# Generalizing Avineil's Trig Identity

In the question Too complex shared by Avineil Jain, he displayed the identity

$\sin 5 \theta = 5 \sin \theta ( 1 - \frac{ \sin ^2 \theta } { \sin ^ 2 p_1 } ) ( 1 - \frac { \sin ^ 2 \theta } { \sin ^ 2 p_2 } )$

for some constants $p_1$ and $p_2$ in the range $( 0, \frac{ \pi}{2} ]$.

I was intrigued by this identity, and did a bit of further investigating. I believe that for all $n$, we have the identity

$\sin ( 2n + 1) \theta = ( 2n + 1 ) \sin \theta \prod_{i = 1 } ^ n \left ( 1 - \frac{ \sin ^2 \theta } { \sin ^2 p_i } \right) .$

Can you figure out what this identity is?

Can you prove it? Note by Calvin Lin
5 years, 11 months ago

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Proposition : $\prod_{k=1}^n \left(1 -\dfrac{\sin^2(\theta)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right) = \dfrac{\sin((2n+1)\theta)}{(2n+1)\sin(\theta)}$

Proof : First, consider the following Lemma,

Lemma :

$\prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}$

Proof : Note that,

$\displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=n+1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right)$

$\displaystyle = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\dfrac{(n+k)\pi}{2n+1}\right)$

$\displaystyle = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \ \left(\because \prod_{k=1}^{n} f(k) = \prod_{k=1}^{n} f(n+1-k) \right)$

$\displaystyle = \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right)$

But,

$\displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}$ (For my proof of this, see here )

$\displaystyle \implies \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}$

Now, let $\displaystyle \text{P} = (2n+1)\sin(\theta) \prod_{k=1}^n \left(1 -\dfrac{\sin^2(\theta)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right)$

$\displaystyle = (2n+1)\sin(\theta) \dfrac{ \displaystyle \prod_{k=1}^n \left( \sin^2\left(\frac{k\pi}{2n+1}\right) - \sin^2(\theta) \right)}{ \displaystyle \prod_{k=1}^{n} \sin^2\left(\frac{k\pi}{2n+1}\right) }$

$\displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^n \left( \cos^2(\theta) - \cos^2\left(\frac{k\pi}{2n+1}\right) \right)$ (Using the Lemma)

$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) + \cos \left(\frac{k\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right)$

$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(2n+1 - k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \ \left(\because \cos (\pi -x) = -\cos x \right)$

$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(n + k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \ \left(\because \prod_{k=1}^{n} f(k) = \prod_{k=1}^{n} f(n+1-k) \right)$

$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=n+1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{ k \pi}{2n+1}\right) \right) \right)$

$\displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right)$

Also,

$\displaystyle U_{n} (x) = 2^{n} \prod_{k=1}^{n} \left(x - \cos \left(\frac{k\pi}{n+1}\right) \right)$

where $\displaystyle U_{n} (x)$ denotes the Chebyshev Polynomial of the Second kind.

$\displaystyle \implies \text{P} = 2^{2n} \sin(\theta) \cdot 2^{-2n} \cdot U_{2n} (\cos \theta)$

$\displaystyle = \sin ((2n+1) \theta) \ \left(\because U_{n} (\cos \theta) = \dfrac{\sin ((n+1) \theta)}{\sin \theta} \right)$

Bonus : We can also use this identity to prove Euler's infinite product for $\dfrac{\sin x}{x}$

In the above proposition, set $(2n+1)\theta = x$ such that $x$ is a constant.

$\displaystyle \implies \sin x = (2n+1)\sin \left( \frac{x}{2n+1} \right) \prod_{k=1}^n \left(1 -\dfrac{\sin^2 \left( \frac{x}{2n+1} \right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right)$

Taking $\displaystyle \lim_{n \to \infty}$ and noting that $x$ is a constant, we get,

$\displaystyle \dfrac{\sin x}{x} = \prod_{k=1}^{\infty} \left( 1 - \dfrac{x^2}{(k\pi)^{2}}\right)$

- 5 years ago

@Calvin Lin If we consider the square of $U_{n}(x)$ instead of $U_{2n}(x)$, we'll get another identity, i.e,

$\sin^2(n\theta) = n^2 \sin^2(\theta) \prod_{k=1}^{n-1}\left(1 - \dfrac{\sin^2 \theta }{\sin^2 \frac{k\pi}{n}}\right)$

But, if we try to prove Euler's infinite product of $\dfrac{\sin x}{x}$, we are getting an incorrect result. Can you kindly help? Also, do you have another way to prove the identity? If so, can you please share that?

- 5 years ago

@Ishan Singh Can you give me a link from where you got that product resulting in the Chebyshev Polynomial ? I can't find it anywhere on any of the wikis. Please tell me if there exist such product for other kind.

