View source for Law of cosines

{{Short description|Property of all triangles on a Euclidean plane}}
{{About|the trigonometric identity|the cosine law of optics|Lambert's cosine law}}
[[Image:Triangle with notations 2.svg|thumb|right|Fig. 1 – A triangle. The angles {{math|''α''}} (or {{math|''A''}}), {{math|''β''}} (or {{math|''B''}}), and {{math|''γ''}} (or {{math|''C''}}) are respectively opposite the sides {{math|''a''}}, {{math|''b''}}, and {{math|''c''}}.]]
In [[trigonometry]], the '''law of cosines''' (also known as the '''cosine formula''' or '''cosine rule''') relates the lengths of the sides of a [[triangle]] to the [[cosine]] of one of its [[angle]]s. For a triangle with sides {{tmath|a}}, {{tmath|b}}, and {{tmath|c}}, opposite respective angles {{tmath|\alpha}}, {{tmath|\beta}}, and {{tmath|\gamma}} (see Fig. 1), the law of cosines states:

<math display=block>\begin{align}
c^2 &= a^2 + b^2 - 2ab\cos\gamma, \\[3mu]
a^2 &= b^2+c^2-2bc\cos\alpha, \\[3mu]
b^2 &= a^2+c^2-2ac\cos\beta.
\end{align}</math>

The law of cosines generalizes the [[Pythagorean theorem]], which holds only for [[right triangle]]s: if {{tmath|\gamma}} is a [[right angle]] then {{tmath|1= \cos\gamma = 0}}, and the law of cosines [[special case|reduces to]] {{tmath|1= c^2 = a^2 + b^2}}.

The law of cosines is useful for [[solution of triangles|solving a triangle]] when all three sides or two sides and their included angle are given.

== Use in solving triangles ==
[[Image:triangle-with-an-unknown-angle-or-side.svg|thumb|right|Fig. 3 – Applications of the law of cosines: unknown side and unknown angle.]]
[[File:SSA triangle ambiguity.png|thumb|Given triangle sides {{mvar|b}} and {{mvar|c}} and angle {{mvar|γ}} there are sometimes two solutions for {{mvar|a}}.]]
The theorem is used in [[solution of triangles]], i.e., to find (see Figure 3):
*the third side of a triangle if two sides and the angle between them is known: <math display="block">c = \sqrt{a^2+b^2-2ab\cos\gamma}\,;</math>
*the angles of a triangle if the three sides are known: <math display="block">\gamma = \arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)\,;</math>
*the third side of a triangle if two sides and an angle opposite to one of them is known (this side can also be found by two applications of the [[law of sines]]):{{efn|Given sides {{tmath|b}}, {{tmath|c}} and angle {{tmath|\gamma}}, {{tmath|\sin \beta}} can be found using the law of sines, leaving up to two possibilities for angle {{tmath|\beta}}. Either choice determines {{tmath|\alpha}} because the three interior angles sum to a straight angle. Finally {{tmath|a}} can be found from {{tmath|c}}, {{tmath|\gamma}}, and {{tmath|\alpha}} by another application of the law of sines.}} <math display="block"> a=b\cos\gamma \pm \sqrt{c^2 -b^2\sin^2\gamma}\,.</math>

These formulas produce high [[round-off error]]s in [[floating point]] calculations if the triangle is very acute, i.e., if {{math|''c''}} is small relative to {{math|''a''}} and {{math|''b''}} or {{math|''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]] {{math|1=''a''<sup>2</sup> − 2''ab'' cos ''γ'' + ''b''<sup>2</sup> − ''c''<sup>2</sup> = 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 {{math|''b'' sin ''γ'' &lt; ''c'' &lt; ''b''}}, only one positive solution if {{math|1=''c'' = ''b'' sin ''γ''}}, and no solution if {{math|''c'' &lt; ''b'' sin ''γ''}}. These different cases are also explained by the [[Congruence (geometry)#Side-side-angle|side-side-angle congruence ambiguity]].

==History==
Book II of Euclid's ''[[Euclid's Elements|Elements]]'', compiled c. 300 BC from material up to a century or two older, contains a geometric theorem corresponding to the law of cosines but expressed in the contemporary language of rectangle areas; Hellenistic trigonometry developed later, and sine and cosine per se first appeared centuries afterward in India.

The cases of [[Acute and obtuse triangles|obtuse triangles and acute triangles]] (corresponding to the two cases of negative or positive cosine) are treated separately, in Propositions II.12 and II.13:<ref name = "Euclid-Heath-Elements Book II Propositions"/>

{{quote|text='''Proposition 12.'''<br>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.|sign=Euclid's ''Elements'', translation by [[Thomas L. Heath]].<ref name = "Euclid-Heath-Elements Book II Propositions">{{cite web|last=Euclid | title = Elements | editor = [[Thomas L. Heath]] | translator = Thomas L. Heath | url = http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A1999.01.0086%3Abook%3D2%3Atype%3DProp
|access-date=24 January 2023}}</ref> }}

