A paper on finding d rsines from aryabhatiya bhasya by bhaskara i

1
Relevance of Ancient Knowledge to the present century:
A Paper on finding D-rsines from Aryabhatiya Bhasya by Bhaskara I
Verification of Aryabhatiya’s (AD 499) Algorithm for finding 24-rsines
Compiled by: Venkatesha Murthy, Honorary Head, Vedic Mathematics Branch,
National Institute of Vedic Sciences ,# 58, Raghavendra Colony,
Chamarajapet, Bangalore – 560 018.
Mobile: 09449425248, gmail ID: tippurgopu@gmail.com
Introduction: -
rÉjÉÉ ÍzÉZÉÉ qÉrÉÔUÉhÉÉÇ lÉÉaÉÉlÉÉÇ qÉhÉrÉÉårÉjÉÉ | iÉ²Sè uÉåSÉ…¡û शा ाणां aÉÍhÉiÉÇ qÉÔkÉïÌlÉ ÎxjÉiÉqÉç ||4||
Like the combs of the peacocks and the crest jewels of the serpents, so does the
tradition of mathematics knowledge stand at the head of all the tradition of Sciences
knowledge forming the auxiliaries of the Vedas. [Refer; 4. (p.27 & 36)]
Indian Place-value System of base ten with the invention of Zero revolutionized
the sciences of the world by helping systematic quantifying of the scientific and
technological concepts.
Aryabhatiya-Vyakhya of Bhaskara-I (629 A.D.), a commentary on Aryabhatiya of
Aryabhata-I explains methods to get the values of 6-rsines, 12-rsines and 24-rsines.
In general D-rsines is ‘n values’ of r sine θn which are the products of radius r
and ‘n number of Sine θn’ where θn is the central angle made by n equal arcs; when
the arc of the first quadrant is divided into D equal arcs in a circle of radius r = 3438
units. These values exactly agree with their present-day values, as radius r = 3438
minutes is an approximate value of a radian.
The radian is the central angle made by an arc of a circle of radius r whose length is
equal to the radius r and it requires the value of the ratio of circumference of the
circle of radius r to its diameter.
The ratio of the circumference of a circle to its diameter , which is now known as π, is
stated in Aryabhatiya of Aryabhata-I thus;
cÉiÉÑUÍkÉMÇü zÉiÉqÉ¹aÉÑhÉÇ ²ÉwÉÌ¹xiÉjÉÉ xÉWûxÉëÉhÉÉqÉç |
ArÉÑiÉ²rÉÌuÉwMüqpÉxrÉÉxÉ³ÉÉå वॄ तmÉËUhÉÉWûÈ ||10||
[Refer; 1. (p.71)]
‘Four more than hundred multiplied by eight and increased by sixty-two thousand is a
near value (AÉxÉ³É) of the perimeter of a circle of diameter twenty-thousand units’.
“[(100 + 4) 8 + 62,000] = 62,832 is a nearer (approximate) value of the perimeter of a
circle of diameter 20,000 units”.
Perimeter in Greek is περιµετερ and its first letter ‘π’ is used to denote ‘the ratio of
the circumference of a circle to its diameter’
( )
62832
π = 3.1416 .
20000
circumference of a circle
app
diameter
= =

2
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ nnθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ n
θ n
1. (i) TRIGONOMETRIC TERMS in Indian Mathematics.
Angles are measured in the clockwise direction from E.
Angle QOG = 90o
= angle made by D equal arcs.
Angle QOB = θn
= angle made by n equal arcs.
OP = r cos θn
= को ट या of n arcs = kn.
Angle BOL = Angle (QOG – QOB)
= (900 - θn)
= Angle made by (D – n) equal arcs.
OP = LB
को ट या of n arcs = kn
= य़ा of (D – n) arcs = j(D- n)
PQ = (OQ – OP) = (r - को ट या of n arcs) = (r - kn)
= उ म या of n arcs = un = शर of n arcs
1. (ii) Etymology of the term ‘sine’
The term ‘sine’ is from the original Hindu word या or जीवा.
The word या or जीवा is used for a half chord in a circle of radius r =1 unit.
Arabs translated या or जीवा as ‘gib.
Arabic word ‘gib’ is similar to the word ‘fold’
‘gib’ was translated into Latin as ‘sinus’ (fold).
Later ‘sinus’ ( या or जीवा ) became ‘sine’.
[Ref: “The Mainstream of Mathematics” by Edna E Kramer, Oxford Univ. Press, Inc.(1951) P.140]
2. Aryabhatiya’s (AD 499) Cryptic Algorithm for finding D-rsines.
Aryabhatiya of Aryabhata-I (b. 476 AD) gives a method cryptically for finding the
values of rsines of n equal arcs (r Sin θn) = Jya of n equal arcs (Jn).
Nåû±MüÌuÉÍkÉlÉÉ erÉÉÅÅlÉrÉlÉqÉç
xÉqÉuÉ×¨ÉmÉËUÍkÉmÉÉSÇ ÍNÇû±ÉiÉç Ì§ÉpÉÑeÉÉŠiÉÑpÉÑïeÉÉŠæuÉ |
xÉqÉcÉÉmÉerÉÉkÉÉïÌlÉ iÉÑ ÌuÉwMüqpÉÉkÉåï rÉjÉå¹ÉÌlÉ ||11||
Divide a quadrant of the circumference of a circle (into as many parts as desired; D).
Then, from (right) triangle and quadrilaterals, one can find as many rsines of equal
arcs as one likes, for a given radius (3438 units).
n (D n)k = j −
( )n nu = r k−

