# Law of cosines

(Redirected from Cosine Law)
Fig. 1 – A triangle. The angles α (or A), β (or B), and γ (or C) are respectively opposite the sides a, b, and c.

In trigonometry, the law of cosines (also known as the cosine formula, cosine rule, or al-Kashi's theorem[1]) relates the lengths of the sides of a triangle to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines states

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos \gamma ,}$

where γ denotes the angle contained between sides of lengths a and b and opposite the side of length c. For the same figure, the other two relations are analogous:

${\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos \alpha ,}$
${\displaystyle b^{2}=a^{2}+c^{2}-2ac\cos \beta .}$

The law of cosines generalizes the Pythagorean theorem, which holds only for right triangles: if the angle γ is a right angle (of measure 90 degrees, or π/2 radians), then cos γ = 0, and thus the law of cosines reduces to the Pythagorean theorem:

${\displaystyle c^{2}=a^{2}+b^{2}.}$

The law of cosines is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known.

## History

Fig. 2 – Obtuse triangle ABC with perpendicular BH

Though the notion of the cosine was not yet developed in his time, Euclid's Elements, dating back to the 3rd century BC, contains an early geometric theorem almost equivalent to the law of cosines. The cases of obtuse triangles and acute triangles (corresponding to the two cases of negative or positive cosine) are treated separately, in Propositions 12 and 13 of Book 2. Trigonometric functions and algebra (in particular negative numbers) being absent in Euclid's time, the statement has a more geometric flavor:

Proposition 12
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

— Euclid's Elements, translation by Thomas L. Heath.[2]

Using notation as in Fig. 2, Euclid's statement can be represented by the formula

${\displaystyle AB^{2}=CA^{2}+CB^{2}+2(CA)(CH).}$

This formula may be transformed into the law of cosines by noting that CH = (CB) cos(π − γ) = −(CB) cos γ. Proposition 13 contains an entirely analogous statement for acute triangles.

Euclid's Elements paved the way for the discovery of law of cosines. In the 15th century, Jamshīd al-Kāshī, a Persian mathematician and astronomer, provided the first explicit statement of the law of cosines in a form suitable for triangulation. He provided accurate trigonometric tables and expressed the theorem in a form suitable for modern usage. As of the 1990s, in France, the law of cosines is still referred to as the Théorème d'Al-Kashi.[1][3][4]

The theorem was popularized in the Western world by François Viète in the 16th century. At the beginning of the 19th century, modern algebraic notation allowed the law of cosines to be written in its current symbolic form.

## Applications

Fig. 3 – Applications of the law of cosines: unknown side and unknown angle.

The theorem is used in triangulation, for solving a triangle or circle, i.e., to find (see Figure 3):

• the third side of a triangle if one knows two sides and the angle between them:
${\displaystyle \,c={\sqrt {a^{2}+b^{2}-2ab\cos \gamma }}\,;}$
• the angles of a triangle if one knows the three sides:
${\displaystyle \,\gamma =\arccos \left({\frac {a^{2}+b^{2}-c^{2}}{2ab}}\right)\,;}$
• the third side of a triangle if one knows two sides and an angle opposite to one of them (one may also use the Pythagorean theorem to do this if it is a right triangle):
${\displaystyle \,a=b\cos \gamma \pm {\sqrt {c^{2}-b^{2}\sin ^{2}\gamma }}\,.}$

These formulas produce high round-off errors in floating point calculations if the triangle is very acute, i.e., if c is small relative to a and b or γ is small compared to 1. It is even possible to obtain a result slightly greater than one for the cosine of an angle.

The third formula shown is the result of solving for a in the quadratic equation a2 − 2ab cos γ + b2c2 = 0. This equation can have 2, 1, or 0 positive solutions corresponding to the number of possible triangles given the data. It will have two positive solutions if b sin γ < c < b, only one positive solution if c = b sin γ, and no solution if c < b sin γ. These different cases are also explained by the side-side-angle congruence ambiguity.

