1

Is my conjecture below true?

If $\alpha, \beta, \gamma$ are the angles of a triangle, then $$\sin n\alpha + \sin n\beta + \sin n\gamma = \pm 4\cos\frac {n\alpha}2\cos\frac{n\beta}2\cos\frac{n\gamma}2,~n\text{ is odd}$$ $$\sin n\alpha + \sin n\beta + \sin n\gamma = \pm 4\sin\frac {n\alpha}2\sin\frac{n\beta}2\sin\frac{n\gamma}2,~n\text{ is even}$$

I've seen the formulas for $n=1$ and $n=2$ pop around here and there. Then recently someone showed a textbook print with the formula for the case $n=3$ (from which book, I have no idea) and now I wonder if this general case holds.

Particular cases have already been asked and answered here on this site. For $n = 1$ a proof of the formula may be found, for instance, here or here. For $n=2$, you may look here. For $n=3$ the expression is equivalent to the one mentioned in this post.

FD_bfa
  • 4,757
Alma Arjuna
  • 6,521
  • This is probably easily shown by angle-addition. Mathematica suggests the correct sign is $+$ if $n\equiv 1,2\bmod 4$ and $-$ if $n\equiv 0,3\bmod 4$. – Integrand Sep 13 '24 at 12:03
  • 1
    The conjecture is true. Let $u,v,w$ be $e^{i\alpha}$, $e^{i\beta}$, $e^{i\gamma}$. Condition $\alpha + \beta + \gamma = \pi$ is equivalent to $uvw = -1$. LHS of your equations equals to $$\Im(u^n + v^n + w^n) = \Im(u^n + v^n + \epsilon (uv)^n) = \epsilon \Im((u^n+\epsilon)(v^n+\epsilon))$$ where $\epsilon = (-1)^{n-1}$. Reexpress it using sine/cosine leads to what you suspect. – achille hui Sep 13 '24 at 13:14

1 Answers1

3

Yes; one way to prove it starts with the following lemma:

Lemma: For complex numbers $x, y, z$ we have $$ \overline {xyz} + \overline xyz + x\overline yz + xy\overline z = 4\mathfrak{Re}(x)\mathfrak{Re}(y)\mathfrak{Re}(z) + 4\mathfrak{Im}(x)\mathfrak{Im}(y)\mathfrak{Im}(z)i.$$

I leave you to prove the lemma (can be done algebraically by expanding each complex number into real and imaginary parts and distributing).

We now choose: $x = e^{n\alpha i/2} = \cos\left(\frac{n\alpha}{2}\right) + \sin\left(\frac{n\alpha}{2}\right)i$, $y = e^{n\beta i/2}= \cos\left(\frac{n\beta}{2}\right) + \sin\left(\frac{n\beta}{2}\right)i$,$ z = e^{n\gamma i/2}= \cos\left(\frac{n\gamma}{2}\right) + \sin\left(\frac{n\gamma}{2}\right)i$ and calculate the following product:

$$xyz = e^{n\alpha i/2}e^{n\beta i/2}e^{n\gamma i/2} = e^{n(\alpha+\beta+\gamma)i/2} = e^{n\pi i/2} = \left(e^{\pi i/2}\right)^n=i^n = \begin{cases}1 & \text{ if } n\equiv 0 \mod 4, \\ i & \text{ if } n\equiv 1 \mod 4, \\ -1 & \text{ if } n\equiv 2 \mod 4, \\ -i & \text{ if } n\equiv 3 \mod 4. \\ \end{cases}$$

We can also calculate $$\overline xyz = e^{-n\alpha i/2}e^{n\beta i/2}e^{n\gamma i/2} = e^{n(\beta+\gamma-\alpha)i/2} = e^{n(\pi-2\alpha)i/2} = e^{(n\pi/2 - n\alpha) i}$$ where

$$e^{(n\pi/2 - n\alpha)i} = \begin{cases}\cos(n\alpha)-\sin(n\alpha)i & \text{ if } n\equiv 0 \mod 4, \\ \sin(n\alpha)+\cos(n\alpha)i & \text{ if } n\equiv 1 \mod 4, \\ -\cos(n\alpha)+\sin(n\alpha)i & \text{ if } n\equiv 2 \mod 4, \\ -\sin(n\alpha)-\cos(n\alpha)i & \text{ if } n\equiv 3 \mod 4. \\ \end{cases},$$ and symmetric results hold for the other two terms. Combining these calculations yields, in the case that $n$ is even (0 or 2 $\mod 4$): $$\left[1 + \cos(n\alpha) + \cos(n\beta) + \cos(n\gamma)\right]-\left[ \sin(n\alpha) + \sin(n\beta) + \sin(n\gamma)\right]i = 4(-1)^{n/2}\left(\cos\left(\frac{n\alpha}{2}\right)\cos\left(\frac{n\beta}{2}\right)\cos\left(\frac{n\gamma}{2}\right) + \sin\left(\frac{n\alpha}{2}\right)\sin\left(\frac{n\beta}{2}\right)\sin\left(\frac{n\gamma}{2}\right)i\right].$$ Matching imaginary parts on both sides yields your identity. Similarly, in the case that $n$ is odd we have $$\left[ \sin(n\alpha) + \sin(n\beta) + \sin(n\gamma)\right] + \left[-1 + \cos(n\alpha) + \cos(n\beta) + \cos(n\gamma)\right]i = 4(-1)^{(n-1)/2}\left(\cos\left(\frac{n\alpha}{2}\right)\cos\left(\frac{n\beta}{2}\right)\cos\left(\frac{n\gamma}{2}\right) + \sin\left(\frac{n\alpha}{2}\right)\sin\left(\frac{n\beta}{2}\right)\sin\left(\frac{n\gamma}{2}\right)i\right]$$ and matching the real parts on both sides yields your identity.

Andrew
  • 1,853