Proposition 13 contains an analogous statement for acute triangles. In his (now-lost and only preserved through fragmentary quotations) commentary, [[Heron of Alexandria]] provided proofs of the [[converse (logic)|converses]] of both II.12 and II.13.<ref>{{cite book |chapter=Introduction |title=The Thirteen Books of Euclid's Elements |last=Heath |first=Thomas |author-link=Thomas Heath (classicist) |edition=2nd |chapter-url=https://archive.org/details/thirteenbooksofe0001eucl_e2h2/page/23 |chapter-url-access=limited |orig-year=1908 |year=1956}}</ref>

[[Image:Obtuse Triangle With Altitude ZP2.svg|thumb|300px|Fig. 2 – Obtuse triangle {{math|''ABC''}} with perpendicular {{math|''BH''}}]]

Using notation as in Fig. 2, Euclid's statement of proposition II.12 can be represented more concisely (though anachronistically) by the formula

<math display=block>AB^2 = CA^2 + CB^2 + 2 (CA)(CH).</math>

To transform this into the familiar expression for the law of cosines, substitute {{tmath|1= AB = c}}, {{tmath|1= CA = b}}, {{tmath|1= CB = a}}, and <math>CH = a \cos(\pi - \gamma)\ </math>{{nobr|<math>\!{} = -a \cos \gamma</math>.}}

Proposition II.13 was not used in Euclid's time for the [[solution of triangles]], but later it was used that way in the course of solving astronomical problems by [[al-Biruni|al-Bīrūnī]] (11th century) and [[Johannes de Muris]] (14th century).<ref>
{{cite journal |last1=Kennedy |first1=E.S. |last2=Muruwwa |first2=Ahmad |year=1958 |title=Bīrūnī on the Solar Equation |journal=Journal of Near Eastern Studies |volume=17 |number=2 |pages=112–121 |doi=10.1086/371451 |jstor=542617}} {{pb}} Johannes de Muris credits an anonymous author for the relevant section of his work ''De Arte Mesurandi''. See {{cite book |last=Van Brummelen |first=Glen |year=2009 |title=The Mathematics of the Heavens and the Earth |publisher=Princeton University Press |pages=240–241}}</ref> Something equivalent to the [[spherical law of cosines]] was used (but not stated in general) by [[Muhammad ibn Musa al-Khwarizmi|al-Khwārizmī]] (9th century), [[al-Battani|al-Battānī]] (9th century), and [[Nilakantha Somayaji|Nīlakaṇṭha]] (15th century).<ref>{{cite book |last=Van Brummelen |first=Glen |year=2012 |title=Heavenly mathematics: The forgotten art of spherical trigonometry |publisher=Princeton University Press |page=98 |bibcode=<!-- 2012hmfa.book.....V --> }}</ref>

The 13th century Persian mathematician [[Nasir al-Din al-Tusi|Naṣīr al-Dīn al-Ṭūsī]], in his {{lang|ar|Kitāb al-Shakl al-qattāʴ}} (''Book on the Complete Quadrilateral'', c. 1250), systematically described how to solve triangles from various combinations of given data. Given two sides and their included angle in a scalene triangle, he proposed finding the third side by dropping a perpendicular from the vertex of one of the unknown angles to the opposite base, reducing the problem to finding the legs of one right triangle from a known angle and hypotenuse using the [[law of sines]] and then finding the hypotenuse of another right triangle from two known sides by the [[Pythagorean theorem]].<ref>{{cite book |title=Traité du quadrilatère attribué a Nassiruddinel-Toussy |chapter=Ch. 3.2: Sur la manière de calculer les côtés et les angles d'un triangle les uns par les autres |page=69 |language=fr |author=Naṣīr al-Dīn al-Ṭūsī |translator-last=Caratheodory |translator-first=Alexandre Pacha |translator-link=Alexander Karatheodori Pasha |chapter-url=https://books.google.com/books?id=gVE7AQAAIAAJ&pg=PA69 |publisher=Typographie et Lithographie Osmanié |year=1891 |quote=On donne deux côtés et un angle. [...] Que si l'angle donné est compris entre les deux côtés donnés, comme l'angle A est compris entre les deux côtés AB AC, abaissez de B sur AC la perpendiculaire BE. Vous aurez ainsi le triangle rectangle [BEA] dont nous connaissons le côté AB et l'angle A; on en tirera BE, EA, et l'on retombera ainsi dans un des cas précédents; c. à. d. dans le cas où BE, CE sont connus; on connaîtra dès lors BC et l'angle C, comme nous l'avons expliqué |trans-quote=Given [...] the angle A is included between the two sides AB AC, drop from B to AC the perpendicular BE. You will thus have the right triangle [BEA] of which we know the side AB and the angle A; in that triangle compute BE, EA, and the problem is reduced to one of the preceding cases; that is, to the case where BE, CE are known; we will thus know BC and the angle C, as we have explained.}}</ref>

