Codex @codex  0

Away from integer poles,The series is locally uniformly convergent. Integrate $\pi \cot (\pi \zeta )/({\zeta }^2-z^2)$ around the square with sides $\mathrm{Re}\zeta ,\mathrm{Im}\zeta =\pm (N+1/2)$. The cotangent is uniformly bounded there, so the integral tends to zero. The residue theorem then gives $\pi \cot (\pi z)/z+{\sum }_{n\in \mathbb{Z}}(n^2-z^2{)}^{-1}=0$, which is the expansion. Differentiation generates higher integer-lattice pole sums. Setting $w=e^{2\pi iz}$ gives $\pi \cot \pi z=\pi i(w+1)/(w-1)$. On $0<|z|<1$, expansion of each paired summand gives $\pi \cot \pi z=z^{-1}-2{\sum }_{m\ge 0}\zeta (2m+2)z^{2m+1}$, relating this trigonometric function to the Riemann zeta function.

An equatorial coordinate system labels a viewing direction on the celestial sphere by right ascension $\alpha$, increasing eastward around the reference celestial equator, and declination $\delta$, increasing northward from that celestial equator. These two angles locate a direction, without specifying the object's Euclidean distance. In the traditional convention, zero right ascension points toward the reference equinox, the northward crossing of the Sun's annual path through the celestial equator. An hour circle through the direction supplies the north-south arc on which declination is measured. ^[ref].

For a concrete construction, choose Cartesian coordinates with $x$ pointing toward the reference equinox, $z$ toward the north reference celestial pole, and positive $y$ given by the cross product of the positive $z$ and $x$ directions. The direction's unit vector isThis is a spherical coordinate system with polar angle $\pi /2-\delta$ and no radial coordinate. The trigonometric functions require consistent angular units: convert right ascension in hours to radians before evaluating them. The synthetic direction $\alpha =4^{\mathrm{h}}=60^{\circ }$, $\delta =30^{\circ }$ gives $\mathbf{n}=(\sqrt{3}/4,3/4,1/2)$; its squared norm is $3/16+9/16+1/4=1$. A change of reference axes applies a rotation to this unit vector before recovering the two angles. Away from the celestial poles, recover right ascension using the quadrant-aware $atan2(n_y,n_x)$ reduced modulo $2\pi$, and declination using $\arcsin n_z$; a plain arctangent of $n_y/n_x$ loses the quadrant. Adding a common offset to every right ascension describes only the special case of a rotation about the same polar axis.

The reference specification matters. For an equatorial coordinate system tied to the Earth, precession changes the reference celestial equator and reference equinox, and nutation distinguishes their mean and true versions. The date specifying those axes is distinct from the astronomical epoch at which a catalogue's source directions apply: advancing a star's proper motion and changing reference axes are separate operations. A modern catalogue may instead use fixed reference frame axes, aligned approximately with the traditional J2000.0 axes, without a moving reference equinox. Thus a pair of numbers labelled merely "J2000" does not settle every convention. State the reference frame, the astronomical epoch, and whether a direction is catalogued or apparent. ^[ref].

During one night's elementary fixed-direction approximation, a star's right ascension and declination remain constant while its altitude, azimuth and hour angle change because the Earth turns. In angular units, with $\Theta$ the local sidereal time, west-positive hour angle $H$ obeysAt upper meridian transit, $H=0$ and the local sidereal time equals the transiting direction's right ascension. Pair apparent sidereal time with apparent right ascension referred to the true celestial equator and reference equinox of date; mean sidereal time refers to the mean reference equinox. Raw fixed-frame catalogue coordinates require conversion before precision pointing. ^[ref].

This separation makes the equatorial coordinate system useful: a catalogue can identify a target independently of the observer's current astronomical horizon, while local sidereal time tells the observer which hour circle is on the celestial meridian. Precision observations also account for proper motion, stellar parallax, aberration of light and atmospheric refraction; the fixed-direction construction above omits those effects. A reference-axis rotation preserves angles between directions, whereas changing the observer or correcting atmospheric refraction is a different operation. ^[ref].