## Proofs

### Using the distance formula

Fig. 4 – Coordinate geometry proof

Consider a triangle with sides of length a, b, c, where θ is the measurement of the angle opposite the side of length c. This triangle can be placed on the Cartesian coordinate system aligned with edge a with origin at C, by plotting the components of the 3 points of the triangle as shown in Fig. 4:

${\displaystyle A=(b\cos \theta ,b\sin \theta ),B=(a,0),{\text{ and }}C=(0,0).}$

By the distance formula,

${\displaystyle c={\sqrt {(a-b\cos \theta )^{2}+(0-b\sin \theta )^{2}}}.}$

Squaring both sides and simplifying

{\displaystyle {\begin{aligned}c^{2}&{}=(a-b\cos \theta )^{2}+(-b\sin \theta )^{2}\\c^{2}&{}=a^{2}-2ab\cos \theta +b^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta \\c^{2}&{}=a^{2}+b^{2}(\sin ^{2}\theta +\cos ^{2}\theta )-2ab\cos \theta \\c^{2}&{}=a^{2}+b^{2}-2ab\cos \theta .\end{aligned}}}

An advantage of this proof is that it does not require the consideration of different cases for when the triangle is acute, right, or obtuse.

### Using trigonometry

Fig. 5 – An acute triangle with perpendicular

Dropping the perpendicular onto the side c through point C, an altitude of the triangle, shows (see Fig. 5)

${\displaystyle c=a\cos \beta +b\cos \alpha .}$

(This is still true if α or β is obtuse, in which case the perpendicular falls outside the triangle.) Multiplying through by c yields

${\displaystyle c^{2}=ac\cos \beta +bc\cos \alpha .}$

Considering the two other altitudes of the triangle yields

${\displaystyle a^{2}=ac\cos \beta +ab\cos \gamma ,}$
${\displaystyle b^{2}=bc\cos \alpha +ab\cos \gamma .}$

Adding the latter two equations gives

${\displaystyle a^{2}+b^{2}=ac\cos \beta +bc\cos \alpha +2ab\cos \gamma .}$

Subtracting the first equation from the last one results in

${\displaystyle a^{2}+b^{2}-c^{2}=ac\cos \beta +bc\cos \alpha +2ab\cos \gamma -(ac\cos \beta +bc\cos \alpha )}$

which simplifies to

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos \gamma .}$

This proof uses trigonometry in that it treats the cosines of the various angles as quantities in their own right. It uses the fact that the cosine of an angle expresses the relation between the two sides enclosing that angle in any right triangle. Other proofs (below) are more geometric in that they treat an expression such as a cos γ merely as a label for the length of a certain line segment.

Many proofs deal with the cases of obtuse and acute angles γ separately.

### Using the Pythagorean theorem

Obtuse triangle ABC with height BH
Cosine theorem in plane trigonometry, proof based on Pythagorean theorem.

#### Case of an obtuse angle

Euclid proved this theorem by applying the Pythagorean theorem to each of the two right triangles in the figure shown (AHB and CHB). Using d to denote the line segment CH and h for the height BH, triangle AHB gives us

${\displaystyle c^{2}=(b+d)^{2}+h^{2},}$

and triangle CHB gives

${\displaystyle d^{2}+h^{2}=a^{2}.}$

Expanding the first equation gives

${\displaystyle c^{2}=b^{2}+2bd+d^{2}+h^{2}.}$

Substituting the second equation into this, the following can be obtained:

${\displaystyle c^{2}=a^{2}+b^{2}+2bd.}$

This is Euclid's Proposition 12 from Book 2 of the Elements.[5] To transform it into the modern form of the law of cosines, note that

${\displaystyle d=a\cos(\pi -\gamma )=-a\cos \gamma .}$

#### Case of an acute angle

Euclid's proof of his Proposition 13 proceeds along the same lines as his proof of Proposition 12: he applies the Pythagorean theorem to both right triangles formed by dropping the perpendicular onto one of the sides enclosing the angle γ and uses the binomial theorem to simplify.

Fig. 6 – A short proof using trigonometry for the case of an acute angle

#### Another proof in the acute case

Using more trigonometry, the law of cosines can be deduced by using the Pythagorean theorem only once. In fact, by using the right triangle on the left hand side of Fig. 6 it can be shown that: {\displaystyle {\begin{aligned}\quad c^{2}&=(b-a\cos \gamma )^{2}+(a\sin \gamma )^{2}\\&=b^{2}-2ab\cos \gamma +a^{2}\cos ^{2}\gamma +a^{2}\sin ^{2}\gamma \\&=b^{2}+a^{2}-2ab\cos \gamma ,\end{aligned}}}

using the trigonometric identity

${\displaystyle \quad \cos ^{2}\gamma +\sin ^{2}\gamma =1.}$

This proof needs a slight modification if b < a cos(γ). In this case, the right triangle to which the Pythagorean theorem is applied moves outside the triangle ABC. The only effect this has on the calculation is that the quantity ba cos(γ) is replaced by a cos(γ) − b. As this quantity enters the calculation only through its square, the rest of the proof is unaffected. However, this problem only occurs when β is obtuse, and may be avoided by reflecting the triangle about the bisector of γ.

