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The standard arcsine distribution is a continuous probability distribution on $[0,1]$ with probability density functionIt is the beta distribution with both parameters equal to $1/2$. The density grows without bound toward both endpoints, yet neither endpoint has any probability mass. Its name comes from the inverse sine in its cumulative distribution function.^[ref]

A useful way to see the shape is to choose a point uniformly around a circle and project it onto a diameter. Equal angular intervals do not give equal horizontal distances: the projection moves slowly near either end. An ideal harmonic oscillator sampled at a uniformly chosen time in one complete period has the same endpoint concentration after rescaling its position. The distribution translates a simple geometric slowdown into a precise probability calculation.

For an original comparison, take equal-length regions: the middle half $[1/4,3/4]$, and the two outside quarters together. ThenA uniform distribution on $[0,1]$ instead assigns each region probability $1/2$. Both distributions have mean $1/2$; the mean alone hides their different concentration.

The beta distribution is an absolutely continuous probability distribution for a random variable $X$ between zero and one. It can describe uncertainty about a probability, such as an unknown coin's chance of heads. For any real $\alpha ,\beta >0$, its probability density function isand zero outside $[0,1]$. The normalizer is the Euler beta function, related to the gamma function by the beta-gamma identity:Thus the probability density function integrates to one. ^[ref].

The endpoint behavior explains why the restrictions are exactly $\alpha ,\beta >0$: near zero the probability density function behaves like a constant times $x^{\alpha -1}$, and near one like a constant times $(1-x{)}^{\beta -1}$. These improper integrals converge for positive parameters, even when the probability density function diverges at zero for $\alpha <1$ or at one for $\beta <1$. There is no probability mass at either endpoint: $\mathbb{P}(X=0)=\mathbb{P}(X=1)=0$. A probability density function's height is not a probability; an interval's probability is the area under it.

Some useful examples of the beta distribution are:

The arcsine distribution curve is plotted only for $0.001\le x\le 0.999$; its finite plotted ends do not represent finite endpoint values of the probability density function. Using the inverse sine, the shaded central interval has exact probability $\mathbb{P}(0.1<X<0.9)=1-4\arcsin (\sqrt{0.1})/\pi \approx 0.5903$, calculated analytically rather than by treating the cropped curve as normalized.

To separate location from spread, write $\kappa =\alpha +\beta$ and $m=\alpha /\kappa$. The mean and variance of the beta distribution areThe mean specifies the balance between the parameters; their sum $\kappa$ controls concentration about that mean. Multiplying both parameters by the same positive number preserves the mean and changes the variance. Larger $\kappa$ puts more area near $m$ and makes the probability density function taller there; the total area stays one. Small $\kappa$ allows much greater spread, sometimes favoring the endpoints rather than the mean itself.

For example, keeping mean $m=0.3$ and taking $\kappa =5,20,80$ gives the beta distributions $\mathrm{Beta}(1.5,3.5)$, $\mathrm{Beta}(6,14)$, and $\mathrm{Beta}(24,56)$. Their variances are $0.035$, $0.01$, and $0.21/81\approx 0.00259259$, respectively. Parameters need not be integers.

The formulas follow from two short moment calculations. Multiplying the probability density function by $x$ or $x^2$ increases the first parameter of the Euler beta function:Apply the Gamma function recurrence to the beta-gamma identity, then subtract $(\mathbb{E}[X]{)}^2$ from $\mathbb{E}[X^2]$ to obtain the variance.

The beta distribution is especially convenient as a prior distribution for the unknown success probability $P$ of Bernoulli trials. The Beta-binomial conjugacy says that $h$ successes and $t$ failures update a prior distribution $\mathrm{Beta}(\alpha ,\beta )$ to a posterior distribution $\mathrm{Beta}(\alpha +h,\beta +t)$. For example, suppose a coin has prior distribution $\mathrm{Beta}(2,2)$ and we observe eight heads and two tails. Its posterior distribution is then $\mathrm{Beta}(10,4)$, with posterior mean $5/7$. ^[ref].

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

A diffraction order is the integer $m$ labeling a principal constructive interference direction of a diffraction grating: adjacent elements differ in path length by $m$ wavelengths. In the normal-incidence transmission geometry described above,Positive and negative diffraction orders lie on opposite sides of the forward normal. The condition expresses the range of the sine of a real angle, so it also limits the propagating diffraction orders.

