Finding $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ given $\sin \alpha+\sin \beta+\sin\gamma=0=\cos\alpha+\cos\beta+\cos\gamma$

Question

I am supposed to find the value of $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ and I have been provided with the information that $\sin \alpha+\sin \beta+\sin\gamma=0=\cos\alpha+\cos\beta+\cos\gamma$.

I tried to approach this using vectors. We can consider three unit vectors that add up to $0$. Unit vectors because the coefficients of the $\sin$ and $\cos$ terms are $1$. For the sake of simplicity, let one of the vectors $\overline{a}$ be along the $x$-axis. Let the angles between $\overline{b}$ and $\overline{c}$ be $\alpha$, between $\overline{a}$ and $\overline{b}$ be $\gamma$ and between $\overline{a}$ and $\overline{c}$ be $\beta$. Then we have:

$$\begin{aligned}\overline{a}&=\left<1,0\right>\\ \overline{b}&=\left<-\cos\gamma, -\sin\gamma\right>\\ \overline{c}&=\left<-\cos\beta, \sin\beta\right>\end{aligned}$$

$$\begin{aligned}\cos\gamma+\cos\beta &=1\\ \sin\beta&=\sin\gamma\end{aligned}$$

Now, $\cos \gamma$ and $\cos\beta$ must have the same sign. So we get $\sin\alpha=-\sqrt{3}/2$, $\sin\beta=\sqrt{3}/2$ and $\sin\gamma=\sqrt{3}/2$. This contradicts with the answer key provided according to which $\sum_{cyc}\sin^2\alpha=3/2$. What am I doing wrong?

This was the picture I had in mind with $\overline{a}$ aligned with the horizontal.

Answer 1

In the case you are dealing with, the three vectors are $(1,0)$, $(-1/2,\sqrt3/2)$ and $(-1/2,-\sqrt3/2)$. The sum of the squares of the $y$-coordinates is $$0^2+\left(\frac{\sqrt3}2\right)^2+\left(-\frac{\sqrt3}2\right)^2=\frac32.$$

The general solution though, has vectors $(\cos\alpha,\sin\alpha)$, $(\cos(\alpha+2\pi/3),\sin(\alpha+2\pi/3))$ $(\cos(\alpha-2\pi/3),\sin(\alpha-2\pi/3))$ so you need to prove the identity $$\sin^2\alpha+\sin^2\left(\alpha+\frac{2\pi}3\right) +\sin^2\left(\alpha-\frac{2\pi}3\right)=\frac32.$$

Answer 2

In fact the cosine equation is not

$$1-\cos\gamma-\cos\beta=0,$$

but

$$\cos\beta\cos\gamma-\sin\beta\sin\gamma-\cos\gamma-\cos\beta=0,$$

by taking the dot-products between the vectors.

The sine equation is obtained from the cross-products,

$$-\sin\beta\cos\gamma-\cos\beta\sin\gamma-\sin\gamma+\sin\beta=0.$$

These are compatible with $\alpha+\beta+\gamma=2\pi.$

Answer 3

From Clarification regarding a question,

$\alpha-\beta=\dfrac{2\pi}3+2\pi a,\beta-\gamma=\dfrac{2\pi}3+2\pi b$

$\implies\alpha-\gamma=\dfrac{4\pi}3+2\pi c=-\dfrac{2\pi}3+2\pi d$

If $S=\sin^2\alpha+\sin^2\beta+\sin^2\gamma$

$$2S=3-(\cos2\alpha+\cos2\beta+\cos2\gamma)$$

Method$\#1:$

$$\cos2\alpha+\cos2\beta=\cos2\left(\gamma-\dfrac{2\pi}3\right)+\cos2\left(\gamma+\dfrac{2\pi}3\right)=2\cos2\gamma\cos\dfrac{2\pi}3=-\cos2\gamma$$

Method$\#2:$

If $\cos3x=\cos3A$

$3x=2n\pi\pm3A\implies x=\dfrac{2n\pi}3\pm A$

Now $\cos3A=\cos3x=4\cos^3x-3\cos x$

$\implies4\cos^3x-3\cos x-\cos3A=0$

$\implies\sum_{r=-1}^1\cos\left(x+r\dfrac{2r\pi}3\right)=\dfrac04$