Taqi al Din Ibn Ma’ruf ‘s Work on Extracting the Cord 2o and Sin 1o

This article by Professor Sevim Tekeli, a leading historian of science in the Ottoman period, deals with an aspect of the work of Taqi al-Din Ibn Ma’ruf in trigonometry, a mathematical discipline which studies the relationships between the sides and the angles of triangles and the trigonometric functions which describe those relationships. Approaching Taqi al-Din’s work through modern methods of notation, his mathematical method in extracting the Cord 1o or Cord 2o is fully disclosed. The author shows the originality of Ibn Ma’ruf’s discovery and states how it constituted a progress with regard to his predecessors, in Greek and Islamic mathematics.

It is easy to obtain the lengths of cords of certain arcs. The formulas needed are Cord 2A, Cord A/2, Cord (A±B). Even by applying all these formulas, it is not possible to get the Cord 1o or Cord 2o.

Ptolemy (ca 150 CE) used an ingenious method of interpolation.(2) He added the greater and the lesser values of Cord 1o and divided the result by 2. This is, of course, approximately equivalent to Cord 1o.

The situation was the same for Sins which began to be used besides the Cords in the Islamic world. This time, the auxiliary theorems are needed for the preparation of the Sins tables, as SinA/2, Sin2A, Sin (A± B). The astronomers who were not content with the approximate values began to follow the subject very closely, such as Abu ‘l Wafa(6) (959-998) and Ibn Yunus(3) (ca 1000 CE). Later, Al-Kâshî (1393-1449) solved this problem as a third equation. Qadizada-i Rûmî(6) (1337-1412), Ulugh Bey(6) (1394-1449), Mirim Chelebi(6) (1524-1525) and Taqi al Din Ibn Ma’ruf (1520-1585) occupied themselves with this problem.

Figure 1: The trigonometric functions of an angle q can be constructed geometrically in terms of a unit circle centered at O.

On the other hand, this problem was not taken into the consideration in the West for a long time. Ptolemy’s value was accepted by the scholars, and the first mathematician who dealt with this problem was Regiomontanus (1436-1476). On his part, Copernicus (1473-1.543) accepted the value of Ptolemy in his The Revolutions of the Heavenly Spheres(4).

Taqi al Din Ibn Ma’ruf Says in his Sidrat al-Muntaha: “As all these methods, finding the Cord (180-A), Cord A/2, Cord 2A and Cord (A±B) are established; it is possible to find many cords. But to complete this work we need the knowledge of the Cord 1o or Cord 2o.

“As we come to the Cord 2o, the ancients could not find a precise way, in consequence of this, they depended on an approximate method which is not worth to describe.

“The late Ulugh Bey said: ‘we had inspiration about extracting Sin 1o. He wrote a text on this subject and explained three ways of finding the Sin 1° and Cord 2°, depending on the geometric theorems related with mathematics.”

Figure 2

One of them, with some additions, is as follows:

AD = Cord 6o = 6p 16 49 7‘” 59“” 8‘”” 56“””” 20‘”””
AB = Cord 2o = x.

According to Ptolemy’s theorem,

AC2 = X2 + X.AD


As the triangle BAF is a right triangle, and AR is a perpendicular drawn to the hypotenuse,

x2 = BR.BF
BF = the diameter of the circle = 2

BR = X2/2
X2 = BR2+ AR2.


On the other hand,

AR2 = 1/4 AC2
X2 – BR2 = 1/4 AC2

X – X4/4 = 1/4 AC2
4X2 – X4 = AC2
4X2 – X4 = X2 + X.AD

(according to II)
(according to I and III)

3X = X3 + AD
X = (X3 + AD)/3

As it does not belong to one of the six equations, so he followed this way:

X = AD/3 = 2p 5 36 22‘” 39“” 42‘”” 58“”” 46‘”” approximately.

In reality,

X = a + 2p 5 36 22‘” 39“” 42‘”” 58“”” 46‘””
X = [( a + 2p 5 36 22‘” 39“” 42‘”” 58“”” 46‘””)3 + AD]/3

By doing this way he gets:

Cord 2o = 2p5 99 26‘” 22“” 29‘”” 32“””.

As we come to obtain the Sin 1o, in the afore-mentioned figure, let X be Sin 1o and BF=1 or 60.

BR is the perpendicular drawn to the hypotenuse of the right triangle ABF.


BF = 1
BR = X2
BR2 = X4.

As the quadrilateral ABCD is drawn in a circle,

AD.X + X2 = AC2
AC2 = 4 AR2
AR2 = ¼ AD.X + ¼ X2


As the triangle ABR is a right triangle,

BR2 = X2 – AR2
BR2 = X2 – ¼ X2 – ¼ AD.X
BR2 = ¾ X2 – ¼ AD.X


BR2 = X4
X4 = ¾ X2 – ¼ AD.X
X = 4/3 X3 + 1/3 AD
1/3 AD = 1p 2 8 11‘” 19“” 51‘”” 29“””” 25“””‘.

To got the precise value,

X = 4/3 (a + 1p 2 8 11‘” 19“”…)3 + 1/3 (1p 2 8 …)


Figure 3

He keeps on doing this till very small change occurs, and at the end he obtains the following value:

Sin 1o = 1p 2 49 43‘” 11“” 14‘”” 44“””16“””‘.