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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.

As an original stipulated low-field model, take the axial magnetic field intensity $H$ in $A/m$ and prescribeAll strain components are dimensionless, and their trace vanishes. This is an even, quasistatic, unsaturated toy law with a fixed measuring axis. It is distinct from the fixed-magnitude directional model in saturated isotropic magnetostrictive strain. In particular, one must not replace local unit magnetization by normalized net magnetization in a many-domain specimen and assume this law follows.

For $H(t)=H_0\sin (2\pi ft)$, $H_0=20kA/m$, and $f=50\mathrm{Hz}$, a trigonometric identity givesThe axial strain ranges from zero to $20\times 10^{-6}$, with mean $10\times 10^{-6}$ and a $100\mathrm{Hz}$ component of amplitude $10\times 10^{-6}$. The cosine coefficient is negative with the chosen time origin. For $L=100\mathrm{mm}$ the length ranges from zero to $2\mu \mathrm{m}$ relative to $H=0$. There is no $50\mathrm{Hz}$ term in this unbiased model.

A static bias $H_b$ changes the conclusion. Expanding $\kappa [H_b+H_0\sin (2\pi ft){]}^2$ produces a fundamental-frequency term $2\kappa H_b{H}_0\sin (2\pi ft)$ as well as the doubled-frequency term. This is an incremental linear response about a state with a chosen magnetic axis; it does not contradict isotropic linear magnetostriction vanishes at an unbiased isotropic reference. Real hysteresis, thermal expansion from Joule heating, and mechanical resonance can add further components or change their phases. Frequency alone cannot identify the cause of motion: Joule heating also contains an even-in-electric current contribution.

Recovering two orthogonal components of a signal by averaging its products with sine and cosine reference signals at the chosen frequency.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

A 2013 experiment by Paul A. Williams and collaborators mounted a 25-mm dielectric mirror on a commercial force-restoration balance. Removing its pan-weight compensation spring allowed horizontal force sensing with a horizontal laser path. They measured mirror optical reflectance and varied incidence from about $7^{\circ }$ to $45^{\circ }$. A 1071-nm laser, capable of 530 W, was switched at 0.1 Hz. Digitized mass-equivalent readings were converted to force using $g$; quadrature demodulation isolated the switched signal from drift. ^[ref]

Measurement calibration used a 20-g mass, with checks down to 1 mg. The optical forces lay below that checked range, limiting the accuracy claim. An angle scan tested the predicted cosine dependence at fixed intercepted power. Replacing the mirror with a carbon nanotube coated absorber increased absorbed power while reducing the predicted optical force; no modulation peak was detected. This bounded gas-heating artifacts under those conditions, rather than proving their universal absence. The fitted slope was about 7% below $2/c$. Inferring laser power requires calibrated force, optical reflectance, radiometric absorptance and angle of incidence, with measurement uncertainty for each. ^[ref]

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

Right ascension $\alpha$ is the eastward angle along the reference celestial equator from the reference equinox to a direction's hour circle, conventionally reduced to $0\le \alpha <24^{\mathrm{h}}$. Its hours label angles: $24^{\mathrm{h}}=360^{\circ }$, $1^{\mathrm{h}}=15^{\circ }$, $1^{\mathrm{m}}=15^{\prime }$ and $1^{\mathrm{s}}=15^{\prime \prime }$. Thus $6^{\mathrm{h}}30^{\mathrm{m}}=97.5^{\circ }=13\pi /24$ radians. A minute of right ascension differs from a minute of arc, and the actual arc on a fixed-declination parallel also contains the cosine factor described under declination.

The hours become useful through sidereal time. At local sidereal time $8^{\mathrm{h}}$, a synthetic direction of right ascension $6^{\mathrm{h}}30^{\mathrm{m}}$ has west-positive hour angle $1^{\mathrm{h}}30^{\mathrm{m}}=22.5^{\circ }$ and has passed its upper meridian transit. Conversely, at $5^{\mathrm{h}}30^{\mathrm{m}}$ it has hour angle $-1^{\mathrm{h}}$ and has yet to cross. Interpret these intervals in sidereal hours; ordinary civil hours are not interchangeable with them. The right ascension of the upper celestial meridian advances as the Earth turns, while the synthetic fixed target's right ascension stays unchanged.