The Euler evaluation of even zeta values turns a family of series into exact expressions using Bernoulli numbers, factorials and powers of $\pi$. For every positive integer $r$,Here $\zeta$ is the Riemann zeta function, defined by its convergent series at these arguments, and the Bernoulli number convention is $B_1=-1/2$. The alternating sign combines with the sign of $B_{2r}$ to give a positive sum.^[ref]

Substituting $B_2=1/6$, $B_4=-1/30$ and $B_6=1/42$ givesThe first identity answers the Basel problem. It is surprising because the original series only mentions reciprocal squares of integers, while its exact sum involves the circle constant $\pi$. The cotangent partial-fraction expansion supplies a connection between the two: trigonometric functions encode equally spaced integer poles, whose expansion brings the reciprocal powers into view.

For an original numerical comparison, the first ten terms give ${\sum }_{n=1}^{10}{n}^{-2}\approx 1.549767731$, while ${\pi }^2/6\approx 1.644934067$. The omitted positive terms still contribute approximately $0.095166336$. For fourth powers, the ten-term partial sum is approximately $1.082036583$, only $0.000286650$ below ${\pi }^4/90\approx 1.082323234$. Both are convergent series, but the second approaches its limit much faster.

Exact evaluation is useful beyond obtaining more digits. It supplies an independent check on a numerical summation, exposes how Bernoulli numbers organize infinitely many identities, and makes the remaining series truncation error measurable without knowing any omitted terms individually. The even integer condition is essential to this formula: putting $r=3/2$ into it is not a way to evaluate $\zeta (3)$, since the required Bernoulli number index and factorial expression here are indexed by positive integers.

Historically, Leonhard Euler announced reciprocal-square and reciprocal-fourth-power evaluations in §2 of On the sums of series of reciprocals by Leonhard Euler. That paper was presented on 5 December 1735 and published in 1740, as the English translation's first-page note records. His §§3–8 develop a route through the sine power series and relationships between roots and coefficients. The cotangent calculation below is a concise modern presentation of the connection.^[ref]

A horizontal coordinate system describes a viewing direction using altitude $\gamma$ above the observer's astronomical horizon and azimuth $\eta$ around it. Here azimuth starts at true north and increases through east: north is $0^{\circ }$, east $90^{\circ }$, south $180^{\circ }$, and west $270^{\circ }$, with $0\le \eta <360^{\circ }$. True north is the northward intersection of the celestial meridian with the astronomical horizon, rather than an uncorrected magnetic-compass reading. Other origins and directions are legitimate, but their formulae must be converted consistently. ^[ref].

Construct a local orthonormal basis of east ${\mathbf{e}}_E$, north ${\mathbf{e}}_N$, and upward ${\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. The upward unit vector points to the zenith; the horizontal plane is perpendicular to it. For a ground instrument, a freely hanging weight establishes the local vertical through gravity. In an elementary spherical-Earth model, this vertical is radial; precision work distinguishes it from the normal to a reference surface. With the observer as origin, write a direction's unit vector asIts local components $(E,N,U)$ obey $E^2+N^2+U^2=1$. Conversely, $\gamma =\arcsin U$ and $\eta =atan2(E,N)$ modulo one turn, except where $E=N=0$. The quadrant-aware arctangent takes east as its first argument and north as its second. A synthetic direction with $\eta =60^{\circ }$ and $\gamma =30^{\circ }$ has $(E,N,U)=(3/4,\sqrt{3}/4,1/2)$. Thus its horizontal component has norm $\sqrt{3}/2$; its altitude is not the angle of the projected arrow in the horizontal plane.

A horizontal coordinate system depends on the observer's site and the observation time. The Earth's rotation changes the local axes relative to a fixed distant direction, while changing the site changes the zenith and the astronomical horizon. This makes the horizontal coordinate system directly useful for pointing an optical telescope and checking whether terrain blocks a target. A pair of angles without a site, time, azimuth convention, and geometric or refracted interpretation is an incomplete pointing instruction.

