Proving the 'boundary' of a function is equal to cosec(x).

[jameslouiskillian]

Hey guys,
For my high school maths folio my teacher has given us an open-ended investigation where we are tasked to investigate the patterns and features of sinx+sin3x+sin5x+...+sin((2n-1)x). I have managed to turn this function into a fraction form and by using Geogebra, I have noticed that the 'boundary' of this function is equal to cosec(x). This will make more sense if you look at this diagram:

< link to objectionable page removed >

How would I go about proving this algebraically?

Some things to note and that may help:
- My teacher said I am not allowed to use calculus.
- When n = 1, the function is simply sinx.
- My teacher said that the proof involves something that he calls the 'boundary function'. Although he stressed that this is simply a made up name he has given to it.

Thank you very much for your help!

 

[ksdhart2]

Your post is a little bit confusing to me, but what I think you're asking is how to prove that the function you were given has an upper bound of csc(x) for x values between 0 and pi. To phrase it another way:

\(\displaystyle \displaystyle \sum _{n=1}^{m}\:sin\left(\left[2n-1\right]x\right)\le csc\left(x\right)\) for \(\displaystyle x \in \left[0,\pi\right]\) and \(\displaystyle m\in \mathbb{Z},\:m\ge 1\)

Now, you've shown that the sum is equivalent to:

\(\displaystyle \displaystyle \frac{1-cos\left(2mx\right)}{2sin\left(x\right)}\)

From here, I'd use what my textbook called the half-angle formula:

\(\displaystyle \displaystyle sin^2\left(x\right)=\frac{1-cos\left(2x\right)}{2}\)

That will make it so the two fractions you're comparing have the same denominator. And from there, proving that one is larger than the other over a given interval shouldn't be too difficult at all.