Is this 3D curve a circle? The following is a curve in $3$ dimensions:
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/3) \\ z & = &  \cos(\theta - 2\pi/3) \end{eqnarray}$$
Is the curve a circle?
If it is, what about this curve in $4$ dimensions?
$$\begin{eqnarray} x & = & \cos(\theta) \\ y & = & \cos(\theta - \pi/4) \\
z & = & \cos(\theta - 2\pi/4) \\ w & = & \cos(\theta - 3\pi/4) \end{eqnarray}$$
I don't know if there is something like a circle in $4$-D. If there is, is this curve the $4$-D version of a circle?
P.S.: Is two-dimensional subspace the generalized plane? I want to learn more about this. What should I read?

$\cos3\theta=cos3(\theta)$
$\cos3(\theta-\frac{\pi}{3})=cos(3\theta-\pi)=-\cos3\theta$
As $-y=-\cos(\theta-\frac{\pi}{3})=cos(\theta+\frac{2\pi}{3})$,
$\cos3(\theta+\frac{\pi}{3})=\cos(2\pi+3\theta)=\cos3\theta$
$\cos3(\theta-\frac{2\pi}{3})=cos(3\theta-2\pi)=\cos3\theta$
Now, $\cos3\theta=4cos^3\theta-3\cos\theta$
If $\cos3\theta=a$ and $\cos\theta=t$
So, $x,-y,z$ are the roots of  $4t^3-3t-a=0$
$=>x+(-y)+z=0$
$=>x(-y)+(-y)z+zx=\frac{3}{4}$
$=>x^2+y^2+z^2=(x+(-y)+z)^2-2(x(-y)+(-y)z+zx)=0+2\frac{3}{4}$
$=>x^2+y^2+z^2=\frac{3}{2}$
Observe that $(x,y,z)$  satisfy a general plane equation $Ax+By+CZ+D=0$   where $A,B,C,D $ constants,  not all zeros.  
Also, satisfies the equation of the general circle in 3-D, $(x-a)^2+(y-b)^2+(z-c)^2=d^2 $.  $a=b=c=0, d^2=\frac{3}{2}$
In case of $x,y,z,w$,
$z=\cos(\theta-\frac{2\pi}{4})=\sin\theta$,
$\sqrt2 y=\cos\theta+\sin\theta$,
$\sqrt2 w=-\cos\theta+\sin\theta$
So,$x^2+z^2=1$  and $w^2+y^2=1$
$x^2+y^2+z^2+w^2=2$
Now $\sqrt2 (y+w)=2\sin\theta=2z=>\sqrt 2z=y+w$
Similarly, $y-w=\sqrt 2x$ 
Observe that   $(x,y,z,w)$  satisfies two general plane equations  $Ax+By+CZ+Dw=E$  where $A,B,C,D,E $ constants, not all zeros. 
Also, satisfies the equation of the general circle in  4-D, $(x-a)^2+(y-b)^2+(z-c)^2+(w-d)^2=e^2 $ .  $a=b=c=d=0, e^2=2$
Again, we know $\cos nx=$Real part of $(\cos x+i\sin x)^n=(\cos x)^n+^nC_2(\cos x)^{n-2}(\sin x)^2+^nC_4(\cos x)^{n-4}(\sin x)^4+...$
Observe there is no term containing  $=(\cos x)^{n-1}$
As $\cos n(2x-\frac{2r_i\pi}{n})=\cos(2nx-2r_i\pi)=\cos 2nx=C(say)$ 
So, $\cos (2x-\frac{2r_i\pi}{n})=R_i$(say), where all $r_i$s are distinct integers with   $0 ≤r_i< n$ are the roots of the equation 
$2^{n-1}y^n+C_1y^{n-2}+...-C=0$
So, $\sum R_i=0$ as the coefficient of $y^{n-1}$ is 0.
If $x_i=\cos (x-\frac{r_i\pi}{n})=>R_i=2(x_i)^2-1$ 
So, $\sum (2(x_i)^2-1)=0 =>\sum (x_i)^2=\frac{n}{2}$
This is another way of generalization("The curve lies on a sphere") already achieved by Rahul Narain. 