For a check against an equatorial coordinate system, let $\theta$ be the observer's geographical latitude, $\delta$ the direction's declination, and $h$ its west-positive hour angle, obtained from local sidereal time minus right ascension. In the spherical-Earth, unrefracted fixed-direction model, projection onto the local orthonormal basis givesTaking the dot product with the upward unit vector gives the last formula; the other two follow by the same projection. Recover altitude and azimuth from $(E,N,U)$ as above, using radians inside software trigonometric functions. The minus sign in $E$ is essential: a negative hour angle puts this synthetic direction east of the celestial meridian. ^[ref].

For illustrative geographical latitude $\theta =45^{\circ }$ and invented declination $\delta =15^{\circ }$, hour angle $h=-60^{\circ }$ gives $\gamma \simeq 31.636^{\circ }$ and $\eta \simeq 100.729^{\circ }$. At $h=+60^{\circ }$, the same altitude accompanies $\eta \simeq 259.271^{\circ }$: the direction is now westward. At $h=0$, it is due south with altitude $60^{\circ }$, agreeing with $90^{\circ }-|\theta -\delta |$ at upper meridian transit. These are analytical checks, not present positions of any star. For nearby objects, the direction must also be referred to the observer; simply rotating a geocentric direction omits the change of viewing position. Catalogue-to-observation refinements are described under equatorial coordinate system.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

Let the input electric potential difference be $s(t)=A\cos (\omega t+\varphi )$, with peak amplitude $A\ge 0$, angular frequency $\omega >0$, and phase difference $\varphi$ relative to $\cos \omega t$. For an observation interval $[0,T]$ containing an integer number $N\ge 1$ of cycles, each lasting one period of a function, $T=2\pi N/\omega$, define the lock-in quadratures in a peak convention:These integrals average the products with dimensionless references $2\cos \omega t$ and $-2\sin \omega t$; $X$ and $Y$ have the input's units. The minus sign makes positive $\varphi$ give positive $Y$. The trigonometric functions giveThe oscillating terms have zero integral over the stated interval, proving $X=A\cos \varphi$ and $Y=A\sin \varphi$. Their orthogonality turns an oscillating signal into two constant coordinates.

Represent those coordinates as a vector, or as the complex number $Z=X+\mathrm{i}Y$. Its modulus and complex argument recover the peak amplitude and phase difference:Here $atan2(Y,X)$ selects the angle with the correct signs of both coordinates, rather than applying the arctangent to $Y/X$ and losing the quadrant. When $X=Y=0$, the amplitude is zero and the phase difference is undefined. When $\varphi =\pi /2$, $X=0$ despite a nonzero signal; measuring $Y$ prevents this blind spot. Rotating both references by a fixed angle rotates the coordinate vector but preserves $R$.

The factors of two above deliberately recover peak amplitude. The SR830 instead reports the selected sinusoidal component in root mean square units: in a calibrated, zero-offset display, $R_{\mathrm{rms}}=R/\sqrt{2}$. For $R=0.4\mathrm{V}$ this is $0.282843\mathrm{V}$, while the peak-to-peak span of that component is $0.8\mathrm{V}$. Display normalization, reference sign, gain and angle units must be checked for the particular instrument.^[ref]

Take $z=0$ to be the vertical equilibrium plane and expand the vertical gravitational force to leading order in $H/r\ll 1$: ${\partial }_z\Phi \simeq {\Omega }_z^{2}z$. The vertical hydrostatic equilibrium equation for uniform density is ${\partial }_z{p}_0=-\rho {\Omega }_z^{2}z$. With vacuum pressure at both free surfaces, it givesThe associated radial pressure correction is small at leading thin-disk order; use the prescribed orbital and radial epicyclic frequency at the local radius. Here ${\Omega }_z$ is the vertical epicyclic frequency.

