If $X_1,X_2\sim\mathcal N(0,1)$ then $\cos(\theta)X_1+\sin(\theta)X_2$, $-\sin(\theta)X_1+\cos(\theta)X_2 \sim\mathcal N(0,1)$? Let $X_1,X_2$ be an independent random variable such that $X_1,X_2\sim\mathcal N(0,1)$.  We pose : $Y_1=\cos(\theta)X_1+\sin(\theta)X_2$, $Y_2=-\sin(\theta)X_1+\cos(\theta)X_2$. Show that $Y_1,Y_2$  be an independent random variable such that $Y_1,Y_2\sim\mathcal N(0,1)$?
We pose $X=(X_1,X_2)$ is a multivariate random variable, the density of
$$f_X(x_1,x_2)=\bigg(\frac {1}{\sqrt {2\pi }}e^{-\frac{x_1^{2}}{2}}\bigg)\bigg(\frac {1}{\sqrt {2\pi }}e^{-\frac{x_2^{2}}{2}}\bigg)=\frac {1}{2\pi}e^{-\frac{\|(x_1,x_2)\|^{2}_2}{2}}$$we have $Y = MX$, with $M=\begin{pmatrix} \cos(\theta)& \sin(\theta)\\
-\sin(\theta)&\cos(\theta)
\end{pmatrix} $ Thus, the vector $Y = MX$ admits for density $y=(y_1,y_2)\to \frac{1}{\det(M)}\frac{1}{2\pi} e^{-\frac{\|M^{-1}y\|^2_2}{2}}$. $M$ is a rotation matrix, so its determinant is $1$, and it is an isometry for the Euclidean norm, which implies that for all $y\in\mathbb{R}^2$, we have $\|M^{-1}y\|_2=\|y\|_2$. The density of $Y$ is $f_X$ is precisely the density of $X$. $Y$ therefore has the same law as $X$, Can we say that the components of $Y$, $Y_1$ and $Y_2$, are independent and $Y_1,Y_2\sim\mathcal N(0,1)$?
 A: Independent normals are jointly normal, so as you note, $X=(X_1,X_2)'$ is standard (bivariate) normal.
Since $Y=MX$ is an affine transform of a normal, it is also normal with covariance matrix $MM'=I$ (using the fact that $M$ is an orthogonal matrix). So $Y$ is standard (bivariate) normal, just as $X$ is. This is not surprising, since $Y$ just rotates $X$ clockwise by angle $\theta$.
Finally since jointly normal and uncorrelated implies independent, we can say $Y_1,Y_2$ are independent. And since any linear combination of components of a multivariate normal vector is itself normal, this implies $Y_1,Y_2$ are each standard normal.
A: You can simply use that any linear combination of normally distributed random variables, is normally distributed. Proof can be found here. So then,
\begin{align}
Y_1 = \cos(\theta)X_1 + \sin(\theta)X_2 \sim \mathcal{N}(\mu_{Y_1},\sigma_{Y_1}^{2}),
\end{align}
where,
\begin{align}
\mu_{Y_1} &= \cos(\theta)\mu_{X_1} + \sin(\theta)\mu_{X_2} = 0\\[1em]
\sigma_{Y_1}^{2}  &\stackrel{\text{ind.}}{=} \cos^2(\theta)\sigma_{X_1}^2 + \sin^2(\theta)\sigma_{X_2}^2 = \cos^2(\theta) + \sin^2(\theta) = 1.
\end{align}
Therefore $Y_1 \sim \mathcal{N}(0,1)$.
The same thing can be done for $Y_2$. Hope this helps.
