Generalizing Avineil's Trig Identity

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

sin5θ=5sinθ(1sin2θsin2p1)(1sin2θsin2p2) \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 p1 p_1 and p2 p_2 in the range (0,π2] ( 0, \frac{ \pi}{2} ] .

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

sin(2n+1)θ=(2n+1)sinθi=1n(1sin2θsin2pi). \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 ago

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Proposition : k=1n(1sin2(θ)sin2(kπ2n+1))=sin((2n+1)θ)(2n+1)sin(θ) \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 :

k=1nsin2(kπ2n+1)=2n+122n \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}

Proof : Note that,

k=12nsin(kπ2n+1)=k=1nsin(kπ2n+1)k=n+12nsin(kπ2n+1) \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)

=k=1nsin(kπ2n+1)k=1nsin((n+k)π2n+1) \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)

=k=1nsin(kπ2n+1)k=1nsin(kπ2n+1) (k=1nf(k)=k=1nf(n+1k)) \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)

=k=1nsin2(kπ2n+1) \displaystyle = \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right)

But,

k=12nsin(kπ2n+1)=2n+122n \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 )

    k=1nsin2(kπ2n+1)=2n+122n \displaystyle \implies \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}

Now, let P=(2n+1)sin(θ)k=1n(1sin2(θ)sin2(kπ2n+1)) \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)

=(2n+1)sin(θ)k=1n(sin2(kπ2n+1)sin2(θ))k=1nsin2(kπ2n+1) \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) }

=22nsin(θ)k=1n(cos2(θ)cos2(kπ2n+1)) \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)

=22nsin(θ)(k=1n(cos(θ)+cos(kπ2n+1)))(k=1n(cos(θ)cos(kπ2n+1))) \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)

=22nsin(θ)(k=1n(cos(θ)cos((2n+1k)π2n+1)))(k=1n(cos(θ)cos(kπ2n+1))) (cos(πx)=cosx) \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)

=22nsin(θ)(k=1n(cos(θ)cos((n+k)π2n+1)))(k=1n(cos(θ)cos(kπ2n+1))) (k=1nf(k)=k=1nf(n+1k)) \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)

=22nsin(θ)(k=n+12n(cos(θ)cos(kπ2n+1)))(k=1n(cos(θ)cos(kπ2n+1))) \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)

=22nsin(θ)k=12n(cos(θ)cos(kπ2n+1)) \displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^{2n} \left( \cos (\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right)

Also,

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

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

    P=22nsin(θ)22nU2n(cosθ) \displaystyle \implies \text{P} = 2^{2n} \sin(\theta) \cdot 2^{-2n} \cdot U_{2n} (\cos \theta)

=sin((2n+1)θ) (Un(cosθ)=sin((n+1)θ)sinθ) \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 sinxx\dfrac{\sin x}{x}

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

    sinx=(2n+1)sin(x2n+1)k=1n(1sin2(x2n+1)sin2(kπ2n+1)) \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 limn \displaystyle \lim_{n \to \infty} and noting that xx is a constant, we get,

sinxx=k=1(1x2(kπ)2) \displaystyle \dfrac{\sin x}{x} = \prod_{k=1}^{\infty} \left( 1 - \dfrac{x^2}{(k\pi)^{2}}\right)

Ishan Singh - 4 years, 1 month ago

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@Calvin Lin If we consider the square of Un(x)U_{n}(x) instead of U2n(x)U_{2n}(x), we'll get another identity, i.e,

sin2(nθ)=n2sin2(θ)k=1n1(1sin2θsin2kπn)\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 sinxx\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?

Ishan Singh - 4 years, 1 month ago

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@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.

Kartik Sharma - 4 years, 1 month ago

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

Ishan Singh - 4 years, 1 month ago

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@Ishan Singh 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(kπn+1)\cos\left(\frac{k\pi}{n+1}\right) are the roots of Un(x)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!

Kartik Sharma - 4 years, 1 month ago

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@Kartik Sharma Can you tell me which definition of Un(x)U_{n}(x) do you use? If you go by the basic definition, i.e, Un(x)=sin(nsin1(x))U_{n}(x) = \sin( n \sin^{-1} (x) ) or another one i.e, Un(cosθ)=sin((n+1)θ)sinθU_{n} (\cos \theta) = \dfrac{\sin((n+1)\theta)}{\sin \theta} then, it is immediately obvious that the roots of Un(x)U_{n}(x) are cos(kπn+1)\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.

Ishan Singh - 4 years, 1 month ago

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@Ishan Singh 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 rir_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.

Kartik Sharma - 4 years, 1 month ago

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@Kartik Sharma 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(kπn+1)x = \cos\left(\dfrac{k\pi}{n+1}\right) the polynomial is zero, so P(cosθ)P( \cos\theta ) should have roots at θ=kπn+1\theta = \dfrac{k\pi}{n+1}. Then, if you're familiar with trigonometric equations, you will realize that P(cosθ)=sin((n+1)θ)sinθP( \cos\theta ) = \dfrac{\sin ( (n+1) \theta)}{\sin \theta}. Further, expand sin((n+1)θ)\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.

Ishan Singh - 4 years, 1 month ago

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An elegant proof involves substitution of θθ

Substituting θ=mπ2n+1 θ = \dfrac{mπ}{2n+1} we find that sin(2n+1)θ=0sin(2n+1)θ = 0

This means that sin2θsin2pi=1 \dfrac{sin^{2} θ}{sin^{2} p_{i}} = 1 .

Logic naturally dictates that mm must be all the positive integers upto 2n+12n+1

Since θ θ is for the range (0,π2](0, \frac{π}{2}] , m must be upto  n~~ n

Thus, pi=mπ2n+1 p_{i} = \dfrac{mπ}{2n+1} for all m such that 0<mn0 < m ≤ n

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

Let the constant be  a~~a .

Substitute θ=π4n+2 θ = \dfrac{π}{4n+2} .

The expression now turns into -

a=(1)n  m=1ntan2mπ2n+1 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 a = (-1)^{n} ~2n+1

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

@Calvin Lin , Your Thoughts about it ?

Avineil Jain - 5 years ago

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

2) I do not understand how substituting in θ=π4n+2 \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 (-1)^n . For example, with n=1 n = 1 , we have

sin3θ=3sinθ4sin3θ=3sinθ(1sin2θ34)=3sinθ(1sin2θsin2π3) \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.

Calvin Lin Staff - 5 years ago

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Ya, I actually made a mistake. We have to substitute θ=π2 θ = \frac{π}{2} . That actually takes care of (1)n (-1)^{n}.

Substituting θ=π4n+2 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.

Avineil Jain - 5 years ago

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@Avineil Jain Thoughts?

Calvin Lin Staff - 5 years ago

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