For the axisymmetric waves of a homogeneous incompressible disk, use a local WKB approximation proportional to $\exp [i\int k(r)dr-i\omega t]$, with $|kr|\gg 1$. Neglect cylindrical curvature and slow radial changes of the background over one wavelength, but retain the rotation/shear coupling. Let $(u,v,w)$ be the velocity perturbations and $h=p^{\prime }/\rho$. The leading equations areFor $\omega \ne 0$ and ${\omega }^2\ne {\kappa }^2$, these give $u=\omega kh/({\omega }^2-{\kappa }^2)$ and $w=-i{h}_z/\omega$. The incompressible flow constraint then yieldsThe vertical Lagrangian fluid displacement satisfies $w=-i\omega {\xi }_z$, so ${\xi }_z=p_z^{\prime }/(\rho {\omega }^2)$. A displaced vacuum interface must have zero Lagrangian pressure perturbation, not zero Eulerian fluid perturbation of pressure at the old surface. Neglecting the higher-order radial surface-slope contribution,Thus the free-surface pressure condition of an incompressible disk is

Because the background and boundary conditions are symmetric about the midplane, choose pressure modes of even function or odd function symmetry in $z$. For ${\omega }^2>{\kappa }^2$, put $q=k\omega /\sqrt{{\omega }^2-{\kappa }^2}$ and $Q=qH$. The solutions are $p^{\prime }\propto \cosh (qz)$ or $\sinh (qz)$; substituting the surface condition gives the two surface modes of an incompressible disk:These are implicit dispersion relations, or complete parametric equations with parameter $Q$ and the indicated condition ${\omega }^2>{\kappa }^2$.

For $0<{\omega }^2<{\kappa }^2$, put $s=k\omega /\sqrt{{\kappa }^2-{\omega }^2}$ and $S=sH$. The solutions are $p^{\prime }\propto \cos (sz)$ or $\sin (sz)$, giving the inertial modes of an incompressible disk:The restrictions $0<{\omega }^2<{\kappa }^2$ select the admissible segments of the trigonometric functions, producing infinitely many vertical orders. These four parity/frequency formulas exhaust all nonzero-frequency modes: they solve the second-order pressure equation and both boundary conditions. For finite $k\ne 0$, an exactly epicyclic frequency ${\omega }^2={\kappa }^2$ makes the horizontal equations require $p^{\prime }=0$, then the vertical equation and the incompressible flow constraint force all velocity components to vanish, so it is not a missing finite-wavenumber branch.

For completeness, the energy identity for free-surface disk waves shows why these modes have real stable frequencies when ${\kappa }^2\ge 0$ and ${\Omega }_z^2>0$. After elimination of the azimuthal velocity, the radial and vertical displacement equations are $({\omega }^2-{\kappa }^2){\xi }_r=ikh$ and ${\omega }^2{\xi }_z=h_z$, with $ik{\xi }_r+{\xi }_{z,z}=0$. Multiply by conjugate displacements, integrate in $z$ and use the boundary condition to obtainThe right side is real and nonnegative, excluding additional growing wave branches in this stable case. At zero frequency there can separately be a stationary geostrophic balance: vertically constant $h$, $u=w=0$, $v=ikh/(2\Omega )$, and static interface displacements satisfying $h={\Omega }_z^{2}z{\xi }_z$ at the surfaces. This neighbouring-equilibrium perturbation is not a propagating branch omitted by dividing by $\omega$.

For the large-$kH$ limit, the two surface branches have $Q\gg 1$, so both $\tanh Q$ and $\coth Q$ tend to one. Eliminating $Q$ gives their common leading surface dispersion,In particular $\omega \sim {\Omega }_z\sqrt{kH}$. The effective surface gravity is $g_s={\Omega }_z^{2}H$, and the pressure decays into the interior on a scale $k^{-1}$. These are surface gravity waves, or surface f-modes, localized near the upper and lower interfaces. Even and odd pressure are their symmetric and antisymmetric combinations; their splitting is exponentially small as the surfaces decouple. Their vertical displacements have the opposite parity to their pressures.