3
Cn
P QR S
BA
D
C Jn
Kn
O
Un
4
3
2
1
5
6
7
8
9
10
11
12
N S
E
W
M L
Figure 1
Cn
P QR S
BA
D
C Jn
Kn
O
Un
4
3
2
1
5
6
7
8
9
10
11
12
M L
Figure 1
Cn
P QR S
BA
D
C Jn
Kn
O
Un
4
3
2
1
5
6
7
8
9
10
11
12
N S
E
W
M L
Figure 1
Cn
P QR S
BA
D
C Jn
Kn
O
Un
4
3
2
1
5
6
7
8
9
10
11
12
M L
Figure 1
[Ref.1, p 77]
Bhaskaracarya-I (AD 629) in his Aryabhatiya bhasya elaborately illustrated methods
to find 6-rsines, 12-rsines and 24-rsines, [6 x 2(t-1) where t = 1, 2 or 3].
uÉxÉÑSWûlÉM×üiÉWÒûiÉÉzÉlÉxÉXçZrÉå ÌuÉwMüqoÉÉkÉåï ÌMürÉimiÉqÉÉhÉÌlÉ erÉÉkÉÉïÌlÉ |
[Ref.1, p 78]
rsines are to be calculated by considering half the diameter of a circle as 3438.
As per the rule Aƒ¡ûÉlÉÉqÉç uÉÉqÉiÉÉåaÉÌiÉÈ ‘(Place-values of) digits (in a numeral of a
number) move (increases) towards left)’.
Combination of words uÉxÉÑSWûlÉM×üiÉWÒûiÉÉzÉlÉ denotes 3438 based on words as numerals;
WÒûiÉÉzÉlÉ M×üiÉ SWûlÉ uÉxÉÑ
3 4 3 8
2. (i) Geometrical Basis for finding the desired number of rsines.
mÉËUkÉå wÉQèpÉÉaÉerÉÉ ÌuÉwMüqpÉÉkÉåïlÉ xÉÉ iÉÑsrÉÉ ||9||
[Refer : Aryabhatiya, with the commentary of Bhaskara-I and Someswara :Edited by K S Shukla,
INSA, New Delhi, (1976), ( based on p.79 for finding rsines)]
The chord of one-sixth of the perimeter (of a circle) is equal to its radius.
Therefore, Aryabhatiya Bhasya (AD 629) of Bhaskara-I has considered a regular
hexagon inscribed in a circle of radius 3438 units to find the desired number of (D),
rsines.
2. (ii) Figure to find D-rsines: -
rÉÉuÉ¨ÉÉuÉimÉëqÉÉhÉmÉËUÎcNû³ÉÌuÉwMüqpÉÉkÉïiÉÑsrÉålÉ MüMïüOûMåülÉ qÉhQûsÉqÉÉÍsÉZrÉ iÉSè ²ÉSzÉkÉÉ ÌuÉpÉeÉåiÉç | iÉå cÉ
²ÉSzÉpÉÉaÉÉ UÉzÉrÉ CÌiÉ mÉËUMüsmÉÉÈ| AjÉ ²ÉSzÉkÉÉ ÌuÉpÉ£åü qÉhQûsÉå mÉÔuÉåïhÉ UÉÍzÉ²rÉÉaÉëÉuÉaÉÉÌWûlÉÏÇ
SÍ¤ÉhÉÉå¨ÉUÉÇ erÉÉMüÉUÉÇ sÉåZÉÉÇ MÑürÉÉïiÉç| LuÉÇ mÉÍ¶ÉqÉpÉÉaÉåÅÌmÉ| LuÉqÉåuÉ SÍ¤ÉhÉÉå¨ÉUpÉÉaÉrÉÉåUÌmÉ cÉ mÉÔuÉÉïmÉUrÉiÉÉÇ
erÉÉÇ MÑürÉÉïiÉç| mÉÑlÉUÌmÉ cÉ mÉÔuÉÉïmÉU SÍ¤ÉhÉÉå-¨ÉUÌS¤ÉÑ iÉjÉæuÉ cÉ UÉÍzÉcÉiÉÑ¹rÉÉaÉëÉuÉaÉÉÌWûlrÉÉå sÉåZÉÉÈ MÑürÉÉïiÉç|
iÉjÉÉ §rÉ´ÉÏ (ÍhÉ) MüiÉïurÉÉ-ÌlÉ| [Ref. (1). P.78-79]
Meaning: -
Draw a circle of radius r (= 3438 minutes)
and divide it into twelve equal parts.
Consider them as ²ÉSzÉpÉÉaÉÉ UÉzÉrÉ. Draw N–S
line passing through the centre
of the circle. Likewise draw E–W line also.
Draw vertical and horizontal lines passing
through the centre of the circle to form two
rectangles (Figure 1), dividing each
quadrant into three equal arcs.
[Ref.1, p 79]
Aryabhatiya is a text on Astronomy, and therefore it has given instructions to draw
²ÉSzÉpÉÉaÉÉ UÉzÉrÉ. (figure 1).
Otherwise drawing a regular hexagon in a circle would be more convenient for
finding required number of rsines, as shown in figure 2

