Law of sines explained

In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law,$$\backslash frac\; \backslash ,=\backslash ,\; \backslash frac\; \backslash ,=\backslash ,\; \backslash frac\; \backslash ,=\backslash ,\; 2R,$$where, and are the lengths of the sides of a triangle, and, and are the opposite angles (see figure 2), while is the radius of the triangle's circumcircle. When the last part of the equation is not used, the law is sometimes stated using the reciprocals;$$\backslash frac\; \backslash ,=\backslash ,\; \backslash frac\; \backslash ,=\backslash ,\; \backslash frac.$$The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ambiguous case) and the technique gives two possible values for the enclosed angle.

The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the law of cosines.

The law of sines can be generalized to higher dimensions on surfaces with constant curvature.^[1]

History

H. J. J. Winter's book Eastern Science states that the 7th century Indian mathematician Brahmagupta describes what we now know as the law of sines in his astronomical treatise Brāhmasphuṭasiddhānta.^[2] In his partial translation of this work, Colebrooke translates Brahmagupta's statement of the sine rule as: The product of the two sides of a triangle, divided by twice the perpendicular, is the central line; and the double of this is the diameter of the central line.^[3]

According to Ubiratàn D'Ambrosio and Helaine Selin, the spherical law of sines was discovered in the 10th century. It is variously attributed to Abu-Mahmud Khojandi, Abu al-Wafa' Buzjani, Nasir al-Din al-Tusi and Abu Nasr Mansur.^[4]

Ibn Muʿādh al-Jayyānī's The book of unknown arcs of a sphere in the 11th century contains the spherical law of sines. The plane law of sines was later stated in the 13th century by Nasīr al-Dīn al-Tūsī. In his On the Sector Figure, he stated the law of sines for plane and spherical triangles, and provided proofs for this law.^[5]

According to Glen Van Brummelen, "The Law of Sines is really Regiomontanus's foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles."^[6] Regiomontanus was a 15th-century German mathematician.

Proof

With the side of length as the base, the triangle's altitude can be computed as or as . Equating these two expressions gives$$\backslash frac\; =\; \backslash frac\backslash ,,$$and similar equations arise by choosing the side of length or the side of length as the base of the triangle.

The ambiguous case of triangle solution

When using the law of sines to find a side of a triangle, an ambiguous case occurs when two separate triangles can be constructed from the data provided (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangles and .

Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous:

• The only information known about the triangle is the angle and the sides and .
• The angle is acute (i.e., < 90°).
• The side is shorter than the side (i.e.,).
• The side is longer than the altitude from angle, where (i.e.,).

If all the above conditions are true, then each of angles and produces a valid triangle, meaning that both of the following are true:$$\text{'}\; =\; \backslash arcsin\backslash frac\; \backslash quad\; \backslash text\; \backslash quad\; =\; \backslash pi\; -\; \backslash arcsin\backslash frac.$$

From there we can find the corresponding and or and if required, where is the side bounded by vertices and and is bounded by and .

Examples

The following are examples of how to solve a problem using the law of sines.

Example 1

Given: side, side, and angle . Angle is desired.

Using the law of sines, we conclude that$$\backslash frac\; =\; \backslash frac.$$$$\backslash alpha\; =\; \backslash arcsin\backslash left(\backslash frac\; \backslash right)\; \backslash approx\; 32.39^\backslash circ.$$

Note that the potential solution is excluded because that would necessarily give .

Example 2

If the lengths of two sides of the triangle and are equal to, the third side has length, and the angles opposite the sides of lengths,, and are,, and respectively then

Relation to the circumcircle

In the identity$$\backslash frac\; =\; \backslash frac\; =\; \backslash frac,$$the common value of the three fractions is actually the diameter of the triangle's circumcircle. This result dates back to Ptolemy.^{[7] [8]}