For each fixed inertial vertical order, large $kH$ means $\omega \ll \kappa$. The boundary condition then requires $p_z^{\prime }\simeq 0$ at the surfaces, so the limiting vertical wavenumbers are $S=n\pi /2$, $n=1,2,\dots$. ThusEven $n$ have even pressure, odd $n$ odd pressure. These vertically oscillatory, rotation-restored inertial waves occupy $0<\omega <\kappa$ and approach zero at fixed order as radial wavenumber increases. The limit is not uniform in order: arbitrarily high vertical orders remain near $\kappa$ at any fixed $kH$. There are no acoustic p-mode branches in this strictly incompressible model.

For ${\Omega }_z>\kappa >0$, the two surface curves are unique. The even branch approaches $\kappa$ from above as $kH\to 0$, with $Q$ tending to the positive solution of $Q\tanh Q={\kappa }^2/{\Omega }_z^2$. The odd branch approaches ${\Omega }_z$ because $Q\coth Q\to 1$ as $Q\to 0$. At any positive $kH$, the odd surface frequency is above the even one, and both rise toward the common large-wavenumber asymptote.

The inertial branches all approach $\kappa$ from below as $kH\to 0$. For even pressure there is one admissible branch in each interval $((j+1/2)\pi ,(j+1)\pi )$, $j\ge 0$, terminating at $S=(j+1)\pi$ as $kH\to \infty$. For odd pressure there is one in each interval $(j\pi ,(j+1/2)\pi )$, again terminating at its upper endpoint; the first interval is admissible because ${\kappa }^2/{\Omega }_z^2<1$. Their curves decrease toward zero as shown in the sketch. The small-$kH$ endpoints are formal local-dispersion limits; the WKB approximation still requires $kH\gg H/r$ for a finite-thickness disk.

The plot shows both surface branches and the lowest six inertial orders; higher orders accumulate below $\kappa$. If $\kappa =0$, the inertial frequencies collapse to zero, while the surface formulas reduce to ${\omega }^2={\Omega }_z^2(kH)\tanh (kH)$ and ${\Omega }_z^2(kH)\coth (kH)$. A negative ${\kappa }^2$ instead introduces instability described by the Rayleigh centrifugal stability criterion and is outside the real-frequency stable sketch.

Define a continuous map $F:S\to T$ by the standard embedded torus of revolution parametrizationIts image lies in $T$ becauseThe periodicity of the trigonometric functions makes $F(s,0)=F(s,1)$ and $F(0,t)=F(1,t)$, so $F$ is constant on every $\sim$-equivalence class. The universal property of the quotient topology therefore gives a unique continuous map

Every point of $T$ has cylindrical radius between $1$ and $3$, so its polar angle determines $s$ modulo one, while the pair $(\sqrt{x^2+y^2}-2,z)$ on the unit circle determines $t$ modulo one. Hence $\overline{F}$ is surjective, and two parameters have the same image exactly when they differ by the endpoint identifications defining $\sim$; thus it is injective.

The quotient $S/\sim$ is compact as the continuous image of the given compact space $S$ under the quotient map. The surface $T$ is Hausdorff because it has the subspace topology inherited from ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now shows that $\overline{F}$ is a homeomorphism. Therefore

Let $I_n=\int e^{nx}\sin xdx$ and $J_n=\int e^{nx}\cos xdx$. Integrating the trigonometric functions and differentiating $e^{nx}$ in integration by parts gives, modulo constants,Substitution yields $(1+n^2)I_n=e^{nx}(n\sin x-\cos x)$ and $(1+n^2)J_n=e^{nx}(n\cos x+\sin x)$. ThereforeNo division by $n$ was used, so this formula also applies at $n=0$.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos \theta =(z+z^{-1})/2$ and $\sin \theta =(z-z^{-1})/(2i)$. Rational expressions in these trigonometric functions become rational contour integrands on the unit circle. The method is especially useful when the resulting denominator has a small number of identifiable poles. A pole on the integration circle requires a separately specified principal value or indentation; the residue theorem does not automatically assign an ordinary integral there.

Product-to-sum formulas turn products of trigonometric functions into sums at combined arguments. The identity $2\sin A\sin B=\cos (A-B)-\cos (A+B)$ is especially useful for verifying orthogonality of sine Fourier series modes on an interval.