About two centuries later, another Persian mathematician, [[Jamshīd al-Kāshī]], who computed the most accurate trigonometric tables of his era, also described the solution of triangles from various combinations of given data in his {{lang|ar|Miftāḥ al-ḥisāb}} (''Key of Arithmetic'', 1427), and repeated essentially al-Ṭūsī's method, now consolidated into one formula and including more explicit details, as follows:<ref>{{cite journal |last=Azarian |first=Mohammad K. |year=2000 |title=Meftab Al-Hesab: A Summary |journal=Missouri Journal of Mathematical Sciences |volume=12 |number=2 |pages=75–95 |doi=10.35834/2000/1202075 |doi-access=free|url=https://projecteuclid.org/journals/missouri-journal-of-mathematical-sciences/volume-12/issue-2/Meftab-Al-Hesab-A-Summary/10.35834/2000/1202075.pdf }}</ref> 

[[File:Law of cosines following al-Kashi.png|thumb|right|upright=1.25|Al-Kashi's version of the law of cosines (case where {{mvar|γ}} is obtuse), expressed with modern algebraic notation.]]
{{quote|text='''Another case''' is when two sides and the angle between them are known and the rest are unknown. We multiply one of the sides by the sine of the [known] angle one time and by the sine of its complement the other time converted and we subtract the second result from the other side if the angle is acute and add it if the angle is obtuse. We then square the result and add to it the square of the first result. We take the square root of the sum to get the remaining side....|sign=Al-Kāshī's {{lang|ar|Miftāḥ al-ḥisāb}},<br>{{in5}}translation by Nuh Aydin, Lakhdar Hammoudi, and Ghada Bakbouk<ref>{{cite book |last1=Aydin |first1=Nuh |last2=Hammoudi |first2=Lakhdar |last3=Bakbouk |first3=Ghada |year=2020 |page=31 |title=Al-Kashi's Miftah al-Hisab, Volume II: Geometry |publisher=Birkhäuser |doi=10.1007/978-3-030-61330-3|isbn=978-3-030-61329-7 }}</ref> }}

Using modern algebraic notation and conventions this might be written

<math display=block>c = \sqrt{(b - a \cos \gamma)^2 + (a \sin \gamma)^2}</math>

when {{tmath|\gamma}} is acute or

<math display=block>c = \sqrt{\left(b + a \left|\cos \gamma\right| \right)^2 + \left(a \sin \gamma\right)^2}</math>

when {{tmath|\gamma}} is obtuse. (When {{tmath|\gamma}} is obtuse, the modern convention is that {{tmath|\cos\gamma}} is negative and <math>\cos(\pi-\gamma) = -\cos\gamma</math> is positive; historically sines and cosines were considered to be line segments with non-negative lengths.) By squaring both sides, [[FOIL method|expanding]] the squared binomial, and then applying the Pythagorean trigonometric identity {{tmath|1= \cos^2\gamma + \sin^2\gamma = 1}}, we obtain the familiar law of cosines:

<math display=block>\begin{align}
c^2
&= b^2 - 2ba\cos \gamma + a^2\cos^2 \gamma + a^2\sin^2\gamma \\[5mu]
&= a^2 + b^2 - 2ab\cos \gamma.
\end{align}</math>

In [[France]], the law of cosines is sometimes referred to as the ''théorème d'Al-Kashi''.<ref name = Pickover>{{Cite book
|url=https://books.google.com/books?id=JrslMKTgSZwC&q=al+kashi+law+of+cosines&pg=PA106
|title=The Math Book: From Pythagoras to the 57th Dimension |last=Pickover |first=Clifford A. |date=2009 |publisher=Sterling Publishing Company, Inc. |isbn=9781402757969 |page=106}}</ref><ref>{{Cite book |title=Programme de mathématiques de première générale |publisher=Ministère de l'Éducation nationale et de la Jeunesse |year=2022 |pages=11,12 |language=fr}}</ref>

The same method used by al-Ṭūsī appeared in Europe as early as the 15th century, in [[Regiomontanus]]'s ''De triangulis omnimodis'' (''On Triangles of All Kinds'', 1464), a comprehensive survey of plane and spherical trigonometry known at the time.<ref>{{cite book |url=https://archive.org/details/regiomontanusont0000regi/ |title=Regiomontanus on Triangles |editor-last=Hughes |editor-first=Barnabas |place=Madison |publisher=University of Wisconsin Press |year=1967 }} English translation of ''De triangulis omnimodis'' by Johann Müller, otherwise known as Regiomontanus, with facsimile of the 1533 Latin edition. [https://archive.org/details/regiomontanusont0000regi/page/98/mode/2up?q=%22two+sides+of+a+triangle+and+their+included+angle%22 Theorem I.49, {{pgs|98–101}}].</ref>

The theorem was first written using algebraic notation 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.<ref>For example in {{cite book |last=Carnot |first=Lazare |title=Géométrie de position |year=1803 |publisher=J.B.M Duprat |url=https://archive.org/details/geometriedeposit00carn/page/202/ |page=202 }}</ref>

==Proofs==

===Using the Pythagorean theorem===
[[Image:Law of cosines in plane trigonometry.svg|thumb|upright=1.25|Proof for the ''Law of cosines'' in acute and obtuse triangles using the  Pythagorean theorem]]
[[Image:Obtuse Triangle With Altitude ZP.svg|thumb|upright=1.25|Obtuse triangle {{math|''ABC''}} with height {{math|''BH''}}]]

