Triple Triple Angle Identity

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.

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

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

Using this for $\sin (9x)$ and $\sin (3x)$ , we get,

$\displaystyle p + q + r = \dfrac{7\pi}{9}$

$\implies A + B = \boxed{16}$

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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,

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

Solution in portuguese:

Fazendo $f(x) = (1 - \frac{\sin^{2} (x)}{sin^{2} (p)}).(1 - \frac{\sin^{2} (x)}{sin^{2} (q)}).(1 - \frac{\sin^{2} (x)}{sin^{2} (r)})$ , temos $f(p) = f(q) = f(r) = 0$ . Pela identidade, podemos escrever:

$f(x) = \frac{\sin (9x)}{3 \sin (3x)}$ .

Em que $sen(3x) ≠ 0$ , ou seja, $x ≠ \frac{kπ}{3}$ . $(I)$

Fazendo $f(x) = 0$ , encontramos como raízes dentro da condição anterior $(I)$ $x = \frac{kπ}{9}$ , onde k não é múltiplo de 3. Desta forma, p, q e r são também escritos nesse formato, mas obedecendo o fato de que $p, q, r ∈ [0, π/2[$ . Sendo assim, temos $\frac{π}{9}, \frac{2π}{9}, \frac{3π}{9} e \frac{4π}{9}$ como possíveis soluções, mas pela condição $(I)$ , temos $\frac{3π}{9}$ como uma solução inválida, restando $(p, q, r) = (\frac{π}{9}, \frac{2π}{9}, \frac{4π}{9})$ , mas não necessariamente nesta ordem. Finalmente,

$p + q + r = \frac{7π}{9}$ .

Logo $A = 7$ e $B = 9$ , resultando $A + B = 7 + 9 = 16$ .

I don't know if this solution is 100% correct, also there is a much faster way to only get the answer, but here it goes the process of the derivation of the identity if we don't know its form:

First let's solve $\dfrac{\sin(9x)}{3\sin(3x)}=0$ . Clearly $\sin(9x)=0$ produces the solutions $x=40^\circ k$ and $x=20^\circ+40^\circ k$ , and $3\sin(3x)=0$ produces $x=120^\circ k$ and $x=60^\circ+120^\circ k$ . We need to exclude these solutions, because they make the denominator $0$ .

So, the solutions are: $x=40^\circ k$ and $x=20^\circ+40^\circ k$ where $x$ is not a multiple of $3^\circ$ .

The next step is to find all the different sines that we can make with all the solutions, and they are: $\sin(20^\circ)$ , $\sin(40^\circ)$ , $\sin(80^\circ)$ , $\sin(200^\circ)$ , $\sin(220^\circ$ ) and $\sin(260^\circ)$ .

Hence, we can factorize the initial expression in this way:

$\dfrac{\sin(9x)}{3\sin(3x)}=a(\sin(x)-\sin(20^\circ))(\sin(x)-\sin(40^\circ))(\sin(x)-\sin(80^\circ))(\sin(x)-\sin(200^\circ))(\sin(x)-\sin(220^\circ))(\sin(x)-\sin(260^\circ))$

Note that $\sin(200^\circ)=-\sin(20^\circ)$ , $\sin(220^\circ)=-\sin(40^\circ)$ and $\sin(260^\circ)=-\sin(80^\circ)$ , hence:

$\dfrac{\sin(9x)}{3\sin(3x)}=a(\sin^2(x)-\sin^2(20^\circ))(\sin^2(x)-\sin^2(40^\circ))(\sin^2(x)-\sin^2(80^\circ))$

$\dfrac{\sin(9x)}{3\sin(3x)}=-a\sin^2(20^\circ)\sin^2(40^\circ)\sin^2(80^\circ)\left(1-\dfrac{\sin^2(x)}{\sin^2(20^\circ)}\right)\left(1-\dfrac{\sin^2(x)}{\sin^2(40^\circ)}\right)\left(1-\dfrac{\sin^2(x)}{\sin^2(80^\circ)}\right)$

Now, note that $\sin(60^\circ)=3t-4t^3$ has solutions $t=\sin(20^\circ),\sin(40^\circ),-\sin(80^\circ)$ . If we let $y=t^2$ , then $t=\sqrt{y}$ and:

$\sin(60^\circ)=3\sqrt{y}-4y\sqrt{y} \implies \sin(60^\circ)=\sqrt{y}(3-4y) \implies \sin^2(60^\circ)=y(3-4y)^2$

After simplifying we get: $64y^3-96y^2+36y-3=0$ , which has roots $y=\sin^2(20^\circ),\sin^2(40^\circ),\sin^2(80^\circ)$ .

Hence, by the factor theorem: $64y^3-96y^2+36y-3=64(y-\sin^2(20^\circ))(y-\sin^2(40^\circ))(y-\sin^2(80^\circ))$

Now, to find $a$ let $x=30^\circ$ :

$-\dfrac{1}{3}=a(1/4-\sin^2(20^\circ)(1/4-\sin^2(40^\circ)(1/4-\sin^2(80^\circ))$

$-\dfrac{1}{3}=a\dfrac{64(1/4)^3-96(1/4)^2+36(1/4)-3}{64}=\dfrac{a}{64}$

Hence, $a=-\dfrac{64}{3}$ . By Vieta's formulas, $\sin^2(20^\circ)\sin^2(40^\circ)\sin^2(80^\circ)=\dfrac{3}{64}$

Finally, our identity is:

$\dfrac{\sin(9x)}{3\sin(3x)}=\left(1-\dfrac{\sin^2(x)}{\sin^2(20^\circ)}\right)\left(1-\dfrac{\sin^2(x)}{\sin^2(40^\circ)}\right)\left(1-\dfrac{\sin^2(x)}{\sin^2(80^\circ)}\right)$

On comparing we get $p=\dfrac{\pi}{9}$ , $q=\dfrac{2\pi}{9}$ and $r=\dfrac{4\pi}{9}$ . So, $p+q+r=\dfrac{7\pi}{9}$ , and $A+B=\boxed{16}$ .