Can the following sum of random variables be normally distributed?

Consider $ N_{1},N_{2} $ independent random variables which both distribute normally, both with zero expectation (and maybe different variance). Let $ \theta\sim U[0,2\pi] $ be uniformly distributed random variable, which is independent of $N_1$ and $N_2$. Define: $$ \begin{cases} X=N_{1}\sin\left(\theta\right)\\ Y=N_{2}\cos\left(\theta\right) \end{cases} $$ Next, define: $$ Z=X+Y\overset{\text{i.e}}{=}N_{1}\sin\left(\theta\right)+N_{2}\cos\left(\theta\right) $$ Does $Z$ distributes normally?

I'd really appreciate a clarification - is it correct or wrong?

I think that its wrong, and I have wrote a proof, but I feel like something's wrong with it and its actually correct. Anyway, here's what I have done:

My work:

It is easy to note that $Z$ has zero expectation, and if we will asssume it has a gaussian distribution, say $ \mathcal{N}\left(0,\sigma^{2}\right) $, then it would have to hold the following moments well known equation:

$$ \mathbb{E}\left[Z^{n}\right]=\begin{cases} 1\cdot3\cdot5\cdot....\cdot\left(n-1\right)\cdot\sigma^{n} & n\thinspace\thinspace\text{even}\\ 0 & n\thinspace\thinspace\text{odd} \end{cases} $$

Now, say $ \mathbb{E}\left[N_{1}^{2}\right]:=\sigma_{1}^{2},\thinspace\thinspace\thinspace\thinspace\mathbb{E}\left[N_{2}^{2}\right]=\sigma_{2}^{2} $

Then $ \mathbb{E}\left[X^{2}\right]=\mathbb{E}\left[N_{1}^{2}\right]\mathbb{E}\left[\sin^{2}\theta\right]=\frac{1}{2}\sigma_{1}^{2} $

And $ \mathbb{E}\left[Y^{2}\right]=\mathbb{E}\left[N_{2}^{2}\right]\mathbb{E}\left[\cos^{2}\theta\right]=\frac{1}{2}\sigma_{2}^{2} $.

Also, since $N_1$ follows the moments equation I wrote, we have $ \mathbb{E}\left[X^{4}\right]=\underset{3\sigma_{1}^{2}}{\underbrace{\mathbb{E}\left[N_{1}^{4}\right]}}\underset{\frac{3}{8}}{\underbrace{\mathbb{E}\left[\sin^{4}\theta\right]}}=\frac{3}{8}\cdot3\sigma_{1}^{2}$ And in the same way:

$ \mathbb{E}\left[Y^{4}\right]=\frac{3}{8}\cdot3\sigma_{2}^{2} $.

Next, note that $ \mathbb{E}\left[XY\right]=\mathbb{E}\left[N_{1}\right]\mathbb{E}\left[N_{2}\right]\mathbb{E}\left[\sin\theta\cos\theta\right]=0 $.

Next, note that

$ Z^{2}=X^{2}+Y^{2}+2XY $ And thus $ \mathbb{E}\left[Z^{2}\right]=\frac{1}{2}\left(\sigma_{1}^{2}+\sigma_{2}^{2}\right) $.

Since $Z$ holds the moments equation we must have $ \mathbb{E}\left[Z^{4}\right]=\frac{3}{4}\left(\sigma_{1}^{2}+\sigma_{2}^{2}\right)^{2} $

But note that:

$ Z^{4}=X^{4}+Y^{4}+2X^{2}Y^{2}+4\left(X^{2}+Y^{2}\right)XY+4X^{2}Y^{2} $

And $ \mathbb{E}\left[4\left(X^{2}+Y^{2}\right)XY\right]=4\mathbb{E}\left[X^{3}Y\right]+4\mathbb{E}\left[Y^{3}X\right]=4\mathbb{E}\left[X^{3}\right]\mathbb{E}\left[Y\right]+4\mathbb{E}\left[Y^{3}\right]\mathbb{E}\left[X\right]=0 $

So that $$ \mathbb{E}\left[Z^{4}\right]=\mathbb{E}\left[X^{4}\right]+\mathbb{E}\left[Y^{4}\right]+2\mathbb{E}\left[X^{2}\right]\mathbb{E}\left[Y^{2}\right]+4\mathbb{E}\left[X^{2}\right]\mathbb{E}\left[Y^{2}\right] $$

$$ =3\cdot\frac{3}{8}\sigma_{1}^{2}+3\cdot\frac{3}{8}\sigma_{2}^{2}+2\cdot\frac{1}{2}\cdot\frac{1}{2}\sigma_{1}^{2}\sigma_{2}^{2}+4\cdot\frac{1}{2}\cdot\frac{1}{2}\sigma_{1}^{2}\sigma_{2}^{2} $$

$$ =\frac{9}{8}\sigma_{1}^{2}+\frac{9}{8}\sigma_{2}^{2}+\frac{3}{2}\sigma_{1}^{2}\sigma_{2}^{2}=\frac{3}{4}\left(\frac{3}{2}\sigma_{1}^{4}+2\sigma_{1}^{2}\sigma_{2}^{2}+\frac{3}{2}\sigma_{2}^{4}\right) $$

Which does not seem equal to $ \frac{3}{4}\left(\sigma_{1}^{2}+\sigma_{2}^{2}\right)^{2} $.

What is wrong with what I have done?

Any help would be appreciated. Thanks in advace.

The correct claim is: $Z$ is normally distributed if and only if $\sigma_1=\sigma_2$.

Your computation is correct, except that $$ \mathsf E\left[X^2 Y^2\right]=\frac{1}{8}\sigma_1^2\sigma_2^2, $$ which gives you instead $$ \mathsf E Z^4=\frac{3}{4}\left(\frac{3}{2}\sigma_1^4+\frac{3}{2}\sigma_2^4+\sigma_1^2\sigma_2^2\right). $$ In fact, you can check by expanding the square that $\mathsf E Z^4=3(\sigma_1^2+\sigma_2^2)^2/4$ if and only if $\sigma_1=\sigma_2$. But reciprocally, if $\sigma_1=\sigma_2=\sigma$, then $Z$ is normally distributed. One computation-free argument would be: for any fixed $\theta\in\mathbb R$, $$ Z(\theta)=N_1 \cos(\theta)+N_2\sin(\theta) $$ has distribution $\mathcal N(0,\sigma^2)$. But now, if you sample $\theta$ from any distribution on $\mathbb R$, the random variable $Z(\theta)$ would still have distribution $\mathcal N(0,\sigma^2)$.