====Case of an obtuse angle====
[[Euclid]] proved this theorem by applying the [[Pythagorean theorem]] to each of the two right triangles in Fig. 2 ({{math|''AHB''}} and {{math|''CHB''}}). Using {{math|''d''}} to denote the line segment {{math|''CH''}} and {{math|''h''}} for the height {{math|''BH''}}, triangle {{math|''AHB''}} gives us
<math display=block>c^2 = (b+d)^2 + h^2,</math>

and triangle {{math|''CHB''}} gives
<math display=block>d^2 + h^2 = a^2.</math>

[[Polynomial expansion|Expanding]] the first equation gives
<math display=block>c^2 = b^2 + 2bd + d^2 +h^2.</math>

Substituting the second equation into this, the following can be obtained:
<math display=block>c^2 = a^2 + b^2 + 2bd.</math>

This is Euclid's Proposition 12 from Book 2 of the ''[[Euclid's Elements|Elements]]''.<ref name=Joyce>[http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII12.html Java applet version] by Prof. [[David E. Joyce (mathematician)|D E Joyce]] of Clark University.</ref> To transform it into the modern form of the law of cosines, note that
<math display=block>d = a\cos(\pi-\gamma)= -a\cos\gamma.</math>

====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 {{math|''γ''}} and uses the square of a difference to simplify.

====Another proof in the acute case====
[[Image:Triangle with trigonometric proof of the law of cosines.svg|thumb|right|upright=1.25|Fig. 6 – A short proof using trigonometry for the case of an acute angle]]

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:

<math display=block>\begin{align}
 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{align}</math>

using the [[trigonometric identity]] {{tmath|1= \cos^2\gamma + \sin^2\gamma = 1}}.

This proof needs a slight modification if {{math|''b'' &lt; ''a'' cos(''γ'')}}. In this case, the right triangle to which the Pythagorean theorem is applied moves ''outside'' the triangle {{math|''ABC''}}. The only effect this has on the calculation is that the quantity {{math|''b'' − ''a'' cos(''γ'')}} is replaced by {{math|''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 {{math|''β''}} is obtuse, and may be avoided by reflecting the triangle about the bisector of {{math|''γ''}}.

Referring to Fig. 6 it is worth noting that if the angle opposite side {{math|''a''}} is {{math|''α''}} then:
<math display=block>\tan\alpha = \frac{a\sin\gamma}{b-a\cos\gamma}.</math>

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

===From three altitudes===
[[Image:Triangle-with-cosines.svg|thumb|right|Fig. 5 – An acute triangle with perpendicular]]
The [[altitude (triangle)|altitude]] through vertex {{mvar|C}} is a segment perpendicular to side {{mvar|c}}. The distance from the foot of the altitude to vertex {{mvar|A}} plus the distance from the foot of the altitude to vertex {{mvar|B}} is equal to the length of side {{mvar|c}} (see Fig.&nbsp;5). Each of these distances can be written as one of the other sides multiplied by the cosine of the adjacent angle,<ref name=ctk>[[Alexander Bogomolny]] credits this proof to teacher John Molokach (2011), but it may be older. {{cite web |last=Bogomolny |first=Alexander |author-link=Alexander Bogomolny |url=http://www.cut-the-knot.org/pythagoras/CosLawMolokach.shtml |title=The Law of Cosines (Independent of the Pythagorean Theorem) |website=Cut the Knot |access-date=2024-01-09 }}</ref>
<math display=block>c=a\cos\beta+b\cos\alpha.</math> 

(This is still true if {{math|''α''}} or {{math|''β''}} is obtuse, in which case the perpendicular falls outside the triangle.)  Multiplying both sides by {{math|''c''}} yields
<math display=block>c^2 = ac\cos\beta + bc\cos\alpha.</math>

The same steps work just as well when treating either of the other sides as the base of the triangle:
<math display=block>\begin{align}
a^2 &= ac\cos\beta + ab\cos\gamma, \\[3mu]
b^2 &= bc\cos\alpha + ab\cos\gamma.
\end{align}</math>

Taking the equation for {{tmath|c^2}} and subtracting the equations for {{tmath|b^2}} and {{tmath|a^2}},
<math display=block>
\begin{align}

c^2 - a^2 - b^2 &= {\color{BlueGreen}\cancel{\color{Black}ac\cos\beta}} + {\color{Peach}\cancel{\color{Black}bc\cos\alpha}} - {\color{BlueGreen}\cancel{\color{Black}ac\cos\beta}} - {\color{Peach}\cancel{\color{Black}bc\cos\alpha}} - 2ab\cos\gamma \\
c^2 &= a^2 + b^2- 2ab\cos\gamma.
\end{align}</math>

This proof is independent of the [[Pythagorean theorem]], insofar as it is based only on the right-triangle definition of cosine and obtains squared side lengths algebraically. Other proofs typically invoke the Pythagorean theorem explicitly, and are more geometric, treating {{math|''a''&thinsp;cos ''γ''}} as a label for the length of a certain line segment.<ref name=ctk/>

Unlike many proofs, this one handles the cases of obtuse and acute angles {{math|''γ''}} in a unified fashion.