Proof

As shown in the figure, let there be a circle with inscribed

\triangle ABC and another inscribed \triangle ADB that passes through the circle's center . The \angle AOD has a central angle of 180 ^{\circ} and thus by Thales's theorem. Since \triangle ABD is a right triangle,$$\backslash sin=\; \backslash frac=\; \backslash frac,$$where $R=\; \backslash frac$ is the radius of the circumscribing circle of the triangle. Angles {\gamma} and {\delta} lie on the same circle and subtend the same chord ; thus, by the inscribed angle theorem, Therefore,$$\backslash sin\; =\; \backslash sin\; =\; \backslash frac.$$

Rearranging yields$$2R\; =\; \backslash frac.$$

Repeating the process of creating

\triangle ADB with other points gives

Relationship to the area of the triangle

The area of a triangle is given by where

\theta is the angle enclosed by the sides of lengths and . Substituting the sine law into this equation gives$$T=\backslash fracab\; \backslash cdot\; \backslash frac\; .$$

Taking

R as the circumscribing radius,

The second equality above readily simplifies to Heron's formula for the area.

The sine rule can also be used in deriving the following formula for the triangle's area: denoting the semi-sum of the angles' sines as we have^[9]

where

R is the radius of the circumcircle:

The spherical law of sines

The spherical law of sines deals with triangles on a sphere, whose sides are arcs of great circles.

Suppose the radius of the sphere is 1. Let,, and be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere,,, and are the angles at the center of the sphere subtended by those arcs, in radians. Let,, and be the angles opposite those respective sides. These are dihedral angles between the planes of the three great circles.

Then the spherical law of sines says:$$\backslash frac\; =\; \backslash frac\; =\; \backslash frac.$$

Vector proof

Consider a unit sphere with three unit vectors, and drawn from the origin to the vertices of the triangle. Thus the angles,, and are the angles,, and, respectively. The arc subtends an angle of magnitude at the centre. Introduce a Cartesian basis with along the -axis and in the -plane making an angle with the -axis. The vector projects to in the -plane and the angle between and the -axis is . Therefore, the three vectors have components:$$\backslash mathbf\; =\; \backslash begin0\; \backslash \backslash \; 0\; \backslash \backslash \; 1\backslash end,\; \backslash quad\backslash mathbf\; =\; \backslash begin\backslash sin\; c\; \backslash \backslash \; 0\; \backslash \backslash \; \backslash cos\; c\backslash end,\; \backslash quad\backslash mathbf\; =\; \backslash begin\backslash sin\; b\backslash cos\; A\; \backslash \backslash \; \backslash sin\; b\backslash sin\; A\; \backslash \backslash \; \backslash cos\; b\backslash end.$$

The scalar triple product, is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle, and . This volume is invariant to the specific coordinate system used to represent, and . The value of the scalar triple product is the determinant with, and as its rows. With the -axis along the square of this determinant isRepeating this calculation with the -axis along gives, while with the -axis along it is . Equating these expressions and dividing throughout by gives$$\backslash frac=\; \backslash frac=\; \backslash frac=\; \backslash frac,$$where is the volume of the parallelepiped formed by the position vector of the vertices of the spherical triangle. Consequently, the result follows.

It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since$$\backslash lim\_\; \backslash frac\; =\; 1$$and the same for and .

Geometric proof

Consider a unit sphere with:$$OA\; =\; OB\; =\; OC\; =\; 1$$

Construct point

D and point E such that \angle ADO = \angle AEO = 90 ^{\circ}

Construct point

A' such that \angle A'DO = \angle A'EO = 90 ^{\circ}

It can therefore be seen that

\angle ADA' = B and \angle AEA' = C

Notice that

A' is the projection of A on plane OBC . Therefore \angle AA'D = \angle AA'E = 90 ^{\circ}

By basic trigonometry, we have:

But

AA' = AD \sin B = AE \sin C

Combining them we have:

By applying similar reasoning, we obtain the spherical law of sine:$$\backslash frac\; =\backslash frac\; =\backslash frac$$

See also: Spherical trigonometry, Spherical law of cosines and Half-side formula.

Other proofs

A purely algebraic proof can be constructed from the spherical law of cosines. From the identity

\sin ^{2} A = 1 - \cos ^{2} A and the explicit expression for \cos A from the spherical law of cosinesSince the right hand side is invariant under a cyclic permutation of a, b, c the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.

