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NPTEL COURSE ON

MATHEMATICS IN INDIA:FROM VEDIC PERIOD TO MODERN TIMES

LECTURE 37

Proofs in Indian Mathematics - Part 2

K. Ramasubramanian

IIT Bombay

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OutlineProofs in Indian Mathematics - Part 2

Volume of a sphere The principle involved Area of a circle Volume of a slice of a given thickness Total volume

A couple of theorems

Jya-sam.varga-nyaya(Theorem on product of chords) Jya-vargantara-nyaya(diff. of the squares of chords) Jya-sam.varga-nyaya Jyotpatti

The cyclic quadrilateral

Expressing diagonals in terms of sides Area in terms of sides Circumradius in terms of sides

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Volume of a sphereThe principle involved

To find the volume of a sphere, it is divided intolarge number(sayn) of

slices ofequal thickness. In the figure the sphereNASBNis divided into slices by planes parallel

to the equatorial planeAGB.

Then the volume of each slice is

obtained. Volume= Area thickess. Area is obtained by finding theaverage

radius of circlesat the top and bottomof the slice.

Ifdis diameter, then the thickess of the

slice= dn

.

Thus, first we need to obtain an

expression for finding the thearea of a

circle.

Sum up the elementary volumesof the

slices, that will add up to the sphere.

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Volume of a sphereObtaining the area of a circular slice

Figure:Circular slice cut into two pieces.

In the figure above we have indicateda circular slice being turned into a

rectangleby appropriately sectioning it, and inserting one half of thecircular slice into the other.

The length of this rectangular strip correspondsto half the

circumferenceC. If the radius is r, then the area of this slice is

Area=

1

2 C

r

. (1)

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Volume of a sphereObtaining the elementary volume of the sphere (= volume of the circular slice

Let r be the radius of the sphere andCthe circumference of a great

circle on this sphere.. The radius of thej-th sliceinto which the sphere has been divided

intocan be conceived of as the half-chordBj (bhuja).

Now, the circumference of this slice is given by

Cjth slice=C

r

Bj

Hence the area of this circular slice is

Ajth slice= 1

2

C

r

Bj Bj

. Therefore, the elementary volume is given by

V = 1

2

C

r

B2j , (2)

where is the thickness of the slices.

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Volume of a sphereSumming up the elementary volumes of the circular slices)

The volume of the sphere is obtained by finding thesum of theelementary volumesconstituted by the slices:

V =

V =n

j=1

1

2

C

r

B2j (3)

In the above expression, the

thickness can be expressed in

terms of the radius as = 2r

n .

IfBjcan also be expressed in termsofr, then we can get V =V(r).

For this we invoke the

Jya-sara-sam.varga-nyayagiven by

Aryabhat.a.

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Jya-sara-sam.varga-nyayaof Aryabhat.aTheorem on the square of chords

In his Aryabhat. ya, Aryabhat.ahas presents the theorem on theproduct of chords as follows (in half arya):

(Aryabhat.ya, Gan. ita17)

The wordsvargaand qsamvargarefer tosquareandproduct

respectively.

Similarly,dhanusand sararefer toarcandarrowrespectively.

Using modern notations the abovenyayamay be expressed as:

product of saras = Rsine2

DE EB = AE2

C

D E B

A

O

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Volume of a sphereSumming up the elementary volumes of the circular slices)

It was shown that the volume of the sphere may be expressed as:

V =

nj=1

12

Cr

B2j

In the figureAP=PB=Bjis the jthhalf-chord, starting fromN.

Applyingjya-sara-sam.varga-nyaya,we have

B2j = AP PB=NP SP

= 1

2

[(NP+ SP)2 (NP2 +SP2)]

= 1

2[(2r)2 (NP2 +SP2)]. (4)

It may be noted in the figure that thej-th RversineNP=jand itscomplementSP= (nj). Hence, while summing the squares of the

RsinesB2j, bothNP2 andSP2 add to the same result.