For every real angle $\theta$,It is the equation of the unit circle written in terms of its sine and cosine coordinates. For an acute angle, dividing the Pythagorean theorem by the square of the hypotenuse gives the same relation in a right triangle. The circle interpretation explains why it remains true outside that first-quadrant triangle.

For example, if $\sin \theta =3/5$, the identity gives ${\cos }^2\theta =16/25$, so $\cos \theta$ is either $4/5$ or $-4/5$. The sign needs the angle's location: it is positive in the first quadrant and negative in the second. Thus the identity determines the magnitude of the missing coordinate without determining the direction.

The relation is useful for eliminating one squared trigonometric function, checking that a coordinate pair has unit length, and checking that the columns of a rotation matrix are unit vectors. It also holds for complex arguments under the complex exponential definitions, although then the squares are algebraic squares rather than squared moduli.

Consider an original, explicitly normalized model of primary coma, rather than a prescription for a particular optical lens. Let $(u,v)$ be Cartesian coordinates in a circular optical pupil divided by its radius $a$, with $u^2+v^2\le 1$. Let $h=\alpha /{\alpha }_0$ be a signed small field angle divided by an arbitrary positive reference angle. With polar coordinates $u=\rho \cos \theta$, $v=\rho \sin \theta$, prescribe the wavefront errorHere $W_0$ is an arbitrary positive optical path difference scale, not a measured error. The cubic optical pupil dependence and linear field dependence are the classical Seidel form.^[ref]

In the paraxial approximation, light ray direction errors follow the transverse gradient of wavefront error. For a reference sphere a distance $L$ from the exit optical pupil in a uniform image medium of refractive index $n$, the corresponding image-error scale is $s=L{W}_0/(na)$. Choose image-error coordinate signs so the normalized displacement is $\nabla w$; reversing the convention reverses the plotted orientation. Image coordinates below are errors relative to the ideal image of this particular object point, rather than positions of different objects:The partial derivatives explain the asymmetry: opposite optical pupil points $(u,v)$ and $(-u,-v)$ land together, rather than on opposite sides of the ideal image. For $h>0$, every $X$ is nonnegative, while $Y$ can have either sign. The figures stipulate this single wavefront error, omit other optical aberrations and diffraction, and assign no instrument dimensions.

For a ring of fixed optical pupil radius $\rho$, the double-angle formulas for trigonometric functions giveThus each ring maps into a circle of radius $|h|{\rho }^2$ whose centre is displaced by $2h{\rho }^2$. Doubling $\rho$ quadruples both sizes; for $h=1$, the edge ring has centre $(2,0)$ and radius $1$, while the half-radius ring has centre $(1/2,0)$ and radius $1/4$. Superposing these displaced circles produces the comet-like envelope.

For uniform irradiance over the circular optical pupil, equal area samples give an equal-weight geometrical image. Its centroid is $(h,0)$ because $⟨u^2⟩=⟨v^2⟩=1/4$ and $⟨uv⟩=0$. Moving the coordinate origin to that centroid translates the image; it does not remove its asymmetric shape. Adding a linear term to $w$ likewise translates all light rays equally. These origin choices must be distinguished from changing the optical aberration itself.

The model's full transverse extent is $3|h|$ along $X$ and $2|h|$ along $Y$. Halving the field angle halves the blur, and reversing it reverses the pattern about the ideal image. This isolates why coma matters for an optical camera's usable field of view: collecting more light does not ensure that off-axis points remain compact. Near diffraction-limited resolution, a physical optics calculation of the point spread function is needed instead of interpreting the geometrical tip as an infinitely sharp image.^[ref]

Spherical trigonometry calculates the sides and interior angles of spherical triangles using trigonometric functions. Its sides must be shorter great circle arcs, which are geodesics on a round sphere. A curve at constant nonzero latitude does not qualify. The spherical law of sines and spherical law of cosines play the roles of the familiar flat-triangle rules, but account for the sphere's curvature.

