Law of sines

Template:Short description Template:About Template:Multiple image Template:Trigonometry 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, $$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }=2R,$$ where Template:Math, and Template:Math are the lengths of the sides of a triangle, and Template:Math, and Template:Math are the opposite angles (see figure 2), while Template:Math 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; $$\frac{\sin \alpha }{a}=\frac{\sin \beta }{b}=\frac{\sin \gamma }{c}.$$ 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.

History

An equivalent of the law of sines, that the sides of a triangle are proportional to the chords of double the opposite angles, was known to the 2nd century Hellenistic astronomer Ptolemy and used occasionally in his Almagest.

Statements related to the law of sines appear in the astronomical and trigonometric work of 7th century Indian mathematician Brahmagupta. In his Brāhmasphuṭasiddhānta, Brahmagupta expresses the circumradius of a triangle as the product of two sides divided by twice the altitude; the law of sines can be derived by alternately expressing the altitude as the sine of one or the other base angle times its opposite side, then equating the two resulting variants. An equation even closer to the modern law of sines appears in Brahmagupta's Khaṇḍakhādyaka, in a method for finding the distance between the Earth and a planet following an epicycle; however, Brahmagupta never treated the law of sines as an independent subject or used it more systematically for solving triangles.

The spherical law of sines is sometimes credited to 10th century scholars Abu-Mahmud Khujandi or Abū al-Wafāʾ (it appears in his Almagest), but it is given prominence in Abū Naṣr Manṣūr's Treatise on the Determination of Spherical Arcs, and was credited to Abū Naṣr Manṣūr by his student al-Bīrūnī in his Keys to Astronomy. Ibn Muʿādh al-Jayyānī's 11th-century Book of Unknown Arcs of a Sphere also contains the spherical law of sines.

The 13th-century Persian mathematician Naṣīr al-Dīn al-Ṭūsī stated and proved the planar law of sines:

In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB : AC = Sin(∠ACB) : Sin(∠ABC)

By employing the law of sines, al-Tusi could solve triangles where either two angles and a side were known or two sides and an angle opposite one of them were given. For triangles with two sides and the included angle, he divided them into right triangles that he could then solve. When three sides were given, he dropped a perpendicular line and then used Proposition II-13 of Euclid's Elements (a geometric version of the law of cosines). Al-Tusi established the important result that if the sum or difference of two arcs is provided along with the ratio of their sines, then the arcs can be calculated.

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." Regiomontanus was a 15th-century German mathematician.

Proof

With the side of length Template:Mvar as the base, the triangle's altitude can be computed as Template:Math or as Template:Math. Equating these two expressions gives $$\frac{\sin \beta }{b}=\frac{\sin \gamma }{c},$$ and similar equations arise by choosing the side of length Template:Mvar or the side of length Template:Mvar 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 Template:Math and Template:Math.

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 Template:Math and the sides Template:Math and Template:Math.
• The angle Template:Math is acute (i.e., Template:Math < 90°).
• The side Template:Math is shorter than the side Template:Math (i.e., Template:Math).
• The side Template:Math is longer than the altitude Template:Math from angle Template:Math, where Template:Math (i.e., Template:Math).

If all the above conditions are true, then each of angles Template:Math and Template:Math produces a valid triangle, meaning that both of the following are true: $$\gamma \text{'}=\arcsin \frac{c\sin \alpha }{a}\text{or}\gamma =\pi -\arcsin \frac{c\sin \alpha }{a}.$$

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

Examples

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

Example 1

Example 1

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

Using the law of sines, we conclude that $$\frac{\sin \alpha }{20}=\frac{\sin (40^{\circ })}{24}.$$ $$\alpha =\arcsin \left(\frac{20\sin (40^{\circ })}{24}\right)\approx 32.39^{\circ }.$$

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

Example 2

Example 2

If the lengths of two sides of the triangle Template:Math and Template:Math are equal to Template:Math, the third side has length Template:Math, and the angles opposite the sides of lengths Template:Math, Template:Math, and Template:Math are Template:Math, Template:Math, and Template:Math respectively then $$\begin{array}{cc}& \alpha =\beta =\frac{180^{\circ }-\gamma }{2}=90^{\circ }-\frac{\gamma }{2}\\ & \sin \alpha =\sin \beta =\sin (90^{\circ }-\frac{\gamma }{2})=\cos \left(\frac{\gamma }{2}\right)\\ & \frac{c}{\sin \gamma }=\frac{a}{\sin \alpha }=\frac{x}{\cos \left(\frac{\gamma }{2}\right)}\\ & \frac{c\cos \left(\frac{\gamma }{2}\right)}{\sin \gamma }=x\end{array}$$

Relation to the circumcircle

In the identity $$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=\frac{c}{\sin \gamma },$$ the common value of the three fractions is actually the diameter of the triangle's circumcircle. This result dates back to Ptolemy.