Referring to Fig. 6 it is worth noting that if the angle opposite side a is α then:

${\displaystyle \tan \alpha ={\frac {a\sin \gamma }{b-a\cos \gamma }}.}$

This is useful for direct calculation of a second angle when two sides and an included angle are given.

### Using Ptolemy's theorem

Proof of law of cosines using Ptolemy's theorem

Referring to the diagram, triangle ABC with sides AB = c, BC = a and AC = b is drawn inside its circumcircle as shown. Triangle ABD is constructed congruent to triangle ABC with AD = BC and BD = AC. Perpendiculars from D and C meet base AB at E and F respectively. Then:

{\displaystyle {\begin{aligned}&BF=AE=BC\cos {\hat {B}}=a\cos {\hat {B}}\\\Rightarrow \ &DC=EF=AB-2BF=c-2a\cos {\hat {B}}.\end{aligned}}}

Now the law of cosines is rendered by a straightforward application of Ptolemy's theorem to cyclic quadrilateral ABCD:

{\displaystyle {\begin{aligned}&AD\times BC+AB\times DC=AC\times BD\\\Rightarrow \ &a^{2}+c(c-2a\cos {\hat {B}})=b^{2}\\\Rightarrow \ &a^{2}+c^{2}-2ac\cos {\hat {B}}=b^{2}.\end{aligned}}}

Plainly if angle B is right, then ABCD is a rectangle and application of Ptolemy's theorem yields the Pythagorean theorem:

${\displaystyle a^{2}+c^{2}=b^{2}.\quad }$

### By comparing areas

One can also prove the law of cosines by calculating areas. The change of sign as the angle γ becomes obtuse makes a case distinction necessary.

Recall that

• a2, b2, and c2 are the areas of the squares with sides a, b, and c, respectively;
• if γ is acute, then ab cos γ is the area of the parallelogram with sides a and b forming an angle of γ′ = π/2γ;
• if γ is obtuse, and so cos γ is negative, then ab cos γ is the area of the parallelogram with sides a and b forming an angle of γ′ = γπ/2.
Fig. 7a – Proof of the law of cosines for acute angle γ by "cutting and pasting".

Acute case. Figure 7a shows a heptagon cut into smaller pieces (in two different ways) to yield a proof of the law of cosines. The various pieces are

• in pink, the areas a2, b2 on the left and the areas 2ab cos γ and c2 on the right;
• in blue, the triangle ABC, on the left and on the right;
• in grey, auxiliary triangles, all congruent to ABC, an equal number (namely 2) both on the left and on the right.

The equality of areas on the left and on the right gives

${\displaystyle \,a^{2}+b^{2}=c^{2}+2ab\cos \gamma \,.}$
Fig. 7b – Proof of the law of cosines for obtuse angle γ by "cutting and pasting".

Obtuse case. Figure 7b cuts a hexagon in two different ways into smaller pieces, yielding a proof of the law of cosines in the case that the angle γ is obtuse. We have

• in pink, the areas a2, b2, and −2ab cos γ on the left and c2 on the right;
• in blue, the triangle ABC twice, on the left, as well as on the right.

The equality of areas on the left and on the right gives

${\displaystyle \,a^{2}+b^{2}-2ab\cos(\gamma )=c^{2}.}$

The rigorous proof will have to include proofs that various shapes are congruent and therefore have equal area. This will use the theory of congruent triangles.

### Using geometry of the circle

Using the geometry of the circle, it is possible to give a more geometric proof than using the Pythagorean theorem alone. Algebraic manipulations (in particular the binomial theorem) are avoided.

Fig. 8a – The triangle ABC (pink), an auxiliary circle (light blue) and an auxiliary right triangle (yellow)

Case of acute angle γ, where a > 2b cos γ. Drop the perpendicular from A onto a = BC, creating a line segment of length b cos γ. Duplicate the right triangle to form the isosceles triangle ACP. Construct the circle with center A and radius b, and its tangent h = BH through B. The tangent h forms a right angle with the radius b (Euclid's Elements: Book 3, Proposition 18; or see here), so the yellow triangle in Figure 8 is right. Apply the Pythagorean theorem to obtain

${\displaystyle c^{2}=b^{2}+h^{2}.}$

Then use the tangent secant theorem (Euclid's Elements: Book 3, Proposition 36), which says that the square on the tangent through a point B outside the circle is equal to the product of the two lines segments (from B) created by any secant of the circle through B. In the present case: BH2 = BC·BP, or

${\displaystyle h^{2}=a(a-2b\cos \gamma ).}$

Substituting into the previous equation gives the law of cosines:

${\displaystyle c^{2}=b^{2}+a(a-2b\cos \gamma ).}$

Note that h2 is the power of the point B with respect to the circle. The use of the Pythagorean theorem and the tangent secant theorem can be replaced by a single application of the power of a point theorem.