For $\lambda =600\mathrm{nm}$ and $d=2\mu \mathrm{m}$, $\sin {\theta }_m=0.3m$. Taking the inverse sine givesThese are seven allowed propagating diffraction orders. Order $m=4$ would require $\sin {\theta }_4=1.2$, and $m=-4$ would require $-1.2$, so neither has a propagating far-field direction. The allowed diffraction orders are equally spaced in $\sin \theta$, while their actual angles become farther apart toward grazing propagation.

Each principal peak has finite width. For the ideal uniform coherent array, the nearest zeros around order $m$ obeyThus the one-sided distance from the order direction to the first zero is $\Delta (\sin \theta )=\lambda /(Nd)$. Near the first diffraction order in our example, $N=10$ puts the zeros at $0.27$ and $0.33$, while $N=40$ puts them at $0.2925$ and $0.3075$. The respective one-sided widths are $0.03$ and $0.0075$; the full zero-to-zero widths are $0.06$ and $0.015$. These are calculated coherent peak widths, with the slit envelope omitted, and do not describe a measured instrument response. ^[ref].

Four times as many illuminated slits therefore give a fourfold narrowing in this coordinate without moving the diffraction order. That sharper directional selection helps distinguish nearby wavelengths in a spectrograph. Actual performance also depends on illumination, optical coherence, and source width; the separate slit-limited resolving power of a grating describes the contribution of the entrance slit.

The finesse of a Fabry–Pérot interferometer is its free spectral range divided by a transmission peak's full width at half maximum. Large finesse means a narrow accepted frequency interval compared with the spacing between repeated orders. Here half maximum means $T=1/2$ of the ideal unit peak, with zero detector background subtracted; it does not mean halfway between the transmission minimum and maximum.

Write $A=4R/(1-R{)}^2$ for the coefficient multiplying ${\sin }^2(\delta /2)$ in Fabry–Pérot transmission from multiple reflections. This coefficient is sometimes called the coefficient of finesse; it is not the finesse $\mathcal{F}$. About a resonance $\delta =2\pi m$, put $ϵ=\delta -2\pi m$. Solving $T=1/2$ givesand, since one free spectral range advances the wave phase by $2\pi$,These are exact for the ideal nondispersive model with constant $R$ and linear frequency dependence of $\delta$. The inverse sine requires $(1-R)/(2\sqrt{R})\le 1$, or $R\ge 3-2\sqrt{2}\simeq 0.171573$. At equality, neighbouring peaks merely touch half maximum at their shared minima: the nominal width is one full free spectral range and $\mathcal{F}=1$. Isolated half-maximum peaks require the strict inequality. Below this threshold the transmission never falls to $1/2$, so this peak-width definition supplies no finesse; at $R=0$ the transmission is flat.

For large reflectivity, the small-angle approximation $\arcsin x\simeq x$ givesThis is a high-reflectivity approximation, not an identity for every $R$. The original dimensionless examples below have exact full width at half maximum values obtained from the inverse sine formula.

The beauty of the narrowing is visible directly in the geometric series: large $R$ preserves contributions from many round trips, so a small phase difference accumulates across many terms and spoils their addition. For the stipulated $10\mathrm{mm}$ vacuum gap, $R=0.95$ gives an ideal full width at half maximum of about $244.792\mathrm{MHz}$. This instrumental width is not a measured spectral line width. A real scan also depends on the source profile, accepted angles, mirror shape, alignment, and scan stability. The University of Washington handout separately discusses mirror-reflectivity, mirror-flatness, and detector-pinhole contributions; increasing $R$ alone cannot remove the others.^[ref]

The inverse sine is one member of the inverse trigonometric functions.

