Law of cosines

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Fig. 1 - A triangle.

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This article is about the law of cosines in Euclidean geometry. For the corresponding theorem in spherical geometry, see law of cosines (spherical). For the cosine law of optics, see Lambert's cosine law.

In trigonometry, the law of cosines (also known as Al-Kashi law or the cosine formula or cosine rule) is a statement about a general triangle which relates the lengths of its sides to the cosine of one of its angles. Using notation as in Fig. 1, the law of cosines states that

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

or, equivalently:

$b^2 = c^2 + a^2 - 2ca\cos(\beta) , \,$

$a^2 = b^2 + c^2 - 2bc\cos(\alpha) , \,$

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

Note that c is the side opposite of angle γ, and that a and b are the two sides enclosing γ. All three of the identities above say the same thing; they are listed separately only because in solving triangles with three given sides one may apply the identity three times with the roles of the three sides permuted.

The law of cosines generalizes the Pythagorean theorem, which holds only in right triangles: if the angle γ is a right angle (of measure 90° or $\scriptstyle\pi/2$ radians), then $\scriptstyle\cos(\gamma)\, =\, 0$ , and thus the law of cosines reduces to

$c^2 = a^2 + b^2 \,$

which is the Pythagorean theorem.

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.

1 History
2 Applications
3 Proofs
4 Vector formulation
5 Isosceles case
6 Analog for tetrahedra
7 References
8 See also
9 External links

[edit] History

Fig. 2 - Obtuse triangle ABC with perpendicular BH

Euclid's Elements, dating back to the 3rd century BC, contains a version of the law of cosines. The case of obtuse triangle and acute triangle (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.[1]

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

$AB^2 = CA^2 + CB^2 + 2 (CA)(CH)\,.$

This formula may be transformed into the law of cosines by noting that CH = a cos(π – γ) = −a cos(γ).

Proposition 13 contains an entirely analogous statement for acute triangles.

It was not until the development of modern trigonometry in the Middle Ages by Muslim mathematicians that the law of cosines evolved beyond Euclid's two theorems. The astronomer and mathematician al-Battani generalized Euclid's result to spherical geometry at the beginning of the 10th century, which permitted him to calculate the angular distances between stars. During the 15th century, al-Kashi in Samarcand computed trigonometric tables to great accuracy and put the theorem into a form suitable for triangulation. In France, the law of cosines is still referred to as the theorem of Al-Kashi.

The theorem was popularised in the Western world by François Viète, who apparently discovered it independently. At the beginning of the 19th century modern algebraic notation allowed the law of cosines to be written in its current form.

[edit] Applications

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

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

the third side of a triangle if one knows two sides and the angle between them:

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

the angles of a triangle if one knows the three sides:

$\,\gamma = \cos^{-1} \frac{a^2+b^2-c^2}{2ab}\,;$

the third side of a triangle if one knows two sides and an angle opposite to one of them (you may also use the Pythagorean Theorem to do this):

$\, a=b\cos(\gamma) \pm \sqrt{c^2 -b^2\sin^2(\gamma)}\,.$

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.

The third formula shown is the result of solving for a the quadratic equation a² − 2ab cos γ + b² − c² = 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(C) < c < b, only one positive solution if c > b or c = b sin(C), and no solution if c < b sin(C). These different cases are also explained by the Side-Side-Angle congruence ambiguity.

[edit] Proofs

[edit] Using the distance formula

Consider a triangle with sides of length a, b, c, where $θ$ is the measurement of the angle opposite the side of length c. We can place this triangle on the coordinate system by plotting $A(b\cos \theta ,\ b\sin \theta ),\ B(a,0),\ \text{and}\ C(0,0).$ By the distance formula, we have $c = \sqrt{(b\cos \theta - a)^2+(b\sin \theta - 0)^2}$ . Now, we just work with this equation:

$\begin{align} c^2 & {} = (b\cos \theta - a)^2+(b\sin \theta - 0)^2 \\ c^2 & {} = b^2 \cos ^2 \theta - 2ab\cos \theta + a^2 + b^2\sin ^2 \theta \\ c^2 & {} = a^2 + b^2 (\sin ^2 \theta + \cos ^2 \theta ) - 2ab\cos \theta \\ c^2 & {} = a^2 + b^2 - 2ab\cos \theta \end{align}$

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

[edit] Using trigonometry

Fig. 4 - An acute triangle with perpendicular

Drop the perpendicular onto the side c to get (see Fig. 4)