Fig. 8b – The triangle ABC (pink), an auxiliary circle (light blue) and two auxiliary right triangles (yellow)

Case of acute angle γ, where a < 2b cos γ. Drop the perpendicular from A onto a = BC, creating a line segment of length b cos γ. Duplicate the right triangle to form the isosceles triangle ACP. Construct the circle with center A and radius b, and a chord through B perpendicular to c = AB, half of which is h = BH. Apply the Pythagorean theorem to obtain

${\displaystyle b^{2}=c^{2}+h^{2}.}$

Now use the chord theorem (Euclid's Elements: Book 3, Proposition 35), which says that if two chords intersect, the product of the two line segments obtained on one chord is equal to the product of the two line segments obtained on the other chord. In the present case: BH2 = BC·BP, or

${\displaystyle h^{2}=a(2b\cos \gamma -a).}$

Substituting into the previous equation gives the law of cosines:

${\displaystyle b^{2}=c^{2}+a(2b\cos \gamma -a)\,.}$

Note that the power of the point B with respect to the circle has the negative value h2.

Fig. 9 – Proof of the law of cosines using the power of a point theorem.

Case of obtuse angle γ. This proof uses the power of a point theorem directly, without the auxiliary triangles obtained by constructing a tangent or a chord. Construct a circle with center B and radius a (see Figure 9), which intersects the secant through A and C in C and K. The power of the point A with respect to the circle is equal to both AB2BC2 and AC·AK. Therefore,

{\displaystyle {\begin{aligned}c^{2}-a^{2}&{}=b(b+2a\cos(\pi -\gamma ))\\&{}=b(b-2a\cos \gamma ),\end{aligned}}}

which is the law of cosines.

Using algebraic measures for line segments (allowing negative numbers as lengths of segments) the case of obtuse angle (CK > 0) and acute angle (CK < 0) can be treated simultaneously.

### Using the law of sines

By using the law of sines and knowing that the angles of a triangle must sum to 180 degrees, we have the following system of equations (the three unknowns are the angles):

${\displaystyle {\frac {c}{\sin \gamma }}={\frac {b}{\sin \beta }},}$
${\displaystyle {\frac {c}{\sin \gamma }}={\frac {a}{\sin \alpha }},}$
${\displaystyle \alpha +\beta +\gamma =\pi .}$

Then, by using the third equation of the system, we obtain a system of two equations in two variables:

${\displaystyle {\frac {c}{\sin \gamma }}={\frac {b}{\sin(\alpha +\gamma )}},}$
${\displaystyle {\frac {c}{\sin \gamma }}={\frac {a}{\sin \alpha }},}$

where we have used the trigonometric property that the sine of a supplementary angle is equal to the sine of the angle.

Using the identity (see Angle sum and difference identities)

${\displaystyle \sin(\alpha +\gamma )=\sin \alpha \cos \gamma +\sin \gamma \cos \alpha }$

${\displaystyle c(\sin \alpha \cos \gamma +\sin \gamma \cos \alpha )=b\sin \gamma ,}$
${\displaystyle c\sin \alpha =a\sin \gamma .}$

By dividing the whole system by cos γ, we have:

${\displaystyle c(\sin \alpha +\tan \gamma \cos \alpha )=b\tan \gamma ,}$
${\displaystyle {\frac {c\sin \alpha }{\cos \gamma }}=a\tan \gamma ,}$
${\displaystyle {\frac {c^{2}\sin ^{2}\alpha }{\cos ^{2}\gamma }}=a^{2}\tan ^{2}\gamma .}$

Hence, from the first equation of the system, we can obtain

${\displaystyle {\frac {c\sin \alpha }{b-c\cos \alpha }}=\tan \gamma }$

By substituting this expression into the second equation and by using

${\displaystyle 1+\tan ^{2}\gamma ={\frac {1}{\cos ^{2}\gamma }}}$

we can obtain one equation with one variable:

