Is the random point $(C,S)$ the same as $(1,0)+(\cos U,\sin U)=(1 + \cos U,\sin U)$, with $U$ a uniform r.v.?

Suppose that $$\theta_1$$ and $$\theta_2$$ are independent and identically distributed (i.i.d.) random variables and that $$\theta_j$$ has probability density function (PDF) $$f_j = \frac{1}{2\pi}$$ ($$i.e.$$, the uniform distribution) for $$j = 1$$ and $$2$$. Next, we define the following random variables $$C = \cos \theta_1 + \cos \theta_2$$ and $$S = \sin \theta_1 + \sin \theta_2$$.

My question is, can I say, based on the rotational invariance, that the random point $$(C,S)$$ is the same as that of $$(1,0)+(\cos U,\sin U)=(1 + \cos U,\sin U)$$, where $$U$$ is uniformly distributed on $$[0,2\pi]$$? If so, how can I show that?

These two scatter plots illustrate the difference, the first is for the points $$(C,S)=(\cos\theta_1+\cos\theta_2,\sin\theta_1+\sin\theta_2)$$, the second for the points $$(1+\cos U,\sin U)=(1+\cos\theta_3,\sin\theta_3)$$, where all angles $$\theta_i$$ are uniformly distributed in $$(0,2\pi)$$.
The second distribution uniformly fills a unit circle with center at $$(1,0)$$. The first distribution fills a disc with radius 2, but not uniformly, the density diverges as $$1/r$$ at a distance $$r$$ from the origin.
To calculate the radial distribution $$P(r)$$, it is helpful to think of a random walk on the plane with unit step size and random orientation. The desired $$P(r)$$ is the distribution of the distance from the origin after two steps.
After the first step the random walker is at some arbitrary point on the unit circle. For the distribution $$P(r)$$ it does not matter where on the unit circle, we may place the point at $$(1,0)$$. Then the second step brings the random walker to the point $$(1+\cos\phi,\sin\phi)$$, with $$\phi$$ uniformly in $$(0,2\pi)$$. The distance from the origin is $$r=\sqrt{2+2\cos\phi}$$, and the probability distribution is $$P(r)=\frac{2/\pi}{\sqrt{4-r^2}},\;\;0 The corresponding density $$\rho(r)=P(r)/2\pi r$$ indeed diverges $$\propto 1/r$$ when $$r\rightarrow 0$$.
For the generalization to $$(C_n,S_n)=(\sum_{i=1}^n\cos\theta_i,\sum_{i=1}^n\sin\theta_i)$$ see this MO posting.