# Show that $\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$

If $\sin\alpha+\sin\beta+\sin\gamma=\cos\alpha+\cos\beta+\cos\gamma=0$, then show that $$\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$$

I am confused how I should start. Hint will do.

thanks in advance!

• What are you summing over? – User0112358 Sep 3 '16 at 6:19
• Perhaps it's the sum over the three variables α, β, γ ? – user326210 Sep 3 '16 at 6:40

## 4 Answers

Let $z_1 = \cos\alpha + i\sin\alpha, z_2 = \cos \beta + i\sin \beta, z_3 = \cos\gamma + i\sin\gamma$. Given that $z_1+z_2+z_3 = 0$. Since the orthocenter of the triangle with vertices $z_1, z_2, z_3$ is $z_1+z_2+z_3$ it follows that for this triangle, orthocenter and circum center coincide and hence the triangle is equilateral. Thus the vertices are of the form $z_1, z_1\omega, z_1\omega^2$, where $\omega$ is a cube root of unity. Thus $\beta = \alpha + \frac{2\pi}{3}$ and $\gamma = \alpha + \frac{4\pi}{3}$. Thus \begin{align*} z_1^2+z_2^2+z_3^2 & = z_1^2(1+\omega^2 + \omega^4) = 0 \end{align*} and hence \begin{align*} \cos 2\alpha+ \cos 2\beta + \cos 2\gamma = 0 \end{align*} Now, \begin{align*} \cos^2\alpha + \cos^2\beta + \cos^2\gamma &= \frac{1+\cos2\alpha}{2}+\frac{1+\cos2\beta}{2}+\frac{1+\cos2\gamma}{2}\\ &= \frac{3}{2} \end{align*} Use $\sin^2 \alpha = 1-\cos^2\alpha$ to get the other result.

$\require{cancel}$ Sketch: You know that there is some $\gamma$ such that $\sin\gamma=-\sin\alpha-\sin\beta$ and $\cos\gamma=-\cos\alpha-\cos\beta$ if and only if

$$\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=1$$

If you run the calcs and use basic trigonometric identities, you'll end up with $$\cos(\beta-\alpha)=\cos(\text{easy angle})$$ i.e. $\beta=\alpha\pm(\text{easy angle})+2k\pi$ which will allow you to write $\cos\beta$ and $\cos\gamma$ as linear combinations of $\cos\alpha$ and $\sin\alpha$. Then you can substitute in the sum of squares. Notice that the sum of the squares of sines follows from the one of cosines.

Hint: By assumption, $$(\cos\alpha+\cos\beta)^2+(\sin\alpha + \sin\beta)^2= (-\cos\gamma)^2+(-\sin\gamma)^2.\tag1$$ Expanding (1), the LHS simplifies to $2+2\cos(\alpha-\beta)$ and the RHS equals $1$. Therefore $\cos(\alpha-\beta)=-\frac12$, which means $\alpha-\beta$ is either $120^\circ$ or $-120^\circ$. By symmetry, the same relationship holds for $\beta-\gamma$ and for $\gamma-\alpha$. Conclude that there exists $\theta$ such that $\alpha,\beta,\gamma$ in some order equals $\theta$, $\theta+120$, $\theta-120$. Now compute $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\cos^2\theta+\cos^2(\theta+120)+\cos^2(\theta-120)$$ using the addition formula for cosines.

I would suggest to use $\cos^2 x = \frac{1}{2} + \frac{\cos 2x}{2}$ for the cosinus sum (the same way with $\sin$).

Of course. sum of the three squared cosinuses is: $\frac{3}{2} + \sum \limits_{x=\alpha,\beta, \gamma} \cos 2x$

Then you need to show that the last sum is zero.

Expressing $\sin x$ and $\cos x$ in exponential form, from the initial conditions we get: $e^{i\alpha} + e^{i \beta} + e^{i \gamma} = 0$ and $e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma} = 0$.

Expressing the $\cos 2x$ ($x= \alpha, \beta, \gamma$) in exponential form and using the equations above, you can prove that $\sum \limits_{x=\alpha,\beta, \gamma} \cos 2x = 0$.

For illustration, if $\alpha + \beta + \gamma = \pi$:

$(e^{i\alpha} + e^{i \beta} + e^{i \gamma})^2 = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} - 2 (e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma})$, so $e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} = 0$

The same way: $(e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma})^2 = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} - 2 (e^{i\alpha} + e^{i \beta} + e^{i \gamma})$, so $e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma}$, and $\sum \limits_{x=\alpha,\beta, \gamma} \cos 2x = 0$

More general, this can be proved not only for $\alpha + \beta + \gamma = \pi$ - let's say that $\alpha + \beta + \gamma = y$. Using the same idea as above:

$0 = (e^{i\alpha} + e^{i \beta} + e^{i \gamma})^2 = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} + 2 (e^{i (y - \alpha)} + e^{i (y - \beta)} + e^{i (y - \gamma)}) = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} + 2 e^{iy} (e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma}) = e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma}$

So $e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma} = 0$, no matter what is the value of $y$.

Also, $0 = (e^{-i\alpha} + e^{-i \beta} + e^{-i \gamma})^2 = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} + 2 (e^{-i (y - \alpha)} + e^{-i (y - \beta)} + e^{-i (y - \gamma)}) = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma} + 2 e^{-iy} (e^{i\alpha} + e^{i \beta} + e^{i \gamma}) = e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma}$

Finally, $\sum \limits_{x=\alpha,\beta, \gamma} \cos 2x = \frac{(e^{2i\alpha} + e^{2i \beta} + e^{2i \gamma})-(e^{-2i\alpha} + e^{-2i \beta} + e^{-2i \gamma})}{2} = 0$

• If $\alpha+\beta+\gamma=(2k+1)\pi$, you have $\sin\alpha+\sin\beta+\sin\gamma=\sin\alpha+\sin\beta+\sin(\alpha+\beta)$, which seems to be $0$ only in very trival cases, like $\sin\alpha\sin\beta=0$ or $\sin\alpha=-\sin\beta$. So, I guess your assumption was not meant to be there. – user228113 Sep 3 '16 at 9:19
• Yes, I also think that this assumption is not necessary, but it makes the proof a little easier. I edited my answer using the same logic to make it more general. – Icarus 369 Sep 3 '16 at 21:00