<s>[INST] Explain the following equalities I am a little stuck on coming up with geometrical explanation for why the following equalities are true. I tried arguing the $\cos(\theta)$ is the projection to the x-axis of a vector $r$ inside a unit circle, so as it goes around by $2 \pi$, the projections on both the positive and negative part of the x-axis cancel out. Could someone please confirm if it is completely correct to think about it this way?
$$\cos(\theta) + \cos(\theta + 2 \pi/3) + \cos(\theta + 4\pi/3) = 0$$
$$\sin(\theta) + \sin(\theta + 2 \pi/3) + \sin(\theta + 4\pi/3) = 0$$
And what about:
$$\cos^2(\theta) + \cos^2(\theta + 2 \pi/3) + \cos^2(\theta + 4\pi/3) = \frac{3}{2}$$
$$\sin^2(\theta) + \sin^2(\theta + 2 \pi/3) + \sin^2(\theta + 4\pi/3) = \frac{3}{2}$$
 [/INST] First set of equations: Here's a geometric reason. Imagine we place $1$ kg weights at the unit circle corresponding to angles of $\theta$, $\theta+\frac{2\pi}{3}$ and $\theta+\frac{4\pi}{3}$. These points form a configuration that is a rotation of 

The center of mass of this configuration is at the origin, and hence the same is true for its rotations. By looking at the $x$ and $y$ coordinates, this implies that. $$\cos(\theta)+\cos(\theta+2\pi/3)+\cos(\theta+4\pi/3)=0$$ and $$\sin(\theta)+\sin(\theta+2\pi/3)+\sin(\theta+4\pi/3)=0.$$

Second set of equations:
Our goal is to show that $$\cos^2(\theta)+\cos^2\left(\theta+\frac{2\pi}{3}\right)+\cos^2\left(\theta+\frac{4\pi}{3}\right)=\sin^2(\theta)+\sin^2\left(\theta+\frac{2\pi}{3}\right)+\sin^2\left(\theta+\frac{4\pi}{3}\right),$$ since we know that the sum of all $6$ terms equals $3$ as these are three unit vectors in the plane. Proceeding by the identity $\cos(2\theta)=2\cos^2(\theta)-1$ and using the first set of equations is the quickest way to prove the result. While I do not have a geometric proof, here is another proof that I am fond of:
For $0<\theta<2\pi/3$ define $$u=\left[\begin{array}{c}
\cos\theta\\
\cos\left(\theta+2\pi/3\right)\\
\cos\left(\theta+4\pi/3\right)
\end{array}\right],\ \ \ v=\left[\begin{array}{c}
\sin\theta\\
\sin\left(\theta+2\pi/3\right)\\
\sin\left(\theta+4\pi/3\right)
\end{array}\right].$$
 Then we know that $\|u\|_{2}^{2}+\|v\|_{2}^{2}=3$
 , and we are trying to show that $\|v\|_{2}=\|u\|_{2}$
 .  By the identity $$\sin(\theta)=\frac{\sin\left(\theta+\frac{\pi}{6}\right)+\sin\left(\theta-\frac{\pi}{6}\right)}{\sqrt{3}}=\frac{-\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)}{\sqrt{3}},$$
 the operator $$T=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}
0 & -1 & 1\\
1 & 0 & -1\\
-1 & 1 & 0
\end{array}\right]$$
 satisfies $T(u)=v$
 . Since $u$
  is in the kernel of the constant matrix by the first part, we see that $$S=T+\frac{1}{3}\left[\begin{array}{ccc}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{array}\right]$$
 satisfies $S(u)=v$
 , and moreover $S^{-1}=S^{T}$
  so that $S$
  is an orthogonal matrix. This implies that $\|u\|_{2}=\|v\|_{2}$
  as $S$
  is then an isometry.
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