$c=a\cos(\beta)+b\cos(\alpha)\,.$

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

$c^2 = ac\cos(\beta) + bc\cos(\alpha)\,.$

By considering the other perpendiculars obtain

$a^2 = ac\cos(\beta) + ab\cos(\gamma)\,,$

$b^2 = bc\cos(\alpha) + ab\cos(\gamma)\,.$

Adding the latter two equations gives

$a^2 + b^2 = ac\cos(\beta) + bc\cos(\alpha) + 2ab\cos(\gamma)\,$

Rearrange to get

$ac\cos(\beta) + bc\cos(\alpha) = a^2 + b^2 - 2ab\cos(\gamma)\,$

Substitute this in the first equation for $c 2$ to yield the law of cosines

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

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 case of obtuse and acute angle γ separately.

[edit] Using the Pythagorean theorem

Fig. 5 - Obtuse triangle ABC with height BH

Case of an obtuse angle. Euclid proves this theorem by applying the Pythagorean theorem to each of the two right triangles in Fig. 5. Using d to denote the line segment CH and h for the height BH, triangle AHB gives us

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

and triangle CHB gives us

$d^2 + h^2 = a^2.\,$

Expanding the first equation gives us

$c^2 = b^2 + 2bd + d^2 +h^2.\,$

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

$c^2 = a^2 + b^2 + 2bd.\,$

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

$d = a\cos(\pi-\gamma)= -a\cos(\gamma).\,$

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 a little more trigonometry, the law of cosines by applying 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:

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

upon using the trigonometric identity

$\cos^2(\gamma) + \sin^2(\gamma) = 1. \,$

Remark. 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. Note. This problem only occurs when β is obtuse, and may be avoided by reflecting the triangle about the bisector of γ.

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

$tan(\alpha)= \frac{a\sin(\gamma)}{b-a\cos(\gamma)}$

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

[edit] 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:

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

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

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

Plainly if angle B is 90 degrees, then ABCD is a rectangle and application of Ptolemy's theorem yields Pythagoras' theorem:

$a^2+c^2=b^2.\quad$

[edit] 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 $\scriptstyle\gamma'\, =\, \pi/2 - \gamma$ ;
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 $\scriptstyle\gamma' \,=\, \gamma - \pi/2$ .

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

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

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

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

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.

[edit] 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 > 2 b 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

$c^2 = b^2 + h^2\,.$

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

$h^2 = a(a - 2b\cos(\gamma))\,.$

Substuting into the previous equation gives the law of cosines:

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

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 < 2 b 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

$b^2 = c^2 + h^2\,.$

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

$h^2 = a(2b\cos(\gamma) - a)\,.$

Substuting into the previous equation gives the law of cosines:

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

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 AB² − BC² and AC·AK. Therefore,

$\begin{align} c^2 - a^2 & {} = b(b + 2a\cos(\pi - \gamma)) \\ & {} = b(b - 2a\cos(\gamma)) \end{align}$

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.

[edit] Vector formulation

The law of cosines is equivalent to the formula

$\vec b\cdot \vec c = \Vert \vec b\Vert\Vert\vec c\Vert\cos \theta$

in the theory of vectors, which expresses the dot product of two vectors in terms of their respective lengths and the angle they enclose.

Fig. 10 - Vector triangle

Proof of equivalence. Referring to Figure 10, note that

$\vec a=\vec b-\vec c\,,$

and so we may calculate:

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

The law of cosines formulated in this context states:

$\Vert\vec a\Vert^2 = \Vert\vec b \Vert^2 + \Vert\vec c \Vert^2 - 2 \Vert \vec b\Vert\Vert\vec c\Vert\cos(\theta) \,,$

which is now visibly equivalent to the above formula from the theory of vectors.

[edit] 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

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

[edit] Analog for tetrahedra

An analogous statement begins by taking $\scriptstyle{\alpha,\ \beta,\ \gamma,\ \delta }$ to be the areas of the four faces of a tetrahedron. Denote the dihedral angles by $\scriptstyle{ \widehat{\beta\gamma}, }$ etc. Then^[1]

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

[edit] References

^ Casey, John (1889). A Treatise on Spherical Trigonometry: And Its Application to Geodesy and Astronomy with Numerous Examples. London: Longmans, Green, & Company, 133..

[edit] See also

[edit] External links

An Excellent Tutorial on the Law of Cosines From PlainMath.Net
Several derivations of the Cosine Law, including Euclid's at cut-the-knot

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