4
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ
nnθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ
n
θ
n
Figure 2
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ
nnθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ
n
θ
n
Figure 2
2. (ii) To calculate 24-rsines from Aryabhatiya-vyakya by Bhaskara I (AD 629).
To calculate 16- rsines
mÉÔuÉïåuÉkÉÉÍsÉÎZÉiÉå ¤Éå§Éå urÉÉxÉÉkÉïqÉåuÉ wÉÉåQûzÉÉlÉÉÇ Mü¸ÉlÉÉÇ [ mÉÔhÉï ] erÉÉ | iÉSkÉïqÉ¹ÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ
1719 (J8= 1719) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ(K8=2977)|
xÉÉ wÉÉåQûzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 2977 (J16= 2977) |
LiÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉqÉ¹MüÉ¸zÉUÈ (U8=461) |
zÉUÉ¹MüÉ¸-erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ A¹ÉlÉÉÇ Mü¸ÉlÉÉÇ
[mÉÔhÉï ] erÉÉ, xÉÉcÉ 2780 | AkÉïqÉxrÉÉÈ cÉiÉÑhÉÉïÇ MüÉ¸ÉlÉÉÇ
erÉÉ, xÉÉcÉ 890 (J4= 890) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ,
pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ (K4=3321) | xÉæuÉ
ÌuÉqzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 3321 (J20= 3321) | LiÉÉÇ
urÉÉxÉÉkÉÉïÌ²-zÉÉåkrÉ zÉåwÉÇ cÉiÉÑÈMüÉ¸zÉUÈ (U4=117) |
zÉUcÉiÉÑwMüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ cÉiÉÑhÉÉïÇ Mü¸ÉlÉÉÇ
[ mÉÔhÉï ] erÉÉ, xÉÉ cÉ 898 (C4 = 898)| AkÉïqÉxrÉÉÈ
MüÉ¸rÉÉåerÉÉï, xÉÉ cÉ 449 (J2= 449) | LwÉÉ pÉÑeÉÉ,
urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ (K2=3409) | xÉæuÉ ²ÉÌuÉÇzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ,
xÉÉcÉ 3409 (J22= 3409) | LiÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉÇ Ì²MüÉ¸zÉUÈ (U2=29) |
zÉUÌ²MüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ MüÉ¸rÉÉåÈ [ mÉÔhÉï ] erÉÉ, xÉÉ cÉ 450| AkÉïqÉxrÉÉÈ MüÉ¸xrÉ
erÉÉ, xÉÉ cÉ 225 (J1= 225) | LwÉÉ pÉÑeÉÉ, urÉÉxÉkÉïÈ MühÉïÈ, pÉÑeÉÉ-MühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ
(K1=3431) | xÉæuÉ §ÉrÉÉåÌuÉqzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 3431 (J23= 3431) |ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ
lÉÉåimÉ±iÉå |
mÉÔuÉïuÉkÉÉÍsÉÎZÉiÉå ¤Éå§Éå urÉÉxÉÉkÉïqÉåuÉ wÉÉåQûzÉÉlÉÉÇ Mü¸ÉlÉÉÇ [ mÉÔhÉï ] erÉÉ | iÉSkÉïqÉ¹ÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ
1719 (J8= 1719) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉï-ÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ(K8=2977)
| xÉÉ wÉÉåQûzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 2977 (J16= 2977) | LiÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç |
zÉåwÉqÉ¹MüÉ¸zÉUÈ (U8=461) | zÉUÉ¹MüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ A¹ÉlÉÉÇ Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ,
xÉÉcÉ 2780 | AkÉïqÉxrÉÉÈ cÉiÉÑhÉÉïÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 890 (J4= 890) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï
MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ (K4=3321) | xÉæuÉ ÌuÉqzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 3321
(J20= 3321) | LiÉÉÇ urÉÉxÉÉkÉÉïÌ²-zÉÉåkrÉ zÉåwÉÇ cÉiÉÑÈMüÉ¸zÉUÈ (U4=117) | zÉUcÉiÉÑwMüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ
MühÉïÈ | xÉ LuÉ cÉiÉÑhÉÉïÇ Mü¸ÉlÉÉÇ [ mÉÔhÉï ] erÉÉ, xÉÉ cÉ 898 (C4 = 898)| AkÉïqÉxrÉÉÈ MüÉ¸rÉÉåerÉÉï,
xÉÉ cÉ 449 (J2= 449) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ
(K2=3409) | xÉæuÉ ²ÉÌuÉÇzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 3409 (J22= 3409) | LiÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç |
zÉåwÉÇ Ì²MüÉ¸zÉUÈ (U2=29) | zÉUÌ²MüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ MüÉ¸rÉÉåÈ [ mÉÔhÉï ] erÉÉ, xÉÉ
cÉ 450| AkÉïqÉxrÉÉÈ MüÉ¸xrÉ erÉÉ, xÉÉ cÉ 225 (J1= 225) | LwÉÉ pÉÑeÉÉ, urÉÉxÉkÉïÈ MühÉïÈ,
pÉÑeÉÉMühÉï-uÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ (K1=3431) | xÉæuÉ §ÉrÉÉåÌuÉqzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 3431
(J23= 3431) |ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ lÉÉåimÉ±iÉå |

