formula may only be applicable in certain conditions and may not be accurate in all scenarios.

Replies
2
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
Replies
16
Views
7K
Replies
6
Views
3K
Replies
3
Views
2K
Replies
2
Views
3K