Is $\pi/\sqrt{2}$ transcendental? I believe that $\frac{\pi}{\sqrt{2}}$ is transcendental but I'm not sure about how to prove it. If $\frac{\pi}{\sqrt{2}}$ was algebraic, there would exist a polynomial $P \in \mathbb{Q}[X]$ such that $P\big( \frac{\pi}{\sqrt{2}} \big) = 0$. By writing $P = \displaystyle \sum_{k=0}^{N} a_{k}X^{k}$, we have :
$$ \sum_{\substack{\text{k even} \\[1mm] k = 2p}} a_{2p} \frac{\pi^{2p}}{2^{p}} +  \frac{\pi}{\sqrt{2}} \sum_{\substack{\text{k odd} \\[1mm] k = 2p+1}} a_{2p+1} \frac{\pi^{2p}}{2^p} = 0. $$
But I'm not sure that helps..
 A: Suppose that $\alpha:=\frac{\pi}{\sqrt{2}}$ would be a root of a $f\in \mathbb Q[X]$.
Then consider $\mathbb Q(\alpha,\sqrt{2})$. This is a field with finite degree over $\mathbb Q$. Since $\pi$ is also contained in this field, $\pi$ must be a root of a polynomial over $\mathbb Q$.
But this would contradict the well-known fact that $\pi$ is transcendental.
A: An elementary argument that $\pi$ transcendental implies $\frac{\pi}{\sqrt{2}}$ transcendental goes like this:
If $\;\frac{\pi}{\sqrt{2}}\;$ is algebraic, then it is a root for some polynomial $P(x)$ over $\mathbb{Q}$.  
This in turn implies $\pi$ is a root of polynomial $\displaystyle\;Q(x) = P\left(\frac{x}{\sqrt{2}}\right)P\left(-\frac{x}{\sqrt{2}}\right)$
Splitting $P(x)$ into its even and odd parts, there exists $E(x), O(x) \in \mathbb{Q}[x]$ such that
$$P(x) = E(x^2) + x O(x^2)
\quad\implies\quad Q(x) = E\left(\frac{x^2}{2}\right)^2 - \frac{x^2}{2}O\left(\frac{x^2}{2}\right)^2$$
This means $Q(x)$ also belongs to $\mathbb{Q}[x]$ and this contradicts with the assumption that $\pi$ is transcendental!
A: Suppose $\dfrac{\pi}{\sqrt{2}}$ were algebraic. Then $\left(\dfrac{\pi}{\sqrt{2}}\right)^2=\dfrac{\pi^2}{2}$ would be algebraic, but this would mean $\pi^2$ is algebraic, which is a contradiction since $\pi$ is transcendental. 
