For some time I have been studying the recurring role of half-angle formulas in geometry, rational parametrization, and integration. Some of that background is collected in The Ubiquitous Half-Angle Formulas (PDF link). Henning Dathe’s Die Halbwinkelsubstitution und ihre Anwendungen (PDF link) gives a broad treatment of the tangent half-angle substitution and its applications. A related paper, A Unified Substitution Method for Integration (see arXiv:2505.03754) (see also the trigonometric/hyperbolic formulas for the roots of quadratic equations that gave rise to the core identities of this method, and how the tangent of half-angle emerges here as well!), develops a branch-conscious framework in which several classical substitutions and identities arise from a common pair of principal-branch exponential identities. It recovers Euler’s first and second substitutions, after the necessary scaling, sign choices, and component-dependent reciprocal reparametrizations and obtains the classical Weierstrass substitution $t=\tan(\omega/2)$ as a local unit-radius corollary of Transform 5. The same identities also contain two notable geometric specializations. For the Cayley/tangent-half-angle form, take the positive hyperbolic parametrization $y=\cosh u$. Then $\arccos(\cosh u)=iu$ and
$$\tan\!\left(\tfrac12\operatorname{arcsec}(\cosh u)\right)=\tanh(u/2),$$so the exterior identity becomes
$$e^{-u}=\frac{1-\tanh(u/2)}{1+\tanh(u/2)}.$$Continuing by $u=i\phi$, with $\tanh(i\phi/2)=i\tan(\phi/2)$, gives the classical Cayley half-angle representation
$$e^{-i\phi}=\frac{1-i\tan(\phi/2)}{1+i\tan(\phi/2)},\qquad e^{i\phi}=\frac{1+i\tan(\phi/2)}{1-i\tan(\phi/2)}.$$This follows directly from Theorems 3.1–3.2 of the arXiv draft and the branch computation used in §4; the derivation is isolated explicitly in this Cayley-type half-angle note. For Lobachevsky’s angle-of-parallelism formula, the same positive hyperbolic parametrization $y=\cosh u$ in Theorem 3.1 identifies $\beta=\operatorname{arccsc}(\cosh u)=\Pi(u)$ and yields
$$\tan\!\left(\frac{\Pi(u)}2\right)=e^{-u};$$equivalently, Theorem 3.2 gives the same result using the reciprocal parametrization $y=\operatorname{sech} u$, for which $\psi=\arcsin(\operatorname{sech} u)=\Pi(u)$. Both derivations are given explicitly in §4, “Lobachevsky’s formula as a geometric specialization,” of the arXiv draft.
The transforms also simultaneously rationalize the associated quadratic radicals and inverse-trigonometric half-angle composites, and for several natural families cancellation between the parametrized radical and the Jacobian reduces the transformed integrand all the way to a finite Laurent polynomial, allowing term-by-term integration (in particular, some quotient-radical integrals collapse directly to Laurent polynomials and require no partial-fraction decomposition; see the arXiv draft, especially Example 4 and the termwise-integration remark).
Recent benchmarks against Mathematica’s native `Integrate` show the strongest gains in the most difficult half-angle families: on mutually solved Transform 1 cases, native `Integrate` was about $194\times$ slower in cold timing and produced expressions about $753\times$ larger on average; for Transform 4 it was about $110\times$ slower cold, with mean closed-form output more than $100\times$ larger and one native result reaching roughly $2.96$ MB. Across the five benchmark families, USM solved and verified every tested example in Transforms 2–5 and produced 95 non-timeout results out of 100 in Transform 1.
The point of this background is not that any of it is needed for the argument below, but rather to illustrate the remarkable range and structural power of half-angle formulas. Against that background, it is especially intriguing to encounter $\tan(\theta/2)$ once again, this time in a formulation equivalent to the Riemann Hypothesis. That recurrence is what made the observation below worth recording.
I want to emphasize that I am not claiming a proof of the Riemann Hypothesis. The RH argument below was generated by ChatGPT 5.6 (OpenAI) in response to my prompts (so take it with a grain of salt).
