UNIT 13: TRIGONOMETRIC SERIES - kkhsou.in€¦ · UNIT 13: TRIGONOMETRIC SERIES UNIT STUCTURE...

Classical Algebra and Trigonometry (Block 2) 239

UNIT 13: TRIGONOMETRIC SERIES

UNIT STUCTURE

13.1 Learning Objectives

13.2 Introduction

13.3 Gregory’s Series

13.3.1 General Theorem on Gregory’s Series

13.4 Summation of Trigonometric Series

13.4.1 C+iS Method

13.4.2 Series Based on Geometric or Arithmetico-Geometric

Series

13.4.3 Sum of a Series of Sines (or Cosines) of Angles in

Arithmetical Progression

13.4.4 Summation of Series using Binomial Series

13.4.5 Summation of Series using of Exponential Series

13.4.6 Summation of Series using Logarithmic Series and

Gregory’s Series

13.4.7 Difference Method

13.5 Let Us Sum Up

13.6 Answers to Check Your Progress

13.7 Further Reading

13.6 Model Questions

13.1 LEARNING OBJECTIVES

After going through this unit, you will be able to:

l know about Gregory’s series

l describe the summation of trigonometric series.

13.2 INTRODUCTION

In previous unit, we discussed DeMovire’s Theorem and its some

important deductions. We will introduce Gregory’s series. Finally, we will

discuss summation of trigonometric series.

Classical Algebra and Trigonometry (Block 2)240

13.3 GREGORY’S SERIES

Statement: If θ lies within the closed interval

ππ−

4,

4 ,

i.e., if 44π≤θ≤π− ,

then ⋅∞⋅⋅+θ−θ+θ−θ=θ 753 tan71

tan51

tan31

tan

Proof: We have )sini(coscos

1)

cossin

i1()tani1( θ+θθ

=θθ+=θ+

θθ= ie.sec

Now, taking logarithm of both sides, we have

)ie.log(sec)tani1log( θθ=θ+ [log(AB) = logA+logB]θ+θ= ielogseclog

θ+θ= iseclog (1)

Now, since θ lies between 4π− and

4π

, θtan lies between –1 and 1,

i.e., θtan is numerically not greater than 1.

We have from (1)

) itanlog(1ilogsec θ+=θ+θ (By Logarithmic series (4.7.1)

∞⋅⋅⋅⋅⋅+θ−θ+θθ= to)tani(41

)tani(31

)tani(21

-itan 432

∞⋅⋅⋅+θ−θ−θ+θ= to4

tan3

tani2

tantani

432

)4

tan2

tan()

3tan

(tani423

⋅∞⋅⋅+θ−θ+⋅∞⋅⋅+θ−θ= (2)

Equating imaginary part on both sides, we get

∞⋅⋅⋅⋅⋅⋅⋅⋅⋅θ+θ−θ=θ5

tan3

tantan

53

(3)

(3) is known as Gregory’s series.

Some Import ant Deduction:

i) Now we put xtan =θ

So that xtan 1−=θ

Then we have from (3)

∞⋅⋅⋅⋅⋅⋅⋅⋅⋅+−=−

5x

3x

xxtan53

1 where 1x1 ≤≤−

Trigonometric SeriesUnit 13


ii) We equate the real parts on both sides of (2), we get

⋅∞⋅⋅⋅⋅−θ+θ−θ=θ 642 tan61

tan41

tan21

logsec

13.3.1 General Theorem on Gregory’ s Series

Statement: If θ lies between π−π41

n and π+π41

n

i.e., π+π≤θ≤π−π41

n41

n ,

then, ⋅∞⋅⋅⋅⋅−θ+θ−θ=π−θ 53 tan51

tan31

tann

Proof: We put α+π=θ n , then π−θ=α n

The given condition reduces to 44π≤α≤π− .

Hence, )ntan(i1tani1 α+π+=θ+

αα+=α+=

cossin

i1tani1

αα+α=

cossinicos

αα= ie.sec

Now, taking logarithm )e.log(sec)tani1log( iαα=θ+

4n

4n

π+π≤θ≤π−πΘ

)4

ntan(tan)4

ntan(π+π≤θ≤π−π∴

1tan1 ≤θ≤−∴

1tani ≤θ

)tani1log( θ+ can be expanded in powers of θtan .

∞⋅⋅⋅⋅⋅⋅+θ−θ+θ−θ=θ+ 432 )tani(41

)tani(31

)tani(21

tani)tani1log(

∞⋅⋅⋅+θ−θ−θ+θ= to4

tan3

tani2

tantani

432

)4

tan2

tan()

3tan

(tani423

⋅∞⋅⋅+θ−θ+⋅∞⋅⋅+θ−θ= .......