Proof

Deriving the ratio of the sine law equal to the circumscribing diameter. Note that triangle Template:Math passes through the center of the circumscribing circle with diameter Template:Math.

As shown in the figure, let there be a circle with inscribed $\mathrm{△}ABC$ and another inscribed $\mathrm{△}ADB$ that passes through the circle's center Template:Math. The $\mathrm{\angle }AOD$ has a central angle of $180^{\circ }$ and thus Template:Nowrap by Thales's theorem. Since $\mathrm{△}ABD$ is a right triangle, $$\sin \delta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{c}{2R},$$ where $R=\frac{d}{2}$ is the radius of the circumscribing circle of the triangle. Angles $\gamma$ and $\delta$ lie on the same circle and subtend the same chord Template:Math; thus, by the inscribed angle theorem, Template:Nowrap Therefore, $$\sin \delta =\sin \gamma =\frac{c}{2R}.$$

Rearranging yields $$2R=\frac{c}{\sin \gamma }.$$

Repeating the process of creating $\mathrm{△}ADB$ with other points gives

$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }=2R.$

Relationship to the area of the triangle

The area of a triangle is given by Template:Nowrap where $\theta$ is the angle enclosed by the sides of lengths Template:Math and Template:Math. Substituting the sine law into this equation gives $$T=\frac{1}{2}ab\cdot \frac{c}{2R}.$$

Taking $R$ as the circumscribing radius,

$T=\frac{abc}{4R}.$

It can also be shown that this equality implies $$\begin{array}{cc}\frac{abc}{2T}& =\frac{abc}{2\sqrt{s(s-a)(s-b)(s-c)}}\\ & =\frac{2abc}{\sqrt{{(a^2+b^2+c^2)}^2-2(a^4+b^4+c^4)}},\end{array}$$ where Template:Math is the area of the triangle and Template:Math is the semiperimeter Template:Nowrap

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 Template:Nowrap we have

$T=4R^2\sqrt{S(S-\sin A)(S-\sin B)(S-\sin C)}$

where $R$ is the radius of the circumcircle: Template:Nowrap

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 Template:Math, Template:Math, and Template:Math be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, Template:Math, Template:Math, and Template:Math are the angles at the center of the sphere subtended by those arcs, in radians. Let Template:Math, Template:Math, and Template:Math 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: $$\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}.$$

Vector proof

Consider a unit sphere with three unit vectors Template:Math, Template:Math and Template:Math drawn from the origin to the vertices of the triangle. Thus the angles Template:Math, Template:Math, and Template:Math are the angles Template:Math, Template:Math, and Template:Math, respectively. The arc Template:Math subtends an angle of magnitude Template:Math at the centre. Introduce a Cartesian basis with Template:Math along the Template:Math-axis and Template:Math in the Template:Math-plane making an angle Template:Math with the Template:Math-axis. The vector Template:Math projects to Template:Math in the Template:Math-plane and the angle between Template:Math and the Template:Math-axis is Template:Math. Therefore, the three vectors have components: $$𝐎𝐀=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right),𝐎𝐁=\left(\begin{array}{c}\sin c\\ 0\\ \cos c\end{array}\right),𝐎𝐂=\left(\begin{array}{c}\sin b\cos A\\ \sin b\sin A\\ \cos b\end{array}\right).$$

The scalar triple product, Template:Math is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle Template:Math, Template:Math and Template:Math. This volume is invariant to the specific coordinate system used to represent Template:Math, Template:Math and Template:Math. The value of the scalar triple product Template:Math is the Template:Math determinant with Template:Math, Template:Math and Template:Math as its rows. With the Template:Math-axis along Template:Math the square of this determinant is $$\begin{array}{cc}(𝐎𝐀\cdot (𝐎𝐁\times 𝐎𝐂){)}^2& ={(\det \left(\begin{array}{ccc}𝐎𝐀& 𝐎𝐁& 𝐎𝐂\end{array}\right))}^2\\ & ={\left|\begin{array}{ccc}0& 0& 1\\ \sin c& 0& \cos c\\ \sin b\cos A& \sin b\sin A& \cos b\end{array}\right|}^2={(\sin b\sin c\sin A)}^2.\end{array}$$ Repeating this calculation with the Template:Math-axis along Template:Math gives Template:Math, while with the Template:Math-axis along Template:Math it is Template:Math. Equating these expressions and dividing throughout by Template:Math gives $$\frac{{\sin }^{2}A}{{\sin }^{2}a}=\frac{{\sin }^{2}B}{{\sin }^{2}b}=\frac{{\sin }^{2}C}{{\sin }^{2}c}=\frac{V^2}{{\sin }^2(a){\sin }^2(b){\sin }^2(c)},$$ where Template:Mvar 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 $$\underset{a\to 0}{\lim }\frac{\sin a}{a}=1$$ and the same for Template:Math and Template:Math.