${\displaystyle c^{2}\sin ^{2}\alpha \left[1+{\frac {c^{2}\sin ^{2}\alpha }{(b-c\cos \alpha )^{2}}}\right]=a^{2}\cdot {\frac {c^{2}\sin ^{2}\alpha }{(b-c\cos \alpha )^{2}}}}$

By multiplying by (bc cos α)2, we can obtain the following equation:

${\displaystyle (b-c\cos \alpha )^{2}+c^{2}\sin ^{2}\alpha =a^{2}.}$

This implies

${\displaystyle b^{2}-2bc\cos \alpha +c^{2}\cos ^{2}\alpha +c^{2}\sin ^{2}\alpha =a^{2}.}$

Recalling the Pythagorean identity, we obtain the law of cosines:

${\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos \alpha .}$

### Using vectors

Denote

${\displaystyle {\overrightarrow {CB}}={\vec {a}},\ {\overrightarrow {CA}}={\vec {b}},\ {\overrightarrow {AB}}={\vec {c}}}$

Therefore,

${\displaystyle {\vec {c}}={\vec {a}}-{\vec {b}}}$

Taking the dot product of each side with itself:

${\displaystyle {\vec {c}}\cdot {\vec {c}}=({\vec {a}}-{\vec {b}})\cdot ({\vec {a}}-{\vec {b}})}$
${\displaystyle \Vert {\vec {c}}\Vert ^{2}=\Vert {\vec {a}}\Vert ^{2}+\Vert {\vec {b}}\Vert ^{2}-2\,{\vec {a}}\cdot {\vec {b}}}$

Using the identity (see Dot product)

${\displaystyle {\vec {u}}\cdot {\vec {v}}=\Vert {\vec {u}}\Vert \,\Vert {\vec {v}}\Vert \cos \angle ({\vec {u}},\ {\vec {v}})}$

${\displaystyle \Vert {\vec {c}}\Vert ^{2}=\Vert {\vec {a}}\Vert ^{2}+{\Vert {\vec {b}}\Vert }^{2}-2\,\Vert {\vec {a}}\Vert \!\;\Vert {\vec {b}}\Vert \cos \angle ({\vec {a}},\ {\vec {b}})}$

The result follows.

## Isosceles case

When a = b, i.e., when the triangle is isosceles with the two sides incident to the angle γ equal, the law of cosines simplifies significantly. Namely, because a2 + b2 = 2a2 = 2ab, the law of cosines becomes

${\displaystyle \cos \gamma =1-{\frac {c^{2}}{2a^{2}}}}$

or

${\displaystyle c^{2}=2a^{2}(1-\cos \gamma ).}$

## Analogue for tetrahedra

An analogous statement begins by taking α, β, γ, δ to be the areas of the four faces of a tetrahedron. Denote the dihedral angles by ${\displaystyle {\widehat {\beta \gamma }}}$ etc. Then[6]

${\displaystyle \alpha ^{2}=\beta ^{2}+\gamma ^{2}+\delta ^{2}-2\left[\beta \gamma \cos \left({\widehat {\beta \gamma }}\right)+\gamma \delta \cos \left({\widehat {\gamma \delta }}\right)+\delta \beta \cos \left({\widehat {\delta \beta }}\right)\right].}$

## Version suited to small angles

When the angle, γ, is small and the adjacent sides, a and b, are of similar length, the right hand side of the standard form of the law of cosines can lose a lot of accuracy to numerical loss of significance. In situations where this is an important concern, a mathematically equivalent version of the law of cosines, similar to the haversine formula, can prove useful:

{\displaystyle {\begin{aligned}c^{2}&=(a-b)^{2}+4ab\sin ^{2}\left({\frac {\gamma }{2}}\right)\\&=(a-b)^{2}+4ab\operatorname {haversin} (\gamma ).\end{aligned}}}

In the limit of an infinitesimal angle, the law of cosines degenerates into the circular arc length formula, c = a γ.

## In spherical and hyperbolic geometry

Spherical triangle solved by the law of cosines.