Let
$$\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s),$$and let $\lambda_n$ denote the Li coefficients. Li’s criterion states
$$\mathrm{RH}\iff \lambda_n\ge 0\qquad\text{for every }n\ge1.$$The Li coefficients also have, in the standard symmetric limiting sense, the zero-sum representation
$$\lambda_n=\sum_\rho\left[1-\left(1-\frac1\rho\right)^n\right].$$Consider the Möbius transformation
$$t(s)=1-\frac1s=\frac{s-1}{s}.$$For finite $s$,
$$\operatorname{Re}(s)=\frac12 \iff |t(s)|=1.$$Thus, assuming RH and writing
$$\rho=\frac12+i\gamma,\qquad \gamma>0,$$we may write
$$1-\frac1\rho=e^{i\theta_\gamma},\qquad 0<\theta_\gamma<\pi.$$A direct calculation gives
$$\cos\theta_\gamma=\frac{4\gamma^2-1}{4\gamma^2+1},\qquad\sin\theta_\gamma=\frac{4\gamma}{4\gamma^2+1},$$and hence
$$\tan\frac{\theta_\gamma}{2}=\frac{\sin\theta_\gamma}{1+\cos\theta_\gamma}=\frac1{2\gamma}.$$Equivalently,
$$\gamma=\frac1{2\tan(\theta_\gamma/2)}.$$Under RH, a conjugate pair $\frac12\pm i\gamma$ contributes
$$2[1-\cos(n\theta_\gamma)]$$to the Li zero sum. If $N(T)$ is the usual Riemann–von Mangoldt zero-counting function, counting upper-half-plane zeros with multiplicity, define
$$\theta(T)=2\arctan\frac1{2T},\qquad g_n(T)=2[1-\cos(n\theta(T))].$$Then the proposed derivation is
$$\lambda_n=\int_0^\infty g_n(T)\,dN(T).$$Stieltjes integration by parts gives
$$\lambda_n=-\int_0^\infty N(T)g_n'(T)\,dT,$$since the boundary term vanishes using
$$N(T)=O(T\log T),\qquad g_n(T)=O_n(T^{-2}).$$Since
$$\theta'(T)=-\frac4{4T^2+1},$$this becomes
$$\lambda_n=8n\int_0^\infty\frac{N(T)\sin(n\theta(T))}{4T^2+1}\,dT.$$Finally, with
$$T=\frac1{2\tan(\theta/2)}=\frac12\cot\frac\theta2,$$we have
$$\frac{dT}{4T^2+1}=-\frac14\,d\theta,$$and therefore, under RH,
$$\boxed{\lambda_n=2n\int_0^\pi N\left(\frac1{2\tan(\theta/2)}\right)\sin(n\theta)\,d\theta}\tag{*}$$for every $n\ge1$.
The integral appears to converge: as $\theta\to0$,
$$\frac1{2\tan(\theta/2)}\sim\frac1\theta,$$so
$$N\left(\frac1{2\tan(\theta/2)}\right)=O\left(\frac{\log(1/\theta)}{\theta}\right),$$while $\sin(n\theta)\sim n\theta$. Thus the integrand is
$$O_n(\log(1/\theta)),$$which is integrable at $0$.
The proposed converse is the following. Define
$$F(\theta)=N\left(\frac1{2\tan(\theta/2)}\right),\qquad 0<\theta<\pi.$$Because $N(T)$ is nonnegative and nondecreasing while $\theta\mapsto1/(2\tan(\theta/2))$ is decreasing, $F$ is nonnegative and nonincreasing.
The elementary lemma being used is:
If $F:(0,\pi)\to[0,\infty)$ is nonincreasing and the improper integral converges, then
$$\int_0^\pi F(\theta)\sin(n\theta)\,d\theta\ge0$$Indeed, putting $h=\pi/n$ and translating the $n$ intervals of length $h$ back to $(0,h)$ gives
$$\int_0^\pi F(\theta)\sin(n\theta)\,d\theta=\int_0^h\left[\sum_{j=0}^{n-1}(-1)^jF(x+jh)\right]\sin(nx)\,dx.$$For fixed $x\in(0,h)$,
$$F(x)\ge F(x+h)\ge F(x+2h)\ge\cdots\ge0,$$so the alternating sum is nonnegative, and $\sin(nx)\ge0$ on $(0,h)$.
Consequently, if $(*)$ holds for the actual Li coefficients for every $n\ge1$ without assuming RH, then
$$\lambda_n\ge0$$for every $n$, and Li’s criterion implies RH. Thus