An instrument makes this geometry tangible. DePauw University's description of McKim Observatory documents a meridian optical telescope with an angular circle, astronomical clocks and a chronograph. The circle measures zenith distance, and the crossing time supplies right ascension. Clock pulses and the observer's electrical key mark a rotating paper record, so crossing time is read relative to the clock marks. One clock kept sidereal time and another solar time. This apparatus separates two measured quantities: when a star crosses the celestial meridian, and at what angle from the zenith. ^[ref].

A teaching reconstruction would fix an optical telescope to sweep the celestial meridian, align its central sighting line, and observe a recognizable reference star followed by a target. Record each central-line crossing against a clock or timestamped camera, and read the angular circle at upper meridian transit. A short sequence shows the image moving across the stationary sighting line; the central crossing supplies a repeatable event. In the ideal model, successive upper crossings of a fixed direction recur once per sidereal day, and directions with increasing right ascension cross later in the same cycle. These are predictions for the reconstruction, rather than reported McKim measurements. Tatum derives the equality between local sidereal time and right ascension at upper meridian transit. ^[ref].

Calibrate the clock's zero by comparing its reference crossing readout $C_r$ with the reference star's predicted apparent right ascension ${\alpha }_r$ in the same date convention: the additive correction is $b={\alpha }_r-C_r$, modulo $24^{\mathrm{h}}$. Then infer target right ascension ${\alpha }_t=C_t+b$ if the clock rate is correct between crossings. In a purely synthetic example, ${\alpha }_r=6^{\mathrm{h}}00^{\mathrm{m}}00^{\mathrm{s}}$ and $C_r=6^{\mathrm{h}}00^{\mathrm{m}}20^{\mathrm{s}}$ imply $b=-20^{\mathrm{s}}$. A target readout $C_t=6^{\mathrm{h}}30^{\mathrm{m}}20^{\mathrm{s}}$ therefore gives ${\alpha }_t=6^{\mathrm{h}}30^{\mathrm{m}}00^{\mathrm{s}}=97.5^{\circ }$. The $20^{\mathrm{s}}$ clock offset is $300^{\prime \prime }$ in right ascension, while its local east-west angular effect at declination $60^{\circ }$ is approximately $150^{\prime \prime }$. The reference and target in this calculation are invented directions, with no claim about any current star's coordinates.

Controls give the inference its meaning. Use independently established geographical latitude and an angular-circle zero for declination; check sighting-line alignment against several reference stars, and bracket the target in time with references to reveal clock drift. Record north or south of the zenith, distinguish upper from lower meridian transit, and account for atmospheric refraction. Convert the observing timestamp to local sidereal time using the site's east-positive longitude correction, rather than a time-zone offset; precise conversion uses the Earth's measured rotation time. A reference star calibrates a zero only after its catalogue direction has been converted to the observation convention. Repeated crossings reduce random measurement error, while an incorrect clock rate, axis alignment or catalogue convention remains a systematic limitation. ^[ref]. These controls are requirements inferred from the geometric and clock model, not a claimed historical observing log.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]

At uniform sampling times $t_n=n/f_s$, adding an integer multiple of the sampling rate to a sinusoid's frequency changes its complex phase by an integer multiple of $2\pi n$. The samples therefore agree. Real cosines also identify the two signs of frequency, giving the displayed representative below the Nyquist frequency. A chosen nonzero sampling-time origin changes the equivalent phase, rather than the alias frequency.

The scalar projection of a real Euclidean vector $\mathbf{a}$ onto a nonzero Euclidean vector $\mathbf{c}$ is its signed component along the unit vector $\mathbf{c}/\Vert \mathbf{c}\Vert$:For $\mathbf{a}=(3,4)$, the scalar projection onto $(2,0)$ is $3$, while that onto $(-2,0)$ is $-3$. Reversing the measuring direction reverses the sign; the scalar projection is not necessarily a nonnegative length. For nonzero $\mathbf{a}$, it also equals $\Vert \mathbf{a}\Vert \cos \theta$, using the angle with $\mathbf{c}$ and its cosine. For the zero vector it is zero, even though that angle is undefined.