Hyperbolic case

In hyperbolic geometry when the curvature is −1, the law of sines becomes$$\backslash frac\; =\; \backslash frac\; =\; \backslash frac\; \backslash ,.$$

In the special case when is a right angle, one gets$$\backslash sin\; C\; =\; \backslash frac$$

which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse.

See also: Hyperbolic triangle.

The case of surfaces of constant curvature

Define a generalized sine function, depending also on a real parameter

\kappa :$$\backslash sin\_\backslash kappa(x)\; =\; x\; -\; \backslash fracx^3\; +\; \backslash fracx^5\; -\; \backslash fracx^7\; +\; \backslash cdots\; =\; \backslash sum\_^\backslash infty\; \backslash fracx^\; =\; \backslash frac\; ^$$

The law of sines in constant curvature

\kappa reads as^[1] $$\backslash frac\; =\; \backslash frac\; =\; \backslash frac\; \backslash ,.$$

By substituting

\kappa =0 , \kappa =1 , and \kappa =-1 , one obtains respectively the Euclidean, spherical, and hyperbolic cases of the law of sines described above.

Let

p _{\kappa(r)} indicate the circumference of a circle of radius r in a space of constant curvature \kappa . Then p _{\kappa(r)=2\pi\sin} _{\kappa(r)} . Therefore, the law of sines can also be expressed as:$$\backslash frac\; =\; \backslash frac\; =\; \backslash frac\; \backslash ,.$$

This formulation was discovered by János Bolyai.^[10]

Higher dimensions

A tetrahedron has four triangular facets. The absolute value of the polar sine of the normal vectors to the three facets that share a vertex of the tetrahedron, divided by the area of the fourth facet will not depend upon the choice of the vertex:^[11]

More generally, for an -dimensional simplex (i.e., triangle, tetrahedron, pentatope, etc.) in -dimensional Euclidean space, the absolute value of the polar sine of the normal vectors of the facets that meet at a vertex, divided by the hyperarea of the facet opposite the vertex is independent of the choice of the vertex. Writing for the hypervolume of the -dimensional simplex and for the product of the hyperareas of its -dimensional facets, the common ratio is$$\backslash frac\; =\; \backslash cdots\; =\; \backslash frac\; =\; \backslash frac.$$

See also

• • Half-side formula for solving spherical triangles
• Law of cosines
• Law of tangents
• Law of cotangents
• Mollweide's formula for checking solutions of triangles
• Solution of triangles
• Surveying

External links

• • The Law of Sines at cut-the-knot
• Degree of Curvature
• Finding the Sine of 1 Degree
• Generalized law of sines to higher dimensions

Notes and References

1. Web site: Generalized law of sines. mathworld.
2. Book: Winter, Henry James Jacques . Eastern Science . . 1952 . 46 .
3. Book: Colebrooke, Henry Thomas . Algebra, with Arithmetic and Mensuration from the Sanscrit of Brahmegupta and Bhascara . London . . 1817 . 299-300 .
4. Sesiano just lists al-Wafa as a contributor. Book: Sesiano, Jacques . Islamic mathematics . 137–157 . Mathematics Across Cultures: The History of Non-western Mathematics. Helaine. Selin. Ubiratan. D'Ambrosio. 2000. Springer. 1-4020-0260-2.
5. Book: Berggren, J. Lennart . The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook . Mathematics in Medieval Islam . . 2007 . 978-0-691-11485-9 . 518 .
6. Book: Van Brummelen, Glen . 2009 . The Mathematics of the Heavens and the Earth: The Early History of Trigonometry . . 259 . 0-691-12973-8.
7. Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967
8. Web site: Law of Sines. www.pballew.net. 2018-09-18.
9. Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.
10. Book: Katok, Svetlana. Svetlana Katok. Fuchsian groups. limited. 1992. University of Chicago Press. Chicago. 0-226-42583-5. 22.
11. Eriksson . Folke . 1978 . The law of sines for tetrahedra and n-simplices . Geometriae Dedicata . 7 . 1 . 71–80 . 10.1007/bf00181352.