5
AjÉ cÉiÉÑhÉÉïÇ MüÉ¸ÉlÉÉÇ erÉÉÇ Ì²zÉÉåkÉrÉåiÉç | zÉåwÉÇ ÌuÉÇwÉiÉåÈ MüÉ¸ÉlÉÉÇ zÉUÈ (U20=2548)| zÉUÌuÉÇzÉÌiÉ-
MüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ ÌuÉÇzÉiÉåÈ Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉcÉ 4186| AkÉïqÉxrÉÉÈ SzÉÉlÉÉÇ
MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 2093 (J10= 2093) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ
MüÉåÌOûÈ (K10= 2727) | xÉæuÉ cÉiÉÑSïzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ cÉ 2727 (J14= 2727) | LiÉÉÇ
urÉÉxÉÉkÉÉïÌ²zÉÉåkrÉ zÉåwÉÇ SzÉMüÉ¸ÉlÉÉÇ zÉUÈ (U10= 711) | zÉUSzÉMüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ| xÉ LuÉ
SzÉÉlÉÉÇ Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉ cÉ 2210| AkÉïqÉxrÉÉÈ mÉgcÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 1105
(J5= 1105) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ | xÉæuÉ LMüÉålÉÌuÉÇzÉiÉåÈ
MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ cÉ 3256 (J19= 3256) | ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ lÉÉåimÉ±iÉå |
AjÉ Ì²MüÉ¸erÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉÇ ²ÉÌuÉÇzÉiÉåÈ MüÉ¸ÉlÉÉÇzÉUÈ (U22=2989) |
zÉU²ÉÌuÉÇzÉÌiÉMüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ| xÉ LuÉ ²ÉÌuÉÇzÉiÉåÈ Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉ cÉ 2585|
AkÉïqÉxrÉÉ LMüSzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 1293 (J11= 1293) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ,
pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ (K11=3186) | xÉæuÉ §ÉrÉÉåSzÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ cÉ 2585
(J13= 3186) | ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ lÉÉåimÉ±iÉå |
SzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉÇ cÉiÉÑSïzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ zÉUÈ (U14=711)|
zÉUcÉiÉÑSïzÉMüÉ¸erÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ cÉiÉÑSïzÉÉlÉÉÇ Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉcÉ 2818|
AkÉïqÉxrÉÉÈ xÉmiÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 1409 (J7= 1409) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ,
pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ | xÉæuÉ xÉmiÉSzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ cÉ 3136 (J17=3136) |
ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ lÉÉåimÉ±iÉå |
LuÉÇ Ì§ÉpÉÑeÉÉSìÉzrÉ¹pÉÉaÉMüÉ¸erÉÉ urÉÉZrÉÉiÉÈ| AjÉ cÉiÉÑpÉÑïeÉÉSè urÉÉZrÉÉxrÉÉqÉÈ|
AliÉÈ xÉqÉcÉiÉÑUxÉëxrÉ¤Éå§ÉxrÉ urÉÉxÉÉkÉïiÉÑsrÉÉ oÉÉWûuÉÈ | iÉrÉÉåuÉïaÉïrÉÉåaÉqÉÔsÉÈ MühÉïÈ | xÉ LuÉ cÉiÉÑ-ÌuÉïÇzÉiÉåÈ
MüÉ¸ÉlÉÉÇ [ mÉÔhÉï ] erÉÉ, xÉÉ cÉ 4862 (C24= 4862) | AkÉïqÉxrÉÉÈ ²ÉSzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ
2431 (J12= 2431) | LiÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉÇ cÉiÉÑSïzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ zÉUÈ|
zÉUcÉiÉÑSïzÉMüÉ¸erÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ ²ÉSzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉcÉ 2631|
AkÉïqÉxrÉÉÈ wÉhhÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 1316 (J6= 1316) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ,
pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ | xÉÉ A¹ÉSzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ cÉ 3177 (J18= 3177) |
LiÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉÇ wÉhhÉÉÇ MüÉ¸ÉlÉÉÇ zÉUÈ| zÉUwÉOèMüÉ¸erÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ wÉhhÉÉÇ
Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉcÉ 1342| AkÉïqÉxrÉÉÈ §ÉrÉÉhÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 671 (J3= 671) |
LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ, pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ | xÉæuÉ LMüÌuÉÇzÉiÉåÈ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ
cÉ 3372 (J21= 3372) | ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ lÉÉåimÉ±iÉå |
AjÉ wÉhhÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉÇ urÉÉxÉÉkÉÉïÌ²zÉÉåkÉrÉåiÉç | zÉåwÉÇ A¹ÉSzÉMüÉ¸ÉlÉÉÇ zÉUÈ|
zÉUÉ¹ÉSzÉMüÉ¸erÉÉuÉaÉïrÉÉåaÉqÉÔsÉÇ MühÉïÈ | xÉ LuÉ A¹ÉSzÉÉlÉÉÇ Mü¸ÉlÉÉÇ [mÉÔhÉï ] erÉÉ, xÉÉcÉ 3820|
AkÉïqÉxrÉÉ lÉuÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉcÉ 1910 (J9=1910) | LwÉÉ pÉÑeÉÉ, urÉÉxÉÉkÉï MühÉïÈ,
pÉÑeÉÉMühÉïuÉaÉïÌuÉzÉåwÉxrÉ qÉÔsÉÇ MüÉåÌOûÈ | xÉæuÉ mÉgcÉSzÉÉlÉÉÇ MüÉ¸ÉlÉÉÇ erÉÉ, xÉÉ cÉ 2859 (J15= 2859) |
ÌuÉzÉqÉiuÉÉSiÉÉå erÉÉ lÉÉåimÉ±iÉå |
LuÉÇ UÉzrÉ¹pÉÉaÉMüÉ¸erÉÉ¶ÉiÉÑÌuÉïÇzÉÌiÉÈ | AlÉålÉæuÉ ÌuÉkÉÉlÉålÉ ÌuÉzMüqpÉÉkÉåï rÉjÉå¹ÉÌlÉ-erÉÉkÉÉïÌlÉ
ÌlÉwmÉÉSÌrÉiÉurÉÉÌlÉ CÌiÉ || 11 ||
[Ref.1, p 81-83]