Geometric proof

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

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

Construct point $A^{\prime }$ such that $\mathrm{\angle }{A}^{\prime }DO=\mathrm{\angle }{A}^{\prime }EO=90^{\circ }$

It can therefore be seen that $\mathrm{\angle }AD{A}^{\prime }=B$ and $\mathrm{\angle }AE{A}^{\prime }=C$

Notice that $A^{\prime }$ is the projection of $A$ on plane $OBC$. Therefore $\mathrm{\angle }A{A}^{\prime }D=\mathrm{\angle }A{A}^{\prime }E=90^{\circ }$

By basic trigonometry, we have: $$\begin{array}{cc}AD& =\sin c\\ AE& =\sin b\end{array}$$

But $A{A}^{\prime }=AD\sin B=AE\sin C$

Combining them we have: $$\begin{array}{cc}\sin c\sin B& =\sin b\sin C\\ \Rightarrow \frac{\sin B}{\sin b}& =\frac{\sin C}{\sin c}\end{array}$$

By applying similar reasoning, we obtain the spherical law of sine: $$\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}$$

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 cosines $$\begin{array}{cc}{\sin }^{2}A& =1-{\left(\frac{\cos a-\cos b\cos c}{\sin b\sin c}\right)}^2\\ & =\frac{(1-{\cos }^{2}b)(1-{\cos }^{2}c)-{(\cos a-\cos b\cos c)}^2}{{\sin }^{2}b{\sin }^{2}c}\\ \frac{\sin A}{\sin a}& =\frac{{[1-{\cos }^{2}a-{\cos }^{2}b-{\cos }^{2}c+2\cos a\cos b\cos c]}^{1/2}}{\sin a\sin b\sin c}.\end{array}$$ Since 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 $$\frac{\sin A}{\sinh a}=\frac{\sin B}{\sinh b}=\frac{\sin C}{\sinh c}.$$

In the special case when Template:Math is a right angle, one gets $$\sin C=\frac{\sinh c}{\sinh b}$$

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

The case of surfaces of constant curvature

Define a generalized sine function, depending also on a real parameter $\kappa$: $${\sin }_{\kappa }(x)=x-\frac{\kappa }{3!}{x}^3+\frac{{\kappa }^2}{5!}{x}^5-\frac{{\kappa }^3}{7!}{x}^7+\cdots =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{\kappa }^n}{(2n+1)!}{x}^{2n+1}.$$

The law of sines in constant curvature $\kappa$ reads as $$\frac{\sin A}{{\sin }_{\kappa }a}=\frac{\sin B}{{\sin }_{\kappa }b}=\frac{\sin C}{{\sin }_{\kappa }c}.$$

By substituting $\kappa =0$, $\kappa =1$, and $\kappa =-1$, one obtains respectively ${\sin }_0(x)=x$, ${\sin }_1(x)=\sin x$, and ${\sin }_{-1}(x)=\sinh x$, that is, 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: $$\frac{\sin A}{p_{\kappa }(a)}=\frac{\sin B}{p_{\kappa }(b)}=\frac{\sin C}{p_{\kappa }(c)}.$$

This formulation was discovered by János Bolyai.

Higher dimensions

A tetrahedron has four triangular facets. The absolute value of the polar sine (Template:Math) 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:

$$\begin{array}{cc}& \frac{|\mathrm{psin}(𝐛,𝐜,𝐝)|}{{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_a}=\frac{|\mathrm{psin}(𝐚,𝐜,𝐝)|}{{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_b}=\frac{|\mathrm{psin}(𝐚,𝐛,𝐝)|}{{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_c}=\frac{|\mathrm{psin}(𝐚,𝐛,𝐜)|}{{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_d}\\ =& \frac{(3{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}_{\mathrm{t}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{n}}{)}^2}{2{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_a{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_b{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_c{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_d}.\end{array}$$

More generally, for an Template:Math-dimensional simplex (i.e., triangle (Template:Math), tetrahedron (Template:Math), pentatope (Template:Math), etc.) in Template:Math-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 Template:Math for the hypervolume of the Template:Math-dimensional simplex and Template:Math for the product of the hyperareas of its Template:Math-dimensional facets, the common ratio is $$\frac{|\mathrm{psin}(𝐛,\dots ,𝐳)|}{{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_a}=\cdots =\frac{|\mathrm{psin}(𝐚,\dots ,𝐲)|}{{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}}_z}=\frac{(nV{)}^{n-1}}{(n-1)!P}.$$

See also

• Template:Annotated link
• Half-side formulaTemplate:Snd for solving spherical triangles
• Law of cosines
• Law of tangents
• Law of cotangents
• Mollweide's formulaTemplate:Snd for checking solutions of triangles
• Solution of triangles
• Surveying

External links

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