Compute correlation between $aX + bY$ and $cX + dY$, for real numbers $a, b, c, d$ and independent $N(0, 1)$ variables $X, Y$ .
I have the following: $\newcommand{\Cov}{\operatorname{Cov}}$ $\newcommand{\Var}{\operatorname{Var}}$ \begin{align*} \Cov(aX + bY, cX + dY)&= E\left[(aX + bY)(cX + dY)\right] - E(aX+bY)E(cX+dY)\\ &= E\left(acX^2 +(ad + bc)XY + bdY^2\right) - (aEX+bEY)(cEX+dEY)\\ &= acE\left(X^2\right) + (ad + bc)E(XY) + bdE(Y^2) \\&\phantom{=}- \left[ac(EX)^2 + (ad + bc)EXEY + bdE\left(Y^2\right)\right]\\ &=ac\Var X + bd\Var Y + (ad + bc)\left[E(XY) - EXEY\right]\\ &= ac + bd \end{align*} where the last step follows from the fact that $X$ and $Y$ are independent $N(0,1)$ variables. Then, \begin{align*} \rho(aX + bY, cX + dY) &= \frac{ac + bd}{\sqrt{\Var(aX + bY)\Var(cX + dY)}}\\ &= \frac{ac + bd}{\sqrt{(a^2\Var X + b^2\Var Y)(c^2\Var X + d^2\Var Y)}}\\ &= \frac{ac + bd}{\sqrt{(a^2+ b^2)(c^2+ d^2)}}\\ \end{align*}
Is there any way to simplify this last expression further? For example to remove the square root, etc.