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

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.

• You are using $\alpha$ etc., not just for the original quantities, but also for the angles between your vectors. – Angina Seng Jun 28 '19 at 6:16
• Exactly, so the sum of the squares of the $y$-coordinates would give the required sum. Thanks for clarifying. :) – Paras Khosla Jun 28 '19 at 6:17
• How do you obtain the last two equations ? – Yves Daoust Jun 28 '19 at 6:17
• 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 :) – Paras Khosla Jun 28 '19 at 6:17

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

• Thanks a lot! :) – Paras Khosla Jun 28 '19 at 6:15
• 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 $\sin\alpha=-\sqrt{3}/2$. – Paras Khosla Jun 28 '19 at 6:29

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 $$\alpha+\beta+\gamma=2\pi.$$

• Yes I am getting $\sin\alpha=-\sqrt{3}/2$ but turns out it contradicts with the solution. Could you clarify this? – Paras Khosla Jun 28 '19 at 6:30
• 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 $2\cos\gamma=2\cos\beta=1$. – Paras Khosla Jun 28 '19 at 6:33
• Since, $\alpha=2\pi-(\beta+\gamma)$ we can find its $\sin$ and $\cos$. – Paras Khosla Jun 28 '19 at 6:35
• $\sin\alpha=-\sqrt{3}/2$ and $|\cos\alpha|=1/2$. So I am getting $\sum_{cyc}\sin^2\alpha=(3/2)^2$. – Paras Khosla Jun 28 '19 at 6:37
• 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? – Paras Khosla Jun 28 '19 at 6:41

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

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

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

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

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