=== Cartesian coordinates ===
[[File:Triangle-law-of-cosines-proof.png|thumb|right|upright=1.25|Fig. 4 – Coordinate geometry proof]]

Consider a triangle with sides of length {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, where {{math|''θ''}} is the measurement of the angle opposite the side of length {{math|''c''}}. This triangle can be placed on the [[Cartesian coordinate system]] with side {{math|''a''}} aligned along the ''x'' axis and angle {{math|''θ''}} placed at the origin, by plotting the components of the 3 points of the triangle as shown in Fig. 4:
<math display=block>A = (b \cos\theta, b \sin\theta), B = (a, 0), \text{ and } C = (0, 0).</math>

By the [[distance formula]],<ref>{{cite book |last= Wylie |first=Clarence Raymond |year=1955 |title=Plane Trigonometry |publisher=McGraw-Hill |at=§9.1 The Law of Cosines, {{pgs|195–198}} |lccn=54-11278 |url=https://archive.org/details/planetrigonometr0000wyli/page/195 |url-access=limited }}</ref>

<math display=block>c = \sqrt{(a - b \cos\theta)^2 + (0 - b \sin\theta)^2}.</math>

Squaring both sides and simplifying 
<math display=block>\begin{align}
c^2 &=  (a - b \cos\theta)^2 + (- b \sin\theta)^2 \\
    &=  a^2 - 2 a b \cos\theta+ b^2 \cos^2 \theta+ b^2 \sin^2 \theta\\
    &=  a^2 + b^2 (\sin^2 \theta+ \cos^2 \theta) - 2 a b \cos\theta\\
    &=  a^2 + b^2  - 2 a b \cos\theta.
\end{align}</math>

An advantage of this proof is that it does not require the consideration of separate cases depending on whether the angle {{mvar|γ}} is acute, right, or obtuse. However, the cases treated separately in ''Elements'' II.12–13 and later by al-Ṭūsī, al-Kāshī, and others could themselves be combined by using concepts of signed lengths and areas and a concept of signed cosine, without needing a full Cartesian coordinate system.

===Using Ptolemy's theorem===
[[Image:Ptolemy cos.svg|thumb|right|Proof of law of cosines using Ptolemy's theorem]]
Referring to the diagram, triangle ''ABC'' with sides {{math|''AB''}} = {{math|''c''}}, {{math|''BC''}} = {{math|''a''}} and {{math|''AC''}} = {{math|''b''}} is drawn inside its circumcircle as shown. Triangle {{math|''ABD''}} is constructed congruent to triangle {{math|''ABC''}} with {{math|''AD''}} = {{math|''BC''}} and {{math|''BD''}} = {{math|''AC''}}. Perpendiculars from {{math|''D''}} and {{math|''C''}} meet base {{math|''AB''}} at {{math|''E''}} and {{math|''F''}} respectively. Then:
<math display=block>\begin{align}
& BF=AE=BC\cos\hat{B}=a\cos\hat{B} \\
\Rightarrow \ & DC=EF=AB-2BF=c-2a\cos\hat{B}.
\end{align}</math>

Now the law of cosines is rendered by a straightforward application of [[Ptolemy's theorem]] to [[cyclic quadrilateral]] {{math|''ABCD''}}:
<math display=block>\begin{align}
& 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{align}</math>

Plainly if angle {{math|''B''}} is [[right angle|right]], then {{math|''ABCD''}} is a rectangle and application of Ptolemy's theorem yields the [[Pythagorean theorem]]:
<math display=block>a^2+c^2=b^2.</math>

===By comparing areas===
[[Image:law of cosines with acute angles.svg|right|thumb|upright=1.25|Fig. 7a – Proof of the law of cosines for acute angle {{math|''γ''}} by "cutting and pasting".]]
[[Image:law of cosines with an obtuse angle.svg|right|thumb|upright=1.25|Fig. 7b – Proof of the law of cosines for obtuse angle {{math|''γ''}} by "cutting and pasting".]]

One can also prove the law of cosines by calculating [[area]]s. The change of sign as the angle {{math|''γ''}} becomes obtuse makes a case distinction necessary.

Recall that
*{{math|''a''<sup>2</sup>}}, {{math|''b''<sup>2</sup>}}, and {{math|''c''<sup>2</sup>}} are the areas of the squares with sides {{math|''a''}}, {{math|''b''}}, and {{math|''c''}}, respectively;
*if {{math|''γ''}} is acute, then {{math|''ab'' cos ''γ''}} is the area of the [[parallelogram]] with sides {{math|''a''}} and {{math|''b''}} forming an angle of {{math|''γ′'' {{=}} {{sfrac|π|2}} − ''γ''}};
*if {{math|''γ''}} is obtuse, and so {{math|cos ''γ''}} is negative, then {{math|−''ab'' cos ''γ''}} is the area of the [[parallelogram]] with sides ''a'' and ''b'' forming an angle of {{math|''γ′'' {{=}} ''γ'' − {{sfrac|π|2}}}}.