6
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ n
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ nθ n
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ n
nθ
2nC
r
C
N S
BA
Q
P Jn
K
O
U
E
W
GH
nn
n
L
θ nθ n
2. (iii) Mathematical Algorithm to find ‘D’- rsines from Aryabhatiya (AD 499)
Divide arc QBG of quadrant QBGO into D equal arcs, where D = 6x(2)( t-1)
Arc QB = arc of n equal arcs. ∠ QOB = θn , PB = (1/ 2)
AB = r sineθn = Jya of n arcs = Jn
(1). AB = C2n = Chord of 2n arcs where 2n =
2D
3
(2). Jn = (1/ 2) C2n
(3). OP= √(r2 – Jn
2) = kotijya of n arcs
= Kn = r cos θn
[Arc BG = arc of (D-n) equal arcs]
OP = LB = r sin (∠BOL)
= r sin (90 - θn) = JD – n
(4). Kn = JD– n
When n or (D - n) = 2m, an even
number,
FQ = (r – K2m) = U2m = Utkrama Jya
(शर) of 2m arcs
(5). U2m = r – K2m
QB = √(J2m
2 + U2m
2) = chord of 2m arcs = C2m
(6). C2m = √(J2m
2 + U2m
2) = chord of 2m arcs
(6) is similar to (1). Repeat steps from (2) to (6)
To find D-rsines: -
I. Find 2m-rsines [where 2m =(2D/ 3)] assuming chord AB as a side of
the inscribed hexagon with 3438 units as radius having 2m equal arcs
= C2m = 3438.
II. Find m-rsines [where m =(D/ 3)] assuming Chord QG as the hypotenuse of
an isosceles right triangle QOG with 3438 as a chord of D arcs = CD = 3438√2
= 4862
2. (iv) To find 24-rsines on the basis of an illustrative example.
Draw an inscribed hexagon having the radius 3438 units (one radian in minutes).
AB, a side of the hexagon
= radius of the circle
= 3438 units.
Angle QOG
= Central angle made by arc QBG = 90o.
Arc QBG is divided into 24 (= D) equal arcs.

