Law of cosines

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 with side a aligned along the x axis and angle θ placed at the origin, 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

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

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.

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

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

Adding the latter two equations gives

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

Substituting the first equation into the last one results in

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

which is equivalent 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.

Euclid proved this theorem by applying the Pythagorean theorem to each of the two right triangles in Fig. 2 (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.[12] To transform it into the modern form of the law of cosines, note that

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

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 square of a difference to simplify.

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:

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

using the trigonometric identity ${\displaystyle \cos ^{2}\gamma +\sin ^{2}\gamma =1.}$

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 b − a 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.

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:

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

Fig. 7a – Proof of the law of cosines for acute angle γ by "cutting and pasting".

Fig. 7b – Proof of the law of cosines for obtuse angle γ by "cutting and pasting".

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/2 − γ;
• 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′ = γ − 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 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

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

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.

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

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

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

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.

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.

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.

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 AB² − BC² 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.

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)

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

leads to

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

${\displaystyle c\sin \alpha =a\sin \gamma .}$

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 (b − c 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:

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

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

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

leads to

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

The result follows.