'''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 {{math|''a''<sup>2</sup>}}, {{math|''b''<sup>2</sup>}} on the left and the areas {{math|2''ab'' cos ''γ''}} and {{math|''c''<sup>2</sup>}} on the right;
*in blue, the triangle {{math|''ABC''}}, on the left and on the right;
*in grey, auxiliary triangles, all [[congruence (geometry)|congruent]] to {{math|''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
<math display=block>a^2 + b^2 = c^2 + 2ab\cos\gamma.</math>

'''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 {{math|''γ''}} is obtuse. We have
*in pink, the areas {{math|''a''<sup>2</sup>}}, {{math|''b''<sup>2</sup>}}, and {{math|−2''ab'' cos ''γ''}} on the left and {{math|''c''<sup>2</sup>}} on the right;
*in blue, the triangle {{math|''ABC''}} twice, on the left, as well as on the right.

The equality of areas on the left and on the right gives
<math display=block>a^2 + b^2 - 2ab\cos(\gamma) = c^2.</math>

The rigorous proof will have to include proofs that various shapes are [[congruence (geometry)|congruent]] and therefore have equal area. This will use the theory of [[congruence (geometry)#Congruence of triangles|congruent triangles]].
{{clear}}

=== Using circle geometry ===
[[Image:law-of-cosines-circle-1.svg|thumb|upright=1.25|Fig. 8a – The triangle {{math|''ABC''}} (pink), an auxiliary circle (light blue) and an auxiliary right triangle (yellow)]]
[[Image:law-of-cosines-circle-2.svg|thumb|right|upright=1.25|Fig. 8b – The triangle {{math|''ABC''}} (pink), an auxiliary circle (light blue) and two auxiliary right triangles (yellow)]]
[[Image:triangle with circle of center B and radius BC.svg|right|thumb|upright=1.25|Fig. 9 – Proof of the law of cosines using the power of a point theorem.]]

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

'''Case of acute angle {{math|''γ''}}, where {{math|''a'' &gt; 2''b'' cos ''γ''}}.''' Drop the [[perpendicular]] from {{math|''A''}} onto {{math|''a''}} = {{math|''BC''}}, creating a line segment of length {{math|''b'' cos ''γ''}}. Duplicate the ''[[triangle#Types of triangle|right triangle]]'' to form the [[triangle#Types of triangle|isosceles triangle]] {{math|''ACP''}}. Construct the [[circle]] with center {{math|''A''}} and radius {{math|''b''}}, and its [[tangent (trigonometric function)|tangent]] {{math|''h'' {{=}} ''BH''}} through {{math|''B''}}. The tangent {{math|''h''}} forms a right angle with the radius {{math|''b''}} (Euclid's ''Elements'': Book 3, Proposition 18; or see [[circle#Tangent|here]]), so the yellow triangle in Figure 8 is right. Apply  the [[Pythagorean theorem]] to obtain
<math display=block>c^2 = b^2 + h^2.</math>

Then use the [[Tangent-secant theorem|''tangent secant theorem'']] (Euclid's ''Elements'': Book 3, Proposition 36), which says that the square on the tangent through a point {{math|''B''}} outside the circle is equal to the product of the two lines segments (from {{math|''B''}}) created by any [[secant line|secant]] of the circle through {{math|''B''}}. In the present case: {{math|''BH''<sup>2</sup> {{=}} ''BC''·''BP''}}, or
<math display=block>h^2 = a(a - 2b\cos\gamma).</math>

Substituting into the previous equation gives the law of cosines:
<math display=block>c^2 = b^2 + a(a - 2b\cos\gamma).</math>

Note that {{math|''h''<sup>2</sup>}} is the [[power of a point|power]] of the point {{math|''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]].

'''Case of acute angle {{math|''γ''}}, where {{math|''a'' &lt; 2''b'' cos ''γ''}}.''' Drop the [[perpendicular]] from {{math|''A''}} onto {{math|''a''}} = {{math|''BC''}}, creating a line segment of length {{math|''b'' cos ''γ''}}. Duplicate the [[triangle#Types of triangle|right triangle]] to form the [[triangle#Types of triangle|isosceles triangle]] {{math|''ACP''}}. Construct the [[circle]] with center {{math|''A''}} and radius {{math|''b''}}, and a [[chord (geometry)|chord]] through {{math|''B''}} perpendicular to {{math|''c'' {{=}} ''AB'',}} half of which is {{math|''h'' {{=}} ''BH''.}} Apply  the [[Pythagorean theorem]] to obtain
<math display=block>b^2 = c^2 + h^2.</math>

Now use the [[intersecting chords theorem|''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: {{math|''BH''<sup>2</sup> {{=}} ''BC''·''BP'',}} or
<math display=block>h^2 = a(2b\cos\gamma - a).</math>

Substituting into the previous equation gives the law of cosines:
<math display=block>b^2 = c^2 + a(2b\cos\gamma - a).</math>

Note that the power of the point {{math|''B''}} with respect to the circle has the negative value {{math|−''h''<sup>2</sup>}}.