7
Central angle made by chord AB [Radius of inscribed hexagon] = angle AOB = 60o.
Central angle made by chord PB =
0
90 n
POB θ
24
nangle = =
= Central angle made by n equal. arcs at centre O.
I. Find 16-rsines [where 2m =(2D/ 3) = 16] assuming Chord AB as a side of
the inscribed hexagon with 3438 units as radius having 16 equal arcs
= C16 = 3438 (= C2m).
II, Find 8-rsines [where m =(D/ 3) = 8] assuming Chord QG as the
hypotenuse of isosceles right triangle QOG as a Chord of 24 arcs
= C24 = 4862 (= CD) = 3438√2]
Next two tables based on the explanation in Aryabhatiya-bhasya of Bhaskara I,
gives a comparison of Aryabhatiya and present-day values of 24-rsines
I. Find 16-rsines [where 2m =(2D/ 3) = 16] assuming Chord AB as a side of
the inscribed hexagon with 3438 units as radius having 16 equal arcs
= C16 = 3438 (= C2m).
C2n Jn
KD-n
Kn Un Jn /3438 θ =
[90n /24]
Sine θ
C16 =3438
C8 = 1780
C4 = 898
C2 = 450
C22 = 4536
J16= 2977
J8= 1719
J4= 890
J20= 3321
J2= 449
J22= 3409
J1 = 225
J23= 3431
J11= 2268
J13= 2584
K8= 2977
K4= 3321
K2= 3409
K22= 446
K1= 3431
K11= 2584
U8= 461
U4= 117
U2= 29
U22= 2992
0.8659
0.5000
0.2589
0.9660
0.1306
0.9916
0.0654
0.9980
0.6596
0.7516
60
30
15
75
7.5
82.5
3.75
86.25
41.25
48.75
0.8660
0.5000
0.2588
0.9660
0.1305
0.9914
0.0654
0.9979
0.6593
0.7518
C20 = 4186
C10= 2290
C14= 3040
J10= 2093
J14= 2727
J5 = 1105
J19 = 3256
J7 = 1520
J17 = 3084
K20= 890
K10= 2727
K5= 3256
K14= 2094
K7 = 3084
U20= 2548
U10= 711
U14= 1344
0.6088
0.7931
0.3214
0.9470
0.4421
0.8970
37.5
52.5
18.75
71.25
26.25
63.75
0.6088
0.7934
0.3214
0.9469
0.4423
0.8969

8
II, Find 8-rsines [where m =(D/ 3) = 8] assuming Chord QG as the
hypotenuse of isosceles right triangle QOG as a Chord of 24 arcs
= C24 = 4862 (= CD) = 3438√2]
C2n Jn
KD-n
Kn Un Jn /3438 θ =
[90n /24]
Sine θ
C24= 4862
C12 =2631
C6 = 1342
C18= 3820
J12= 2431
J6= 1316
J3 = 671
J21 = 3372
J18 = 3176
J9 = 1910
J15= 2859
J24= 3438
K12= 2431
K6= 3176
K3= 3372
K21= 670
K18= 1316
K9= 2859
U12= 1007
U6= 262
U18= 2122
0.7071
0.3828
0.1952
0.9808
0.9238
0.5556
0.8316
1.0000
45
22.5
11.25
78.75
67.5
33.75
56.25
90
0.7071
0.3827
0.1951
0.9808
0.9239
0.5556
0.8315
1.0000
2. (v) Reason for considering 24-rsines - Suryadeva yajvan (b. A.D. 1191)
“cÉiÉÑÌuÉïÇzÉÌiÉkÉÉ cÉÉmÉZÉhQûlÉå M×üiÉå mÉëjÉqÉerÉÉkÉïÇ cÉÉmÉÇ cÉ iÉÑsrÉxÉXçZrÉÇ eÉÉiÉqÉç|”
[Ref.: Aryabhatiya, with the commentary of Suryadeva Yajvan :
Edited by K V Sharma, INSA, New Delhi, (1976) p. 47]
‘On dividing a quadrant into 24 equal parts,
the first rsine and the corresponding arc are same’.
Jya of 1 arc = J1 =225 = 3438 x sin 3.750 = 3438 x 0.0654 = 224. 86 = 225 (app.),
where [(90x1)/ 24] = 3.750 = 225 minutes.
This is almost equivalent to sin θ = θ, when θ is very small.
2. (vi) Explanation for taking 3438 as the circum-radius - Suryadeva yajvan
Suryadeva yajvan has explained the reason for taking 3438 as the circum-radius for
all scientific calculations.
rÉÌS USuÉxÉÑrÉqÉsÉUxÉ (62,832) ÍqÉiÉmÉËUkÉåUç ArÉÑiÉ²rÉ (20,000) urÉÉxÉÈ
ZÉZÉwÉ†ûlÉ (21,600) ÍsÉmiÉÉiqÉMü ÍqÉiÉmÉËUkÉå¶É¢üxrÉ MüÉå urÉÉxÉÈ CÌiÉ|
‘If 62,832 is the perimeter of a circle of diameter 20.000 units,
What is the circum-diameter of a circle of perimeter 21,600 units?’
ArÉÑiÉ²rÉ (20,000) cÉ¢üMüsÉÉ (21600)-ÍpÉWïûiuÉÉ USuÉxÉÑrÉqÉsÉUxÉ (62,832)-xrÉ ÌuÉpÉerÉ sÉokÉÇ
cÉ¢üurÉÉxÉÈ| iÉSkÉïÇ cÉ¢üurÉÉxÉÉkÉïÇ uÉxuÉÉÎalÉuÉåSUÉqÉ (3438) xÉXçZrÉqÉç| AlÉålÉ urÉÉxÉÉkÉåïlÉ
zÉÉx§ÉÏrÉxxÉMüsÉÉå urÉuÉWûÉUÈ|
[Ref.: Aryabhatiya, with the commentary of Suryadeva Yajvan : Edited by K V Sharma,
INSA, New Delhi, (1976) p. 47]