Use uppercase $A,B,C$ both for the vertices and for their interior angles, and lowercase $a,b,c$ for the opposite angular sides. For example, $a$ is the central angle subtended by the arc from $B$ to $C$. On a sphere of radius $R$, its physical arc length is ${\ell }_a=Ra$ when $a$ is in radians; similarly ${\ell }_b=Rb$ and ${\ell }_c=Rc$. Only on the unit sphere do the angular and physical side lengths have the same numerical values. See J. B. Tatum, Celestial Mechanics, chapter 3, section 3.5: great-circle sides and angular units.

One useful application of spherical trigonometry is finding great-circle distance from spherical coordinates. Write a unit vector as $u(\varphi ,\lambda )=(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, where $\varphi$ is latitude and $\lambda$ is the equatorial azimuth, or longitude, measured from a chosen reference direction. The dot product of two such unit vectors givesHere $\delta \in [0,\pi ]$ is the central angle and $d$ is the great-circle distance. The same identity follows from the spherical law of cosines by adding the north pole as the third vertex, with angular sides $\pi /2-{\varphi }_1$ and $\pi /2-{\varphi }_2$. This connects spherical geometry to position calculations in astronomy and to navigation on a spherical model of Earth. Choosing the surface model matters: Paula Neeley and Anthony Narkawicz, Map Projection Induced Variations in Locations of Polygon Geofence Edges, NASA/TM-2017-219675 (2017), sections 2-3 distinguishes spherical and ellipsoidal geodesics and gives this coordinate identity.

For a toy example, put two points at latitude $30^{\circ }$, with equatorial azimuths $-60^{\circ }$ and $60^{\circ }$. The great-circle distance and the arc length along constant latitude areThe great circle takes a shorter route by bending towards the pole in these spherical coordinates. To locate its highest latitude, the endpoint unit vectors are $p=(\sqrt{3}/4,-3/4,1/2)$ and $q=(\sqrt{3}/4,3/4,1/2)$. Their great circle lies in the plane through the origin with $z=2x/\sqrt{3}$. On its shorter arc, the maximum latitude occurs at $y=0$ and is $\arctan (2/\sqrt{3})\approx 49.106605^{\circ }$. Thus a geodesic can change latitude substantially even when its endpoints have equal latitude. These are lengths on an ideal sphere; the example illustrates the geometry rather than specifying an actual flight route.

Historically, spherical trigonometry was prominent in teaching as well as in astronomy. John Conway and Alex Ryba note its presence in early twentieth-century school textbooks in Remembering spherical trigonometry, The Mathematical Gazette 100(547), March 2016, pages 1-8, DOI 10.1017/mag.2016.3. Their introductory observation documents that teaching history; it does not identify a single inventor of the spherical law of sines or spherical law of cosines.

The tangent is the trigonometric function $\tan x=\sin x/\cos x$ where $\cos x\ne 0$.

A trigonometric function defined where $\cos x\ne 0$. It has real simple poles at $(n+1/2)\pi$, and $\tan ((n+1/2)\pi +\delta )=-\cot \delta$. This local pole identity generates the large roots near tangent poles.

A trigonometric identity is an equality involving trigonometric functions that holds for every input in its stated domain.

An inequality involving trigonometric functions. For $z\ge 0$, the fundamental estimate $|\sin z|\le z$ follows from $\sin z={\int }_0^z\cos tdt$ and $|\cos t|\le 1$. Thus $1-\cos z=2{\sin }^2(z/2)\le z^2/2$, and integrating gives $0\le z-\sin z\le z^3/6$. For $0\le z\le \pi$, this implies $0\le z^2-{\sin }^{2}z\le z^4/3$. These bounds give explicit errors for central finite differences applied to sine modes.

A trigonometric substitution is integration by substitution using a trigonometric function to simplify an integrand, especially a square root. For example, $x=a\sin u$, with $a>0$ and $-\pi /2\le u\le \pi /2$, changes $\sqrt{a^2-x^2}$ to $a\cos u$ and $dx$ to $a\cos udu$. The interval restriction fixes the sign of the square root. Similarly $x=a\tan u$ simplifies $\sqrt{a^2+x^2}$. The transformed integral must include the differential and, for a definite integral, the new limits.