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I wouldn't really call this sophisticated, but maybe not so far from the 20th century. There are several half-angle formulas for tangent, among them: $$ \tan \frac{x}{2}=\frac{\sin x}{1+\cos x}=\frac …

Write $a_k$ for your integral. If we define $g(s)=\zeta(s)-\frac{1}{s-1}$, then $\left(\frac{d}{ds}\right)^n|_{s=1}g(s)=(-1)^n\gamma_n$. Your observation can be written $a_k=(-1)^k\left(\frac{d}{ds} …

More generally, there is an equality of power series in two variables $$ \prod_\limits{m=0}^\infty \dfrac{1}{(1-t x^{5m+1})}=\sum_\limits{i=0}^\infty \dfrac{t^i x^i}{\prod_\limits{j=1}^i (1-x^{5j})}. …

There is an isomorphism $$ \begin{align*} \varphi:G_{(p,q)}&\to\mathbb{Z}[1/q],\\ g_i&\mapsto\frac{p^i}{q^i}. \end{align*} $$ To check surjectivity: for any $\frac{a}{q^n}\in\mathbb{Z}[1/q]$, we can …

For (Q1): A finite dimensional $\mathbb{C}$-algebra $A$ is Artinian, so $A$ is a product of Artin local algebras. The units of a product of algebras is the product of the units, so we may assume $A$ i …

Here's one way to derive the identity: Suppose $\tan(\gamma)=\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)+\cos(\beta)}$. Multiplying this equation through by $\cos(\gamma)$ gives an expression for $\s …