9
AB
s2n
C jn
EW
N
R
O
kn
D
un
AB AB
s2ns2n
C jnC jn
EW
N
R
O
EW
N
R
EW
N
R
EW
N
R
O
knkn
D
un
D
unun
‘When 20.000 multiplied by (the unit equivalent to) central angle made by the
perimeter in minutes (21,600) and the product divided by the perimeter (62,832)
gives the circum-diameter’. Half of this is used for all scientific calculations.
( )
20000
20000 21000
6876 .
62832
Perimeter of circle of diameter units
app
×
= =
Half of this 6876 = 3438 (app,) units is used for all scientific calculations.
The radian is the central angle made by an arc of length equal to the radius of the
circle. Radian measure requires ‘the ratio of perimeter of the circle to its diameter.’
( )
180 60
= =3437.746771= 3438 .
π
One radian app
×
Aryabhata-I used a circle of radius 3438 units to find D-rsines. 3438 is the measure
of ‘one radian’ in minutes approximately.
3. Calculating the Aryabhatiya circumference of a circle of diameter 20000 units
by adapting the algorithm to find D-rines ( of Bhaskar I)
3. (i) Algorithm to double the number of sides ‘n’ of an inscribed Regular Polygon
1) AB = a side of n-sided regular
polygon = sn
2) pn = Perimeter of the polygon
= nAB = nsn
3) AC = (1 / 2) AB
= (1/2) sn = jn
4) OC = √(OA2 – AC2)
= √(r2 – jn
2) = kn
5) CD = (OD – OC)
= (R - kn ) = un
6) AD = √(AC2 +CD2)
= √(jn
2 +un
2 ) = s2n
= side of 2n-sided
regular polygon.
Steps 1 to 6 ( except step 2) converts a side of n-sided regular polygon to a side of
2n-sided regular polygon. Step 2 gives perimeter of inscribed regular polygon of n
sides, pn = nsn
3. (ii) The following table shows the Aaryabhata’s approximation corresponding
to the regular 6 x 2t-1–sided regular polygon, when t = 1;
i.e., of a regular hexagon inscribed in a circle of diameter 20,000 units.

