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:
$$\cos^2(\theta) + \cos^2(\theta + 2 \pi/3) + \cos^2(\theta + 4\pi/3) = \frac{3}{2}$$
$$\sin^2(\theta) + \sin^2(\theta + 2 \pi/3) + \sin^2(\theta + 4\pi/3) = \frac{3}{2}$$
 A: 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.
A: The answer by @Eric Naslund gives really good insight to first bit the problem. My answer is in the direction of the second bit, but doesnt present it as elegantly (as I cant seem to find a physical phenomena that makes use of those equations). It uses the first bit in the explaination and proof so hopefully, you shall have a clearer picture at the end of this. 
From initial observation, we gather that $\cos^2(\theta)+\cos^2(\theta+2\pi/3)+\cos^2(\theta+4\pi/3)$ is definitely larger than $0$, because sum of three positive numbers has to be positive. It is also less than 3, because $\cos$ and thus, $\cos^2$ of any angle is less than 1.
In the first bit, we saw that the locus of points with coordinates $(\cos(\theta),\sin(\theta))$ is a circle centred at origin and radius = $1$ unit. This locus spans all the 4 quadrants. The locus of points with coordinates $(\cos^2(\theta),\sin^2(\theta))$ is a straight line through $(1,0)$ and $(0,1)$. 

This makes sense, because $\cos^2(\theta)+\sin^2(\theta)=1$ at all values of $\theta$. As $\theta$ varies from $[0,\pi/2]$, one goes from $(0,1)$ to $(1,0)$ along that line. Moving $\theta$ from $[\pi/2,\pi]$, one goes back from $(1,0)$ to $(0,1)$ along that line. This cycle continues for the subsequent quadrants. The projection to the x-axis is always positive and this proves that the sum will be greater than 0. Also, since the three angles differ from each other by $2\pi/3$, the sum will be less than 3. 
Lets presume our $\theta$ lies in the first quadrant. Thus, $\theta+2\pi/3$ is in second or third quadrant and $\theta+4\pi/3$ is in the third or fourth quadrant.  If one were to plug in values and see, one will find that as $\theta$ varies from $[0,\pi/2]$, we have:
$$ \cos^2(\theta) \in [0,1] \;\;\;\text{  blue region} \\ \cos^2(\theta+2\pi/3) \in [0.25,1] \;\;\;\text{  red region}\\ \cos^2(\theta+4\pi/3) \in [0,0.75] \;\;\;\text{  green region}$$

The plot shows $\cos^2(\theta)$ v/s $\theta$. The function in red and blue region, both start at $0.25$ and end at $0.75$. However, each moves in a different direction, letting red region to touch $1.00$ and the green region to touch $0.00$. The sum of the function in the two regions, it can be observed, is increasing. One can also note that $0.5\leq\cos^2(\theta+2\pi/3)+\cos^2(\theta+4\pi/3) \leq 1.5$. We see that as the function decreases in the blue region, the sum of function in the red and green regions increases.  
At the very least, we can conclude $0.5\leq\cos^2(\theta)+\cos^2(\theta+2\pi/3)+\cos^2(\theta+4\pi/3)\leq2.5$. If you have been quick to notice, then you would already have concluded that the inequality limits can be made tighter. But, lets presume we don't have that knowledge yet. We only know how the terms vary in their ranges, that the total sum should be positive and as blue region decreases the red and green increase. To see what the sum exactly should be, we make use of trigonometric identities. 
We know that:
$$
\cos^2(\theta) = \frac{1}{2}(\cos(2\theta)+1)
$$
Thus, we have:
$$
\cos^2(\theta)+\cos^2(\theta+2\pi/3)+\cos^2(\theta+4\pi/3) = \\ 
\frac{1}{2}\left ( \cos(2\theta)+\cos(2\theta+4\pi/3)+\cos(2\theta+8\pi/3)+3 \right )
$$
Now, since $\cos(2\theta+8\pi/3) = \cos(2\theta+2\pi/3+2\pi)=\cos(2\theta+2\pi/3)$, the right side of the above equations becomes 
$$
= \frac{1}{2}\left ( \cos(2\theta)+\cos(2\theta+4\pi/3)+\cos(2\theta+2\pi/3)+3 \right ) \\
= \frac{1}{2}\left ( 0+3 \right )
$$
Which is what you want. The $\sin$ part is then easily derived from this. 
A: We have $\cos(\theta) + \cos(\theta + 2 \pi/3) + \cos(\theta + 4\pi/3) =
\operatorname{Re}(e^{i\theta}+e^{i\theta+i2\pi/3}+e^{i\theta+i4\pi/3}) $ and $\sin\theta+ \sin(\theta + 2 \pi/3) + \sin(\theta + 4\pi/3)=\operatorname{Im}(e^{i\theta}+e^{i\theta+i2\pi/3}+e^{i\theta+i4\pi/3})$, but:
$$
e^{i\theta}+e^{i\theta+i2\pi/3}+e^{i\theta+i4\pi/3}=
e^{i\theta}(1+e^{i2\pi/3}+e^{i4\pi/3})=0,
$$
because $1+e^{i2\pi/3}+e^{i4\pi/3}$ is the sum of the solutions of equation $x^3-1=0$.
If we now recall that $[\operatorname{Re}(z)]^2={1\over2}[|z|^2+\operatorname{Re}(z^2)]$ and $[\operatorname{Im}(z)]^2={1\over2}[|z|^2-\operatorname{Re}(z^2)]$, we have
$$
\cos^2(\theta) + \cos^2(\theta + 2 \pi/3) + \cos^2(\theta + 4\pi/3)=\\
{1\over2}\bigl(|e^{i\theta}|^2+|e^{i\theta+i2\pi/3}|^2+|e^{i\theta+i4\pi/3}|^2+\operatorname{Re}(e^{2i\theta}+e^{2i\theta+i4\pi/3}+e^{i\theta+i8\pi/3})\bigr)=\\
{3\over2}+\operatorname{Re}\bigl(e^{2i\theta}(1+e^{i4\pi/3}+e^{i2\pi/3})\bigr)={3\over2}
$$
and a similar equality for the sine.
A: Hint you can use the formula $cosA+cosB+cosC=cosAcosBcosC[1-tanAtanB-tanBtanC-tanAtanC]$ and $sinA+sinB+sinC=cosAcosBcosC[tanA+tanB+tanC-tanAtanBtanC$ Note I have given it as a trigonometric proof for your formula. Its tidious work but you will surely get a trigonometric proof for it.Hope it helps you.