The secant function is the reciprocal of the cosine, $\sec x=1/\cos x$, where the denominator is nonzero. Its derivative is $\sec x\tan x$, and ${\sec }^{2}x=1+{\tan }^{2}x$ follows from the Pythagorean trigonometric identity.

A signed spatial filter model of a simple cell assigns a weight $K_{ij}$ to each sampled location and takes its dot product with the visual contrast image $C_{ij}$. Positive weights favor bright features; negative weights favor dark features. The signed driveis a linear function of the image, while the rectified linear unit output $R$ is nonlinear. $D$ and $R$ below are arbitrary model units. To obtain a predicted firing rate one would need a calibrated gain, background response and possibly a different output function. The model does not generate a spike train; a linear-nonlinear-Poisson cascade model adds a stochastic event-generation assumption.

For the first diagram, every one of five rows has weightsA bright central column, with $C_{i3}=1$ and zero elsewhere, gives $D=5/5=1$. A bright middle row gives $D=-1/20-1/20+1/5-1/20-1/20=0$; a uniformly bright patch also gives $D=0$, since every row sums to zero. Move the bright column one site right: $D=5(-1/20)=-1/4$, so $R=0$. Make that shifted column dark instead: $D=1/4$ and $R=1/4$. Thus an OFF-like weight can increase output for darkness. A negative weight is a descriptive sensitivity, not evidence for a particular inhibitory synapse.

For a smooth example use the real, even Gabor filterThe Gaussian function localizes the weights; the cosine supplies alternating polarity. Choose ${\sigma }_x=1/2$, ${\sigma }_y=1$ and $q=\pi$ in arbitrary spatial units. The stripes' spatial frequency is $q/(2\pi )=1/2$ cycles per unit. Their axis makes angle $\theta$ with horizontal, while brightness changes across the perpendicular axis. $\phi$ is their spatial phase. Normalize the full-plane integral of $K{C}_{\theta ,\phi }$ by its value at $\theta =\pi /2$, $\phi =0$. The Gaussian integral and the product identity for cosines givewhereAt $\phi =0$, horizontal stripes give $R\approx 0.004159$, $60^{\circ }$ stripes give $R\approx 0.286741$, and vertical stripes give $R=1$. At the same vertical orientation, $\phi =\pi /2$ gives $D=R=0$, while reversing the polarity with $\phi =\pi$ gives $D=-1$ and $R=0$. The preferred axis persists while the response changes dramatically with position. Shifting the vertical stripes by a quarter of their wavelength changes phase by $\pi /2$ up to sign, without rotating them.

These are two different toy filters, not discretizations of the same fitted receptive field. The discrete filter exactly rejects a uniform contrast patch; this Gabor filter need not reject uniform contrast because its weight integral need not vanish. Neither toy includes stimulus history, recurrent interactions, saturation or a measured response baseline. Their value is a transparent prediction: orientation selectivity can emerge from spatial summation, and it does not by itself imply tolerance to phase.

For a real angle $\theta$, $\sin \theta$ is the vertical coordinate on the unit circle reached by a counterclockwise rotation through $\theta$ radians. For an acute angle in a right triangle, it is the opposite side divided by the hypotenuse. In particular,The first-quarter rise and second-quarter fall are visible in the circle projection; the lower semicircle gives negative values.

Reflecting the unit circle across the horizontal axis gives $\sin (-\theta )=-\sin \theta$, so sine is an odd function. A full turn gives $\sin (\theta +2\pi )=\sin \theta$, making it a periodic function. For real arguments, its range is $[-1,1]$ and its zeros of a function occur at $\theta =k\pi$ for integers $k$. The cosine describes the perpendicular coordinate of the same point, and the Pythagorean trigonometric identity keeps their two values tied together.