Snell's law is $n_a\sin {\theta }_a=n_b\sin {\theta }_b$, with both angles measured from the interface normal vector. Let $\psi$ be the meridional ray's angle to the optical axis inside the core and let $n_0$ be the incident medium's refractive index. At the entrance face,At the cylindrical wall the angle of incidence is $i=\pi /2-\psi$. For $n_{\mathrm{inner}}>n_{\mathrm{outer}}$, the critical angle for total internal reflection satisfies $\sin i_c=n_{\mathrm{outer}}/n_{\mathrm{inner}}$. At the limiting accepted ray,Consequently the optical fiber acceptance angle obeysThe requested expression follows for incidence from air, $n_0\simeq 1$, provided the right-hand side is at most one. Strict total internal reflection occurs inside this limiting cone. The derivation uses the standard meridional ray model; skew launches need additional geometry.

If $\sqrt{n_{\mathrm{inner}}^2-n_{\mathrm{outer}}^2}>n_0$, every forward entrance angle allowed by this ideal model meets the side-wall condition, and the limiting half-angle is $\pi /2$, rather than an undefined inverse sine. For example, $n_{\mathrm{inner}}=1.6$, $n_{\mathrm{outer}}=1.1$, $n_0=1$ gives $\sqrt{1.35}>1$. Thus the unsaturated printed formula is not valid for every possible pair of refractive indices.

The PDF specifies decay as $y\to \infty$; the local TeX drops this limit from its displayed boundary data. Take $a=4\pi$, so the conformal map is $w=\sin (z/4)$. Its components areFor $-2\pi <x<2\pi$ and $y>0$, $\eta >0$. The bottom edge maps monotonically to $(-1,1)$, and the two vertical sides map to the real rays $(-\infty ,-1)$ and $(1,\infty )$. The derivative $\cos (z/4)/4$ never vanishes in the interior. Injectivity follows from the identities for equal sines: the alternatives $z_1/4=z_2/4+2\pi n$ and $z_1/4=\pi -z_2/4+2\pi n$ give only the identical point within this strip. Surjectivity can be seen directly: for fixed $\xi$ and $v=y/4>0$, solve $\sin (x/4)=\xi /\cosh v$; thenOn $v>\max (0,\mathrm{arcosh}|\xi |)$, with the lower endpoint read as $0$ when $|\xi |\le 1$, the right side increases strictly from zero to infinity. Thus every point of the upper half-plane has a unique preimage.

By conformal invariance of harmonicity, write the transformed harmonic function as $\Phi (\xi ,\eta )$. Its real-axis boundary data areThe inverse sine on $(-1,1)$ is its real principal branch. For bounded integrable data, take the Fourier transform convention $\hat{\Phi }(k,\eta )={\int }_{\mathbb{R}}\Phi (\xi ,\eta )e^{-ik\xi }d\xi$. The Laplace equation becomes ${\hat{\Phi }}_{\eta \eta }-k^2\hat{\Phi }=0$, with the bounded decaying solution $\hat{\Phi }=\hat{F}(k)e^{-|k|\eta }$. Inverting the Fourier transform givesConsequently the Poisson kernel for the upper half-plane yieldsThe decay is uniform across the strip: $|w{|}^2={\sin }^2(x/4)+{\sinh }^2(y/4)\ge {\sinh }^2(y/4)$ and, for $|w|>1$, $|\Phi |\le \Vert F{\Vert }_{L^1}|w|/(\pi (|w|-1{)}^2)\to 0$. This verifies the printed decay condition. Boundary limits give the prescribed values at continuity points away from the corners.

For the given sine data,Values assigned at $s=\pm 1$ do not alter the integral. There is no solution continuous on the entire closed half-strip for these data: the bottom edge has corner limits $\pm 1$, but the side edges have corner limits zero. The formula is the bounded Dirichlet problem solution with the specified limits away from those two corners. This is the usual interpretation of the printed problem; boundedness excludes additional boundary-singular harmonic functions, while decay alone without a regularity or boundedness condition does not establish uniqueness.

The Gram matrix representation and $Y_{ii}=1$ give $\Vert v_i{\Vert }_2=1$. For a standard Gaussian random vector $Z$, each $⟨v_i,Z⟩$ is a standard normal scalar, so the zero event has probability zero. Choose either sign convention at zero; the resulting vector is almost surely a feasible sign vector. Consequently $y^{T}Ay\le v^{\ast }$ almost surely.

The Gaussian sign-correlation identity givesThe last equality uses symmetry of the real matrices. HenceThe inverse sine is applied entrywise, not through spectral matrix functional calculus.

