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?

Here is a simple treatment of the $n$-dimensional case, where the point on the curve is $\vec x = (x_0, \ldots, x_{n-1})$ with $x_i = \cos(\theta-\pi i/n)$.

1. The curve lies on a sphere.

   We have $x_i^2 = \cos^2(\theta-\pi i/n) = \frac12+\frac12\cos(2\theta-2\pi i/n)$. So $$\|\vec x\|^2 = \sum x_i^2 = \frac n2 + \frac12 \sum \cos(2\theta-2\pi i/n).$$ The latter term is zero because it is the sum of $n$ equally spaced sinusoids (it is equivalently the $x$-component of the sum of $n$ unit vectors equally spaced along the unit circle, or the real part of the sum of the $n$th roots of $e^{2n\theta\sqrt{-1}}$; in either case, the entire thing is zero by symmetry). So $\|\vec x\|^2$ is a constant, $\frac n2$, independent of $\theta$.

2. The curve lies on a two-dimensional subspace.

   We have $x_i = \cos(\theta-\pi i/n) = a_i\cos\theta + b_i\sin\theta$ for some fixed $a_i$ and $b_i$ independent of $\theta$. Then $\vec x = \vec a\cos\theta + \vec b\sin\theta$. So $\vec x$ lies on the two-dimensional subspace spanned by $\vec a$ and $\vec b$.

Thus, $\vec x$ lies on the intersection of a sphere in $n$ dimensions and a two-dimensional subspace, i.e. a sphere in two dimensions, also known as a circle.

If you use basic trigonometric identities, you can show the first set of three equations is a piece of this plane: $y-z=x$.

Using trigonometric identities, we get $$\cos(\theta-\frac\pi3) = \cos\theta\cos\frac\pi3+\sin\theta\sin\frac\pi3=\frac12\cos\theta+\frac12\sqrt3\sin\theta,$$ $$\cos(\theta-\frac{2\pi}3) = \cos\theta\cos\frac{2\pi}3+\sin\theta\sin\frac{2\pi}3=-\frac12\cos\theta+\frac12\sqrt3\sin\theta.$$ Thus, if we write $\gamma=(x,y,z)$ we get $\gamma = v_1\cos\theta + v_2\sin\theta$ with $v_1=(1,\frac12,-\frac12)$ and $v2=(0,\frac12\sqrt3,\frac12\sqrt3)$. Thus, we have at least an ellipse. Moreover, it is easy to check that $v1\cdot v2=0$. Furthermore, $v_1^2 = 1 + \frac14+\frac14=\frac32$ and $v_2^2=\frac34+\frac34=\frac32$; thus the vectors are orthogonal and of equal length, and therefore it is indeed a circle.

In the general case, we have $(v_1)_k = \cos \frac{(k-1)\pi}d$ and $(v_2)_k=\sin \frac{(k-1)\pi}d$, where $d$ is the dimension of the vector space ($d=4$ in your second case). Thus, $$v_1\cdot v_2 = \sum_{k=1}^d\cos\frac{(k-1)\pi}d\sin\frac{(k-1)\pi}d = \frac12\sum_{k=1}^d\sin\frac{(k+1)2\pi}{d} = 0$$ and $$v_2^2-v_1^2 = \sum_{k=1}^d\left(\cos^2\frac{(k+1)\pi}d-\sin^2\frac{(k+1)\pi}d\right) = \sum_{k=1}^d\cos\frac{(k+1)2\pi}{d} = 0,$$ The construction thus gives a circle in any dimension.

For at least the three-dimensional case, here's a more mechanical method (i.e. less enlightening than Rahul's nice answer) to verify if your curve is a circle:

We try to evaluate the curvature $\kappa$ and torsion $\tau$ of the given curve. By the fundamental theorem of space curves, a space curve is uniquely determined (up to rigid motions) by $\kappa(s)$ and $\tau(s)$; if, in addition, $\tau(s)=0$ (i.e. the curve is flat) and $\kappa(s)$ is a constant greater than zero, then we know that the space curve is indeed a circle.

Using formula 26 here for the curvature, we have

$$\begin{align*} \kappa&=\frac{\|\mathbf r^\prime\times \mathbf r^{\prime\prime}\|}{\|\mathbf r^\prime\|^3}\\ &=\frac{\left\|\langle-\sin\,\theta,\cos\left(\theta+\frac{\pi}{6}\right),\cos\left(\theta-\frac{\pi}{6}\right)\rangle\times\langle-\cos\,\theta,-\sin\left(\theta+\frac{\pi}{6}\right),\sin\left(\frac{\pi}{6}-\theta\right)\rangle\right\|}{\left\|\langle-\sin\,\theta,\cos\left(\theta+\frac{\pi}{6}\right),\cos\left(\theta-\frac{\pi}{6}\right)\rangle\right\|^3}\\ &=\sqrt\frac23 \end{align*}$$

Using formula 3 here for the torsion, we have

$$\begin{align*} \tau&=\frac{\mathbf r^\prime\times \mathbf r^{\prime\prime}\cdot\mathbf r^{\prime\prime\prime}}{\kappa^2}\\ &=\frac1{2/3}\begin{vmatrix}-\sin\,\theta&\cos\left(\theta+\frac{\pi}{6}\right)&\cos\left(\theta-\frac{\pi}{6}\right)\\-\cos\,\theta&-\sin\left(\theta+\frac{\pi}{6}\right)&\sin\left(\frac{\pi}{6}-\theta\right)\\\sin\,\theta&-\cos\left(\theta+\frac{\pi}{6}\right)&-\cos\left(\theta-\frac{\pi}{6}\right)\end{vmatrix}\\ &=0 \end{align*}$$

(Note that the determinant is easily seen to be zero, since the third row is a multiple of the first.)

Since $\tau=0$, the curve is flat; in addition, since $\kappa=\sqrt{2/3}$, we find that our space curve is a circle with radius $1/\kappa=\sqrt{3/2}$.