'''Case of obtuse angle {{math|''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 {{math|''B''}} and radius {{math|''a''}} (see Figure 9), which intersects the [[secant line|secant]] through {{math|''A''}} and {{math|''C''}} in {{math|''C''}} and {{math|''K''}}. The [[power of a point|power]] of the point {{math|''A''}} with respect to the circle is equal to both {{math|''AB''<sup>2</sup> − ''BC''<sup>2</sup>}} and {{math|''AC''·''AK''}}. Therefore,
<math display=block>\begin{align}
c^2 - a^2 & {} = b(b + 2a\cos(\pi - \gamma)) \\
& {} = b(b - 2a\cos\gamma),
\end{align}</math>

which is the law of cosines.

Using algebraic measures for line segments (allowing [[negative numbers]] as lengths of segments) the case of obtuse angle ({{math|''CK'' &gt; 0}}) and acute angle ({{math|''CK'' &lt; 0}}) can be treated simultaneously.

===Using the law of sines===
The law of cosines can be proven algebraically from the [[law of sines]] and a few standard trigonometric identities.<ref>{{cite journal |last=Burton |first=L. J. |year=1949 |title=The Laws of Sines and Cosines |journal=The American Mathematical Monthly |volume=56 |number=8 |pages=550–551 |doi=10.1080/00029890.1949.11999439 |jstor=2305533}}</ref> To start, three angles of a triangle sum to a [[straight angle]] (<math>\alpha + \beta + \gamma = \pi</math> radians). Thus by the angle sum identities for sine and cosine,

<math display=block>\begin{alignat}{3}
\sin \gamma &= \phantom{-}\sin(\pi - \gamma)
&&= \phantom{-}\sin(\alpha + \beta)
&&= \sin\alpha\,\cos\beta + \cos\alpha\,\sin\beta, \\[5mu]
\cos \gamma &= -\cos(\pi - \gamma)
&&= -\cos(\alpha + \beta)
&&= \sin\alpha\,\sin\beta - \cos\alpha\,\cos\beta.
\end{alignat}</math>

Squaring the first of these identities, then substituting <math>\cos\alpha\,\cos\beta = {}</math><math>\sin\alpha\,\sin\beta - \cos \gamma</math> from the second, and finally replacing <math>\cos^2 \alpha + \sin^2 \alpha = {}</math><math> \cos^2 \beta + \sin^2 \beta = 1,</math> the [[Pythagorean trigonometric identity]], we have:

<math display=block>\begin{align}
\sin^2 \gamma
&= (\sin\alpha\,\cos\beta + \cos\alpha\,\sin\beta)^2 \\[3mu]
&= \sin^2\alpha\,\cos^2\beta + 2\sin\alpha\,\sin\beta\,\cos\alpha\,\cos\beta + \cos^2\alpha\,\sin^2\beta \\[3mu]
&= \sin^2\alpha\,\cos^2\beta + 2\sin\alpha\,\sin\beta(\sin\alpha\,\sin\beta - \cos \gamma) + \cos^2\alpha\,\sin^2\beta \\[3mu]
&= \sin^2\alpha(\cos^2\beta + \sin^2\beta) + \sin^2\beta(\cos^2\alpha + \sin^2\alpha) - 2\sin\alpha\,\sin\beta\,\cos \gamma \\[3mu]
&= \sin^2\alpha + \sin^2\beta - 2\sin\alpha\,\sin\beta\,\cos \gamma.
\end{align}</math>

The law of sines holds that
<math display=block>\frac{a}{\sin \alpha\vphantom{\beta}} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma\vphantom{\beta}} = k,</math>

so to prove the law of cosines, we multiply both sides of our previous identity by {{tmath|k^2}}:

<math display=block>\begin{align}
\sin^2 \gamma \frac{c^2}{\sin^2 \gamma}
&= \sin^2 \alpha \frac{a^2}{\sin^2 \alpha} + \sin^2 \beta \frac{b^2}{\sin^2 \beta} - 2\sin\alpha\,\sin\beta\,\cos\gamma\frac{ab}{\sin\alpha\,\sin\beta\vphantom{\sin^2}} \\[10mu]
c^2 &= a^2 + b^2 - 2ab\cos\gamma.
\end{align}</math>

This concludes the proof.

=== Using vectors ===
[[File:Dot product cosine rule.svg|100px|thumb|Triangle with vector edges '''a''' and '''b''', separated by angle ''θ''.]]

Denote

<math display=block>
\overrightarrow{CB}=\vec{a},\ \overrightarrow{CA}=\vec{b},\ \overrightarrow{AB}=\vec{c}
</math>

Therefore,
<math display=block>\vec{c} = \vec{a}-\vec{b}</math>

Taking the [[dot product]] of each side with itself:
<math display=block>\begin{align}
\vec{c}\cdot\vec{c} &= (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) \\
\Vert\vec{c}\Vert^2 &= \Vert\vec{a}\Vert^2 + \Vert\vec{b}\Vert^2 - 2\,\vec{a}\cdot\vec{b}
\end{align}</math>

Using the identity

<math display=block>
\vec{u}\cdot\vec{v} = \Vert\vec{u}\Vert\,\Vert\vec{v}\Vert \cos\angle(\vec{u},\ \vec{v})
</math>

leads to

<math display=block>\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})</math>

The result follows.

