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?

enter image description here

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

  • 2
    $\begingroup$ You are using $\alpha$ etc., not just for the original quantities, but also for the angles between your vectors. $\endgroup$ Jun 28, 2019 at 6:16
  • $\begingroup$ Exactly, so the sum of the squares of the $y$-coordinates would give the required sum. Thanks for clarifying. :) $\endgroup$ Jun 28, 2019 at 6:17
  • $\begingroup$ How do you obtain the last two equations ? $\endgroup$
    – user65203
    Jun 28, 2019 at 6:17
  • $\begingroup$ The vectors add up to $0$. That was a careless mistake. I did not prefix the negative sign with $\cos \beta$ in the vector definition. Thanks for pointing out @YvesDaoust :) $\endgroup$ Jun 28, 2019 at 6:17

3 Answers 3


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 $$\sin^2\alpha+\sin^2\left(\alpha+\frac{2\pi}3\right) +\sin^2\left(\alpha-\frac{2\pi}3\right)=\frac32.$$

  • $\begingroup$ Thanks a lot! :) $\endgroup$ Jun 28, 2019 at 6:15
  • $\begingroup$ I'm still a bit unclear about calling $\sin\alpha=0$. I'm not sure how that would be because $\alpha=2\pi-(\beta+\gamma)$ which gives $\sin\alpha=-\sqrt{3}/2$. $\endgroup$ Jun 28, 2019 at 6:29

In fact the cosine equation is not




by taking the dot-products between the vectors.

The sine equation is obtained from the cross-products,


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

  • $\begingroup$ Yes I am getting $\sin\alpha=-\sqrt{3}/2$ but turns out it contradicts with the solution. Could you clarify this? $\endgroup$ Jun 28, 2019 at 6:30
  • $\begingroup$ Since $\left|\sin\gamma\right|=\left|\sin\beta\right|\implies \left|\cos\gamma\right|=|\cos\beta|$. Now, $\cos\gamma$ and $\cos\beta$ must also have the same sign for them to add up to $1$, so that means $2\cos\gamma=2\cos\beta=1$. $\endgroup$ Jun 28, 2019 at 6:33
  • $\begingroup$ Since, $\alpha=2\pi-(\beta+\gamma)$ we can find its $\sin$ and $\cos$. $\endgroup$ Jun 28, 2019 at 6:35
  • $\begingroup$ $\sin\alpha=-\sqrt{3}/2$ and $|\cos\alpha|=1/2$. So I am getting $\sum_{cyc}\sin^2\alpha=(3/2)^2$. $\endgroup$ Jun 28, 2019 at 6:37
  • $\begingroup$ I don't think the resolution is correct somehow. To avoid resolving $\overline{a}$ because it's angle didn't fit in well with the axes, I let it be $\left<0,1\right>$. Could that be the problem? $\endgroup$ Jun 28, 2019 at 6:41

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





If $\cos3x=\cos3A$

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

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

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



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