For completeness, the correlation coefficient identity has a geometric proof. If the angle between two unit vectors is $\theta$, the isotropic Gaussian random vector direction in their two-dimensional span gives opposite signs on angular sectors with probability $\theta /\pi$. Thus the sign product has expectation $1-2\theta /\pi =(2/\pi )\arcsin (\cos \theta )$. Parallel and antiparallel pairs give the endpoint values $1$ and $-1$ directly. This is Gaussian hyperplane rounding, and needs no independence between the rounded coordinates.

The Krivine rounding constant has $t=c_K\pi /2=\log (1+\sqrt{2})<\pi /2$. Since $|X_{ij}|\le 1$, the principal real inverse sine satisfieson every cross-block entry. The matrix $A$ has zero diagonal blocks, so those cross blocks are the only contributors to its Frobenius inner product with $\arcsin [Y]$. Explicitly,No analogous identity is needed for the diagonal blocks involving $\sinh$.

Combining with the preceding expectation formula and optimality of $X$ gives the bipartite sign rounding boundAt least one feasible rounded outcome attains at least this expectation. This is an expectation-based approximation guarantee for the specific bipartite sign problem; no claim is made that $c_K$ is the largest possible constant. The inequalities remain valid when the optimal objective is zero.

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this isequivalently $\{Q,P{\}}_{q,p}=1$.

Suppose a type-two generating function for a canonical transformation $F(q,P)$ definesThenEquality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.

Forthe fundamental theorem of calculus and differentiation under the integral sign giveThusand in particular $P=(p^2+q^2)/2$.

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply $H=P$. Hamilton's equations becomeThereforeand transformation back gives the familiar solution

Now consider the weak quartic oscillatorChoose the modified generating functionIt givesso the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the Taylor series givesDifferentiation with respect to $P$ therefore yieldswhere

Set $\theta =t-t_0$ andWriting $q=q_0+ϵq_1+O({ϵ}^2)$ and expanding the inverse sine relation at $q_0$ givesExpanding $p^2=2E-q^2-2ϵq^4$ similarly givesConsequentlySince $\theta =t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.

Set $x=2\sin \theta$ with $-\pi /2<\theta <\pi /2$, so $dx=2\cos \theta d\theta$ and $\sqrt{4-x^2}=2\cos \theta$. The integration by substitution givesThis uses the principal inverse sine; the chosen angle interval keeps the square root positive.

Again $C=\sin (2\pi /3)=\sqrt{3}/2$, but the principal inverse sine would give $\pi /3$ at $x=1$. The required branch is $v=\pi -\arcsin (Cx)$, corresponding to $k=1$. ThereforeAt $x=1$ the bracket is $2\pi /3$, as required. Both the initial condition and the original differential equation are thus satisfied on the stated branch.

The initial condition gives $C=\sin (\pi /3)=\sqrt{3}/2$. Since $\pi /3$ belongs to the principal inverse sine range, choose $k=0$ in part (i). The particular solution isThe upper endpoint is excluded because $y/x\to \pi /2$ there and the original tangent ceases to be defined.

Use the substitution $v=y/x$ for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, it givesWhere $\sin v\ne 0$, this separable differential equation integrates to $\ln |\sin v|=\ln |x|+C$, hence $\sin v=C_{1}x$. One can obtain the same relation without division by $\sin v$: the equation gives $(\sin v{)}^{\prime }=(\sin v)/x$, so $\sin v=C_{1}x$, including $\sin v\equiv 0$.

All real branches of the inverse sine are captured byEach branch is used on an interval with $x\ne 0$ and $|Cx|<1$, with fixed $k$. The strict inequality excludes zeros of $\cos (y/x)$, where the original tangent is undefined. When $C=0$, the formula includes all straight-line solutions $y=k\pi x$. Keeping both families of inverse sine branches is essential for the two different initial conditions below.

For a nondegenerate convex minor spherical triangle, the spherical law of cosines relates two angular sides $b,c$, their included interior angle $A$, and the opposite angular side $a$:Relabelling the vertices gives the corresponding formulas for $b,B$ and $c,C$. As with the spherical law of sines, physical side lengths must first be divided by the sphere's radius. The spherical law of cosines determines $a\in (0,\pi )$ uniquely from $b,c,A$ when the data describe such a spherical triangle. It also recovers each interior angle from three sides, for exampleUsing the cosine and the range $0<A<\pi$ selects a unique angle, avoiding the supplementary-angle ambiguity of an inverse sine calculation.