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Volume of a sphereSumming up the elementary volumes of the circular slices)

Thus the expression for the volume of the sphere given by

V n

j=1

12

Cr

B2j ,

reduces to

V C

2r1

2 [n.(2r)22r

n2

2.[12

+22

+...n2

] 2r

n

.

It was known to Kerala mathematicians, thatfor largen

12 +22 +...n2 = n3

3 .

Using this, the expression for the volume of the sphere becomes

V =

C

2r

4r3

8

3r3

= C6d2. (5)

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Jyasam.varga-nyayaandJya-vargantara-nyayaTheorem on product of chords and the difference of the square of chords

In the figureDMis the perpendicular

from the vertexDonto the diagonal

ACof the cyclic quadrilateral.

Let us consider the two triangles

AMD and CMD, that is formed by DM

in the triangleADC. It can be easily

seen that

AD2 DC2 =AM2 MC2. (6)

In other words, the difference in the

squares of thejyasAD, DCis equal

to the difference in the square of the

base segments (abadhas)AM, MC.

This result may be noted downfor later useas

jyavargantara= abadhavargantara

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Jyasam.varga-nyayaandJya-vargantara-nyayaTheorem on product of chords and the difference of the square of chords

In order to derive a certain property, we

rewrite (6) as

AD2 DC2 = (AM+MC)(AM MC) (7)

It may be noted that

AM+MC jyaof sum of the arcsAM MC jyaof diff. of the arcs

Denoting the chordsADandDCasJ1 andJ2, and the arcs associated

with them asc1 andc2, we may rewrite (7) as

J21 J22 =Jya(c1+ c2) Jya(c1 c2)

In otherwords,

=

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Jyasam.varga-nyayaandJya-vargantara-nyayaTheorem on product of chords and the difference of the square of chords

Recalling the equation,

J21 J22 =Jya(c1+ c2) Jya(c1 c2) (8)

It can also be shown that

J1J2 = Jya

c1+ c2

2 2

Jya

c1 c2

2 2

(9)

In otherwords,

=

-

The two equations (8) and (9) are equivalent to the trigonometric

relations:

sin2(1) sin2(2) = sin(1+ 2) sin(1 2),

sin(1)sin(2) = sin2

(1+ 2)

2

sin2

(1 2)

2

.

with our convention that c1>

c2

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The cyclic quadrilateral & the third diagonal

Consider the equation

sin 1 sin2 =sin2

(1+ 2)

2

sin2

(1 2)

2

.

If we put 1 = (n+1), and 2 = (n 1)in the above equation,then we immediately obtain the following equation

sin(n+1)= sin2 n sin2

sin(n 1)

This is precisely the equation that is presented in the following

verse given by Sankarain hisKriyakramakar:

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The cyclic quadrilateral & the third diagonal

Notation:

the arcs (AB, BC, CD, DA) c1, c2, c3, c4

the chords (AB, BC, CD, DA)

J1,

J2,

J3,

J4 the diagonals (BD,AC,DG) K1, K2, K3

Having drawn a quadrilateral,obviously we can have only two

diagonals. Gis a point chosen such that

BC=AG.

By swapping any two sides could be adjacent, or oppositewe would be affectingonly oneof the two diagonal of thequadrilateral, while the otherdiagonal remains fixed.

Thisnewdiagonal that getsgenerated due to this swappingof sides is referred to as thethird diagonal or bhavikarn. a

A I

B

CD

G

H

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The elegant results we would like to prove

The three diagonals of the cyclic quadrilateral would be referred to as

1. (is. t.akarn. a) chosen/first diagonal

2. (itarakarn. a) other/second diagonal

3.

(bhavikarn. a) future/third diagonal

The results that we would prove:

1. =

2.

=

3. =

The above results essentially express the product of the diagonals in

terms of the sum of the product of the sidesjyas.

Making use of them we express the diagonals in terms of sides.

Then by making use of yet another result

product of two sides of a triangle

circum-diameter =the altitude, (10)

we show that thearea can be expressed in terms of the diagonals and

in turn, in terms of the sides.

Th k !

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Thanks!

THANK YOU

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