==Isosceles case==
When {{math|''a'' {{=}} ''b''}}, i.e., when the triangle is [[triangle#Types of triangle|isosceles]] with the two sides incident to the angle {{math|''γ''}} equal, the law of cosines simplifies significantly. Namely, because {{math|''a''<sup>2</sup> + ''b''<sup>2</sup> {{=}} 2''a''<sup>2</sup> {{=}} 2''ab''}}, the law of cosines becomes
<math display=block>\cos\gamma = 1 - \frac{c^2}{2a^2}</math>

or
<math display=block>c^2 = 2a^2 (1 - \cos\gamma).</math>

==Analogue for tetrahedra==
Given an arbitrary [[tetrahedron]] whose four faces have areas {{mvar|A}}, {{mvar|B}}, {{mvar|C}}, and {{mvar|D}}, with [[dihedral angle]] {{tmath|\varphi_{ab} }} between faces {{mvar|A}} and {{mvar|B}}, etc., a higher-dimensional analogue of the law of cosines is:<ref>{{cite book
|last=Casey
|first=John
|title=A Treatise on Spherical Trigonometry: And Its Application to Geodesy and Astronomy with Numerous Examples
|location=London
|publisher=Longmans, Green, & Company
|year=1889
|page=133}}</ref>
<math display=block>
A^2 = B^2 + C^2 + D^2 - 2\left(BC \cos \varphi_{bc} + CD \cos\varphi_{cd} + DB\cos \varphi_{db}\right).
</math>

==Version suited to small angles==
When the angle, {{math|''γ''}}, is small and the adjacent sides, {{math|''a''}} and {{math|''b''}}, are of similar length, the right hand side of the standard form of the law of cosines is subject to [[catastrophic cancellation]] in numerical approximations. 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:
<math display=block>\begin{align}
c^2 &= (a - b)^2 + 4ab\sin^2\left(\frac{\gamma}{2}\right) \\
& = (a - b)^2 + 4ab\operatorname{haversin}(\gamma).
\end{align}</math>

In the limit of an infinitesimal angle, the law of cosines degenerates into the [[circular arc length]] formula, {{math|''c'' {{=}} ''a'' ''γ''}}.

==In non-Euclidean geometry==
{{Main article|Spherical law of cosines|Hyperbolic law of cosines}}
[[Image:Law-of-haversines.svg|right|thumb|Spherical triangle solved by the law of cosines.]]

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

A triangle is defined by three points {{math|'''u'''}}, {{math|'''v'''}}, and {{math|'''w'''}} on the unit sphere, and the arcs of [[great circle]]s connecting those points. If these great circles make angles {{math|''A''}}, {{math|''B''}}, and {{math|''C''}} with opposite sides {{math|''a''}}, {{math|''b''}}, {{math|''c''}} then the [[spherical law of cosines]] asserts that all of the following relationships hold:

<math display=block>\begin{align}
\cos a &= \cos b\cos c + \sin b\sin c\cos A\\
\cos A &= -\cos B\cos C + \sin B\sin C\cos a\\
\cos a &= \frac{\cos A + \cos B\cos C} {\sin B \sin C}.
\end{align}</math>


In [[hyperbolic geometry]], a pair of equations are collectively known as the [[hyperbolic law of cosines]]. The first is
<math display=block>\cosh a = \cosh b\cosh c - \sinh b \sinh c \cos A</math>

where {{math|sinh}} and {{math|cosh}} are the [[hyperbolic functions|hyperbolic sine and cosine]], and the second is
<math display=block>\cos A = -\cos B \cos C + \sin B\sin C\cosh a.</math>

The length of the sides can be computed by:

<math display=block>\cosh a = \frac{\cos A + \cos B\cos C} {\sin B \sin C}. </math>

==Polyhedra==
The law of cosines can be generalized to all [[polyhedra]] by considering any polyhedron with vector sides and invoking the [[divergence Theorem]].<ref>{{cite journal |last1=Collins |first1=L |last2=Osler |first2=T |year=2011 |title=Law of Cosines Generalised for any Polygon and any Polyhedron |journal=The Mathematical Gazette |volume=95 |number=533 |pages=240–243 |doi=10.1017/S0025557200002953 |jstor=23248682}}</ref>

==See also==
*[[Half-side formula]]
*[[Law of sines]]
*[[Law of tangents]]
*[[Law of cotangents]]
*[[List of trigonometric identities]]
*[[Mollweide's formula]]

==Notes==
{{notelist}}
==References==
{{Reflist}}

==External links==
{{Commonscat}}
* {{springer|title=Cosine theorem|id=p/c026660}}
* [http://www.cut-the-knot.org/pythagoras/cosine.shtml Several derivations of the Cosine Law, including Euclid's] at [[cut-the-knot]]
* [http://www.mathwarehouse.com/trigonometry/law-of-cosines-formula-examples.php Interactive applet of Law of Cosines]

{{DEFAULTSORT:Law Of Cosines}}
[[Category:Trigonometry]]
[[Category:Angle]]
[[Category:Articles containing proofs]]
[[Category:Theorems about triangles]]