To derive the spherical law of cosines, let $u,v,w$ be the vertex unit vectors at $A,B,C$, so $u\cdot v=\cos c$, $u\cdot w=\cos b$ and $v\cdot w=\cos a$. The outgoing unit tangent vectors at $u$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their dot product is both $\cos A$ and $(\cos a-\cos b\cos c)/(\sin b\sin c)$, proving the identity. This gives a geometric explanation: spherical trigonometry compares angles in the tangent plane using vectors in the surrounding Euclidean space.

For a worked example on the unit sphere, take the verticesAt $A$, the outgoing unit tangent vectors are $(1,0,0)$ and $(0,1,0)$, so the interior angle is a right angle. The vertex dot products give $b=c=\pi /3=60^{\circ }$. The spherical law of cosines then reduces to the right-angle identity $\cos a=\cos b\cos c$, givingTo find $B$, use the relabelled spherical law of cosines:The physical side lengths on a sphere of radius $R$ are ${\ell }_a\approx 1.318116R$ and ${\ell }_b={\ell }_c=\pi R/3\approx 1.047198R$. For example, on a toy sphere with $R=10$ metres they are approximately $13.18116$, $10.47198$ and $10.47198$ metres; the interior angles stay the same. Applying the planar Pythagorean theorem directly to the angular sides would instead give $\sqrt{60^2+60^2}\approx 84.852814^{\circ }$. That flat right triangle calculation does not give the great-circle distance between $B$ and $C$.

The familiar law of cosines reappears as a local approximation. With $a,b,c$ all small in radians, substitute the small-angle approximations $\sin x\approx x$ and $\cos x\approx 1-x^2/2$ into the spherical law of cosines and keep terms through second order:Multiplication by $R^2$ gives the planar law of cosines for the physical side lengths. For a fixed small surface triangle on a progressively larger sphere, the angular sides shrink and this approximation improves. Thus the curved and flat formulas fit together: the Pythagorean theorem is recovered locally when $A$ is a right angle.

For a nondegenerate convex minor spherical triangle, with angular sides $a,b,c$ opposite interior angles $A,B,C$, the spherical law of sines statesThe sines of sides refer to central angles, so a physical side length ${\ell }_a$ on a sphere of radius $R$ must enter as $\sin ({\ell }_a/R)$. All six angles lie in $(0,\pi )$, and the denominators are positive. The reciprocal form, $\sin a/\sin A=\sin b/\sin B=\sin c/\sin C$, is equivalent. This is Tatum's sine formula, equation 3.5.1, in Celestial Mechanics, section 3.5: spherical sine and cosine formulas.

A short proof explains the symmetry of the spherical law of sines. Let $u,v,w$ be the vertex unit vectors at $A,B,C$. At $u$, the outgoing unit tangent vectors towards $v,w$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their angle is $A$, so the magnitude of their cross product is $\sin A$. Since both tangent vectors are orthogonal to $u$, the scalar triple product obeysDividing by $\sin a\sin b\sin c$ proves the spherical law of sines. Its beauty is that the same three-dimensional scalar triple product can be calculated at any vertex, producing all three ratios at once.

For the spherical triangle worked out under the spherical law of cosines, $A=90^{\circ }$, $b=c=60^{\circ }$ and $\cos a=1/4$. Hence $\sin a=\sqrt{15}/4$, and the spherical law of sines givesThe inverse sine gives $B\approx 63.434949^{\circ }$, but the supplementary angle $116.565051^{\circ }$ has the same sine. The spherical law of sines alone therefore leaves two candidates for this angle. The spherical law of cosines, using the known three sides, gives $\cos B=1/\sqrt{5}>0$ and selects $63.434949^{\circ }$; likewise $C=B$. When only two sides and an opposite angle are given, two distinct valid spherical triangles can remain. A cosine relation resolves an inverse sine branch only when enough additional data are known; it does not make every ambiguous data set unique.