10
AB
O
C
D
jn
kn
un
s2n
r
AB
O
C
D
jn
kn
un
s2n
r
To find circumference of a circle of diameter 20,000 units in Excel Programer
Tally the lengths of different parts in the figure and in
the table.
62831.85307
3.141592654 = 3.1416
20000
π = =
3. (iii) Value of π (to 1000 digits)
π = 3.141592653 58979323846264338327950288419716939937510582097
494459230781640628620899862803482534211706798214808651328230
664709384460955058223172535940812848111745028410270193852110
555964462294895493038196442881097566593344612847564823378678
316527120190914564856692346034861045432664821339360726024914
127372458700660631558817488152092096282925409171536436789259
036001133053054882046652138414695194151160943305727036575959
195309218611738193261179310511854807446237996274956735188575
272489122793818301194912983367336244065664308602139494639522
473719070217986094370277053921717629317675238467481846766940
513200056812714526356082778577134275778960917363717872146844
090122495343014654958537105079227968925892354201995611212902
196086403441815981362977477130996051870721134999999837297804
995105973173281609631859502445945534690830264252230825334468
503526193118817101000313783875288658753320838142061717766914
730359825349042875546873115956286388235378759375195778185778
0532171226806613001927876611195909216420199....
t n = 6*2^(t-1) Sn Pn=n*Sn Jn=(Sn/2) Kn=sqrt(1000^2-Jn^2) Un=(1000-Kn) S2n=sqrt(Jn^2+Un^2) (Pn/20000)
1 6 10000 60000 5000 8660.254038 1339.745962 5176.380902 3
2 12 5176.380902 62116.57082 2588.19045 9659.258263 340.7417371 2610.523844 3.105828541
3 24 2610.523844 62652.57227 1305.26192 9914.448614 85.55138626 1308.062585 3.132628613
4 48 1308.062585 62787.00406 654.031292 9978.589232 21.41076761 654.3816564 3.139350203
5 96 654.3816564 62820.63902 327.190828 9994.645875 5.354125236 327.2346325 3.141557608
6 192 327.2346325 62829.04945 163.617316 9998.661379 1.338620904 163.6227921 3.141452472
7 384 163.6227921 62831.15216 81.811396 9999.665339 0.334660826 81.81208052 3.141557608
8 768 81.81208052 62831.67784 40.9060403 9999.916334 0.083665556 40.90612582 3.141583892
9 1536 40.90612582 62831.80926 20.4530629 9999.979084 0.020916411 20.45307361 3.141590463
10 3072 20.45307361 62831.84212 10.2265368 9999.994771 0.005229104 10.22653814 3.141592619
11 6144 10.22653814 62831.85033 5.11326907 9999.998693 0.001307276 5.113269237 3.141592517
12 12288 5.113269237 62831.85239 2.55663462 9999.999673 0.000326819 2.55663464 3.141592619
13 24576 2.55663464 62831.8529 1.27831732 9999.999918 8.17048E-05 1.278317322 3.141592645
14 49152 1.278317322 62831.85303 0.63915866 9999.99998 2.04262E-05 0.639158662 3.141592651
15 98304 0.639158662 62831.85306 0.31957933 9999.999995 5.10655E-06 0.319579331 3.141592654
16 196608 0.319579331 62831.85307 0.15978967 9999.999999 1.27664E-06 0.159789665 3.141592653
17 393216 0.159789665 62831.85307 0.07989483 10000 3.1916E-07 0.079894833 3.141592654
18 786432 0.079894833 62831.85307 0.03994742 10000 7.979E-08 0.039947416 3.141592654
19 1572864 0.039947416 62831.85307 0.01997371 10000 1.9947E-08 0.019973708 3.141592654
20 3145728 0.019973708 62831.85307 0.00998685 10000 4.98585E-09 0.009986854 3.141592654
21 6291456 0.009986854 62831.85307 0.00499343 10000 1.24601E-09 0.004993427 3.141592654
22 12582912 0.004993427 62831.85307 0.00249671 10000 3.11047E-10 0.002496714 3.141592654
23 25165824 0.002496714 62831.85307 0.00124836 10000 7.82165E-11 0.001248357 3.141592654
24 50331648 0.001248357 62831.85307 0.00062418 10000 2.00089E-11 0.000624178 3.141592654
25 100663296 0.000624178 62831.85307 0.00031209 10000 0 0.000312089 3.141592654
26 201326592 0.000312089 62831.85307 0.00015604 10000 0 0.000156045 3.141592654
62831.67784
3.141583892 = 3.1416
20000
π = =

11
Perimeter of an inscribed regular polygon of sides 20, 13, 26,592 in a circle of
diameter 20000 units = 62831.85307 - by Aryabhatiya Algorithm
Reference: -
1. “Aryabhatiya of Aryabhata with the commentary of Bhaskara-I and
Somesvara” – critically edited by K S Shukla, The Indian National Science
Academy, Bahadur Shaw Zafar Marg, New Delhi – 1 (1976).
2. “Aryabhatiya of Aryabhata with the commentary of Suryadeva Yajvan” –
critically edited by K V Sharma, The Indian National Science Academy,
Bahadur Shaw Zafar Marg, New Delhi – 1 (1976).
3. “Vatesvara Sidhanta and Gola of Vatesvara” (part I) – critically edited by K
S Shukla, The Indian National Science Academy, Bahadur Shaw Zafar Marg,
New Delhi – 1 (1976).
4. Vedanga jyotisa of Lagadha – the translation and notes of Prof. T.S.
Kuppanna Sastry, critically edited by K.V. Sarma, Indian National Science
Academy, New Delhi 1985.
5. “A Concise History of Science in India”; D. M. Bose, S. N. Sen, B. V.
Subbarayappa, Editors] INSA, New Delhi
6. “From One to Zero - A Universal History of Numbers”, Georges Ifra, Viking
Penguin Inc. New York (1985).]
7. “pÉÉUiÉÏrÉ aÉÍhÉiÉqÉç- Indian Mathematics in Sanskrit : Concepts and Achievements” -
Venkatesha Murthy, Rashtriya Sanskrit Vidyapeetha (Deemed University)
Tirupati - 517 507 (2005).
8. “Bharatiya Ganita Darpana”: Venkatesha Murthy, National Institute of Vedic
Sciences, # 58, Raghavendra Colony, Chamarajapet, BANGALORE – 560
058. (2010).
9. CRYPTIC NUMERALS in SANSKRIT TEXTS : Venkatesha Murthy,
National Institute of Vedic Sciences, #58, Raghavendra Colony,
Chamarajapet, Bangalore, 560 018, Karnataka (2013)
22-02-2015
62831.85307
= 3.141592654
20000
π =

A paper on finding d rsines from aryabhatiya bhasya by bhaskara i

More Related Content

Similar to A paper on finding d rsines from aryabhatiya bhasya by bhaskara i (20)

Recently uploaded (20)

A paper on finding d rsines from aryabhatiya bhasya by bhaskara i