- 4 years, 11 months ago

$\left\{\cos\left(\dfrac{k\pi}{n+1}\right)\right\}_{k=1}^{n}$ are the roots of $U_{n} (x)$ and since we can write any polynomial as its leading coefficient times $\displaystyle \prod \ (x - \text{roots})$, hence the factorization. As for the leading coefficient of $U_{n} (x)$ and the fact that its roots are $\left\{\cos\left(\dfrac{k\pi}{n+1}\right)\right\}_{k=1}^{n}$, you can prove it via induction.

- 4 years, 11 months ago

Yeah, I know about the Weirestrass and all but I was just asking how to guess. I mean the first step - how can one guess that $\cos\left(\frac{k\pi}{n+1}\right)$ are the roots of $U_n(x)$. Of course, we can prove by induction but where will one get the statement.

So, the only way is to look, prove and remember. And I am asking where to look?

BTW, I am sorry to bother you. I will try to look it up somewhere on internet. Thanks!

- 4 years, 11 months ago

Can you tell me which definition of $U_{n}(x)$ do you use? If you go by the basic definition, i.e, $U_{n}(x) = \sin( n \sin^{-1} (x) )$ or another one i.e, $U_{n} (\cos \theta) = \dfrac{\sin((n+1)\theta)}{\sin \theta}$ then, it is immediately obvious that the roots of $U_{n}(x)$ are $\cos \left( \dfrac{k\pi}{n+1} \right)$ you don't even need induction. As for the link, I used this to study about some basic properties of chebyshev polynomials. Rest I derived myself.

- 4 years, 11 months ago

It's not about how to prove that the roots are such. It is all about how you would guess the function known the roots? Okay I would like to find a Weirestrass definition with roots $r_i$, it's actually that one needs to know the result at hand.

BTW, I will contradict myself on that as I also believe that that is the whole aim of mathematics - guessing the statement from the result. And, it is all observation about how to make the function outta roots.

- 4 years, 11 months ago

Oh, so you meant how to find the polynomial when roots are given. That's also quite easy. First you see that at $x = \cos\left(\dfrac{k\pi}{n+1}\right)$ the polynomial is zero, so $P( \cos\theta )$ should have roots at $\theta = \dfrac{k\pi}{n+1}$. Then, if you're familiar with trigonometric equations, you will realize that $P( \cos\theta ) = \dfrac{\sin ( (n+1) \theta)}{\sin \theta}$. Further, expand $\sin ((n+1) \theta)$ using Euler's identity and prove every other property of chebyshev polynomial of the second kind, without induction or 'guessing' any result. Similarly for the first kind.

Also, I differ from the statement that the aim of mathematics is guessing the statement from the result and just observations. It's about deductive reasoning, exploring and discovering results and identities from basic theorems. It's not about using maths as a tool or some bag of tricks that you use in a situation, but doesn't work in some other situation. Every branch in maths is interlinked and interdependent. It's all about the mathematical spirit.

- 4 years, 11 months ago

An elegant proof involves substitution of $θ$

Substituting $θ = \dfrac{mπ}{2n+1}$ we find that $sin(2n+1)θ = 0$

This means that $\dfrac{sin^{2} θ}{sin^{2} p_{i}} = 1$.

Logic naturally dictates that $m$ must be all the positive integers upto $2n+1$

Since $θ$ is for the range $(0, \frac{π}{2}]$ , m must be upto$~~ n$

Thus, $p_{i} = \dfrac{mπ}{2n+1}$ for all m such that $0 < m ≤ n$

As far as the constant multiplied to the expression is concerned, there is another neat trick involved -

Let the constant be$~~a$ .

Substitute $θ = \dfrac{π}{4n+2}$.

The expression now turns into -

$a = (-1)^{n}~~\displaystyle\prod_{m=1}^n tan^{2} \dfrac{mπ}{2n+1}$ which is a standard identity for tangent.

Thus, $a = (-1)^{n} ~2n+1$

The only Problem I am facing is that there is a $(-1)^{n}$ popping up in the solution.

- 5 years, 11 months ago

1) Good way to justify what the values of $p$ must be, assuming that the functional form is correct.

2) I do not understand how substituting in $\theta = \frac{ \pi } { 4n+2}$ leads to the result. Can you explain in more detail?

3) I am quite certain that there is no $(-1)^n$. For example, with $n = 1$, we have

$\sin 3 \theta = 3 \sin \theta - 4 \sin ^3 \theta = 3 \sin \theta ( 1 - \frac{ \sin^2 \theta } { \frac{3}{4} } ) = 3 \sin \theta ( 1 - \frac{ \sin ^ 2 \theta } { \sin ^2 \frac{ \pi } { 3} })$

4) After finding the possible functional form, you still have to prove that it is correct.

Staff - 5 years, 11 months ago

Ya, I actually made a mistake. We have to substitute $θ = \frac{π}{2}$ . That actually takes care of $(-1)^{n}$.

Substituting $θ = \dfrac{π}{4n+2}$ is a mistake. My bad. And I still have to think of a good way to actually prove

that this form is correct.

- 5 years, 11 months ago

@Avineil Jain Thoughts?

Staff - 5 years, 11 months ago