Versions similar to the law of cosines for the Euclidean plane also hold on a unit sphere and in a hyperbolic plane. In spherical geometry, a triangle is defined by three points u, v, and w on the unit sphere, and the arcs of great circles connecting those points. If these great circles make angles A, B, and C with opposite sides a, b, c then the spherical law of cosines asserts that both of the following relationships hold:

{\displaystyle {\begin{aligned}\cos a&=\cos b\cos c+\sin b\sin c\cos A\\\cos A&=-\cos B\cos C+\sin B\sin C\cos a.\end{aligned}}}

In hyperbolic geometry, a pair of equations are collectively known as the hyperbolic law of cosines. The first is

${\displaystyle \cosh a=\cosh b\cosh c-\sinh b\sinh c\cos A}$

where sinh and cosh are the hyperbolic sine and cosine, and the second is

${\displaystyle \cos A=-\cos B\cos C+\sin B\sin C\cosh a.}$

As in Euclidean geometry, one can use the law of cosines to determine the angles A, B, C from the knowledge of the sides a, b, c. In contrast to Euclidean geometry, the reverse is also possible in both non-Euclidean models: the angles A, B, C determine the sides a, b, c.

## Unified formula for surfaces of constant curvature

Defining two functions ${\displaystyle \cos _{R}}$ and ${\displaystyle \sin _{R}}$ as

${\displaystyle \cos _{R}(x)=\cos(x/R)\quad }$ and ${\displaystyle \quad \sin _{R}(x)=R\cdot \sin(x/R)}$

allows to unify the formulae for plane, sphere and pseudosphere into:

${\displaystyle \cos _{R}(BC)=\cos _{R}(AB)\cdot \cos _{R}(AC)+{\frac {1}{R^{2}}}\sin _{R}(AB)\cdot \sin _{R}(AC)\cdot \cos({\widehat {BAC}}).}$

In this notation ${\displaystyle R}$ is a complex number, representing the surface's radius of curvature.

• For ${\displaystyle R\in \mathbb {R} }$ the surface is a sphere of radius ${\displaystyle R}$, and its constant curvature equals ${\displaystyle 1/R^{2};}$
• for ${\displaystyle R=iR'\colon R'\in \mathbb {R} }$ the surface is a pseudosphere of (imaginary) radius ${\displaystyle R,}$ with constant curvature equal to ${\displaystyle 1/R^{2}=-1/R'^{2};}$
• for ${\displaystyle R\to \infty }$ : the surface tends to a Euclidean plane, with constant zero curvature.

Verifying the formula for non-Euclidean geometry

In the first two cases, ${\displaystyle \cos _{R}}$ and ${\displaystyle \sin _{R}}$ are well-defined over the whole complex plane for all ${\displaystyle R\neq 0}$, and retrieving former results is straightforward.

Hence, for a sphere of radius ${\displaystyle 1}$

${\displaystyle \cos(BC)=\cos(AB)\cdot \cos(AC)+\sin(AB)\cdot \sin(AC)\cdot \cos({\widehat {BAC}})}$.

Likewise, for a pseudosphere of radius ${\displaystyle i}$

${\displaystyle \cosh(BC)=\cosh(AB)\cdot \cosh(AC)-\sinh(AB)\cdot \sinh(AC)\cdot \cos({\widehat {BAC}}).}$

Indeed, ${\displaystyle \cosh(x)=\cos(x/i)}$ and ${\displaystyle \sinh(x)=i\cdot \sin(x/i).}$

Verifying the formula in the limit of Euclidean geometry

In the Euclidean plane the appropriate limits for the above equation must be calculated:

${\displaystyle \cos _{R}(x)=\cos(x/R)=1-{\frac {1}{2}}\cdot {\frac {x^{2}}{R^{2}}}+o\left({\frac {1}{R^{4}}}\right)}$

and

${\displaystyle \sin _{R}(x)=R\cdot \sin(x/R)=x+o\left({\frac {1}{R^{3}}}\right)}$.

Applying this to the general formula for a finite ${\displaystyle R}$ yields:

{\displaystyle {\begin{aligned}1-{\frac {BC^{2}}{2R^{2}}}+o\left[{\frac {1}{R^{4}}}\right]={}&\left[1-{\frac {AB^{2}}{2R^{2}}}+o\left({\frac {1}{R^{4}}}\right)\right]\cdot \left[1-{\frac {AC^{2}}{2R^{2}}}+o\left({\frac {1}{R^{4}}}\right)\right]+\\[5pt]&{}+{\frac {1}{R^{2}}}\left[AB+o\left({\frac {1}{R^{3}}}\right)\right]\cdot \left[AC+o\left({\frac {1}{R^{3}}}\right)\right]\cdot \cos({\widehat {BAC}})\end{aligned}}}

Collecting terms, multiplying with ${\displaystyle -2R^{2},}$ and taking ${\displaystyle R\to \infty }$ yields the expected formula:

${\displaystyle BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot \cos({\widehat {BAC}}).}$
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