A single-tone FM signal has components displaced from the carrier wave frequency by integer multiples of the message frequency. These are temporal sidebands; the spatial sidebands used to perturb a spatial pattern are a related but different setting. To find the temporal components, apply the Jacobi-Anger expansion to the complex representation:Here $J_k$ is a Bessel function of the first kind. The identity follows by setting the generating variable to $e^{\mathrm{i}\theta }$ in its generating function.^[ref]

Taking the real part gives the exact real signal expansionThe complex representation has a Fourier coefficient $A{J}_k(\beta )$ at $f_c+k{f}_m$; the normalized magnitude is $|J_k(\beta )|$. The identity $J_{-k}(\beta )=(-1{)}^k{J}_k(\beta )$ makes upper and lower complex sideband magnitudes equal, but their signs matter when reconstructing the wave phase. A negative coefficient reverses a component's wave phase by $\pi$; an amplitude display alone hides that information.

For $\beta =5$, original evaluation gives $J_0(5)\approx -0.17760$, $J_1(5)\approx -0.32758$, $J_5(5)\approx 0.26114$ and $J_6(5)\approx 0.13105$. Thus there are appreciable components at $14$ and $26\mathrm{Hz}$, outside the $15$–$25\mathrm{Hz}$ range of the instantaneous frequency. There is no contradiction: instantaneous frequency describes the local rate of the combined wave phase, while the frequency spectrum describes a global superposition of constant-frequency components.

For the real signal, negative frequencies must also be accounted for: a cosine at $-f$ equals one at $f$. When $f_c/f_m=20$, the normalized positive-frequency cosine coefficient at $n{f}_m$, $n>0$, is exactly $J_{n-20}(\beta )+J_{-n-20}(\beta )$. The second term is extremely small over the plotted range, but the figure includes it. The chosen complex representation need not have a frequency spectrum supported only at positive frequencies, even when its specified instantaneous frequency is everywhere positive.

For a nonzero FM modulation index, the ideal frequency spectrum has infinitely many sidebands. Many become tiny, but drawing only visible lines does not make the exact signal bandwidth finite. A useful engineering width must state a threshold or a retained-power fraction. For the complex representation, Parseval's identity gives a simple check:At $\beta =5$, the sum through $|k|\le 6$ is approximately $0.99356$. Retaining those complex components captures over $99\%$ of their squared magnitudes and spans $12\mathrm{Hz}$ from $14$ to $26\mathrm{Hz}$. This is a stated truncation criterion, not an exact cutoff or a universal bandwidth rule. It also explains why a diminished carrier component need not imply a diminished envelope: its contribution has been redistributed among sidebands.

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.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

With thermal diffusivity $\xi$, a Fourier mode of angular wavenumber $k$ decays at rate $\beta =\xi k^2$. To avoid confusing this $k$ with thermal conductivity, use $\alpha =\xi$ and angular wavenumber $K$ below: the same rate is $\beta =\alpha K^2$, with units ${\mathrm{s}}^{-1}$.

Consider an insulated bar $0\le x\le L$ with constant thermal diffusivity, no sources and no heat flux through either end or the sides. An original initial temperature profile and its exact evolution areThe cosine has zero temperature gradient at both ends, satisfying the insulated Neumann boundary conditions. Its spatial mean is zero, so conservation of energy leaves the mean temperature $\overline{T}$ unchanged while the nonuniform part disappears. A short derivation substitutes $T=\overline{T}+A(t)\cos (K_{n}x)$ into the heat equation: the second spatial derivative supplies $-K_n^2$, hence $A^{\prime }=-\alpha K_n^{2}A$. The exponential decay time is ${\tau }_n=1/(\alpha K_n^2)$; doubling angular wavenumber quarters this time. If cycles-per-length spatial frequency $f_x$ is used instead, $K=2\pi f_x$ and the rate is $4{\pi }^2\alpha f_x^2$.

For the two model materials in thermal diffusivity and $L=0.010\mathrm{m}$, ${\tau }_{1,A}\approx 5.07\mathrm{s}$ and ${\tau }_{1,B}\approx 20.3\mathrm{s}$. After $10\mathrm{s}$, the remaining amplitude fractions are approximately $0.139$ and $0.610$. A narrow hot-and-cold pattern disappears faster than a broad one because its temperature curvature, and hence the imbalance in conductive heat flux, is larger.