α+α= i)log(sec

Equating the imaginary parts on both sides, we have

Trigonometric Series Unit 13


∞⋅⋅⋅⋅⋅⋅−θ+θ−θ=α 53 tan51

tan31

tan

Or, ∞⋅⋅⋅⋅⋅⋅−θ+θ−θ=π−θ 53 tan51

tan31

tann

Example 1: Prove that:

⋅⋅⋅⋅+−+− 322 3.7

13.51

31

1.32 π=

Solution: L.H.S

⋅⋅⋅⋅+−+−= 322 3.7

13.51

31

1.32

( ) ( ) ( )

⋅⋅⋅⋅+−+−= 642

3.7

1

3.5

1

33

11.32

( ) ( ) ( )

⋅⋅⋅⋅+−+−= 753

37

1

35

1

33

1

3

13.3.2

= −

3

1tan6 1

6.6π=

π= = R.H.S

Example 2: Sum the series

i) ⋅∞⋅⋅⋅⋅−+− 42 4.51

.4.31

1

ii) ∞⋅⋅⋅⋅+−+− 222 3.71

3.51

31

1

Solution: i) The given series–

⋅∞⋅⋅⋅⋅−+− 42 4.51

.4.31

1

⋅∞⋅⋅⋅⋅−+−= 53 4.5

1.4.3

141

4

41

tan4 1−= (By Gregory’s series )

ii) The given series

∞⋅⋅⋅⋅+−+− 322 3.71

3.51

31

1

∞⋅⋅⋅⋅+−+−= 32 3.71

3.51

3.31

1



( ) ( ) ( )

⋅⋅⋅⋅+−+−= 753

3.7

1

3.5

1

3.3

1

3

13 (Gregory’s series)

= −

3

1tan3 1

6.3π= ( )

63π=

Example 3: Prove that:

⋅⋅⋅⋅−

++

+−

+=π

5533 71

32

51

71

32

31

71

32

4

Solution: R.H.S, ⋅⋅⋅⋅−

++

+−

+ 5533 7

132

51

71

32

31

71

32

⋅⋅⋅⋅−+−+

⋅⋅⋅⋅−+−= 5353 7

1.

51

71

.31

71

31

.51

31

.31

31

2

71

tan31

tan2 11 −− += (Gregory’s series)

71

tan

91

1

32

tan 11 −− +−

=

71

tan43

tan 11 −− +=

−

+= −

71

.43

1

71

43

tan 1

1tan 1−=

4π= = R.H.S

Example 4: Prove that

( ) ⋅⋅⋅⋅⋅+

+

−−+⋅⋅⋅⋅⋅+

×−=π −−

+n21n1

1n

79.32

1n21

34381713

1217

4

Solution: The nth term of the series is given by

( )

+

−−= −−

+n21n1

1n

n 79.32

1n21

T

( )

+

−−= −−

+

1n21n2

1n

71

31

.21n2

1



+=∴

71

32

T1 ,

+−= 332 7

132

31

T ,

+= 553 7

132

51

T

+−= 774 7

132

71

T , and so on.

Hence, ⋅∞⋅⋅++⋅⋅⋅⋅+++= n321 TTTTS

⋅∞⋅⋅++−+

⋅∞⋅⋅+−+−= 73753 7.7

17.31

71

3.71

3.51

3.31

31

2

71

tan31

tan2 11 −− +=

71

tan

31

1

31

.2tan 1

2

1 −− +−

=

71

tan43

tan 11 −− +=

71

.43

1

71

43

tan 1

−

+= −

1tan 1−=

4π= = L.H.S

CHECK YOUR PROGRESS

Q.1: Prove that: ⋅∞⋅⋅⋅⋅+++=π11.91

7.51

3.11

8

Q.2: Prove that:

⋅∞⋅⋅+

++

+−

+=π

5533 71

32

51

71

32

31

71

32

4

Q.3: Prove that: 41

tan44.51

4.31

1 142

−=∞⋅⋅⋅⋅−+−

Q.4: If θ lies between 0 and 2π

, prove that

⋅∞⋅⋅−θ+θ−θ=

θ+θ−−

2tan

51

2tan

31

2tan

cos1cos1

tan 10621



13.4 SUMMATION OF TRIGONOMETRICAL SERIES

Here, we shall discuss important methods for summing up

trigonometric series which may be finite or infinite. There are two important

methods for summation. These are (a) iSC + method, (b) the difference

method.

13.4.1 iSC + Method

Consider the series:

⋅⋅⋅+β+α+β+α+α= )2cos(a)cos(acosaC 210 (1)

and

⋅⋅⋅⋅+β+α+β+α+α= )2sin(a)sin(asinaS 210 (2)

The above series may be finite or infinite.The coefficients

210 a,a,a ... and βα, ,... may be any numbers real or complex.

In the series (1), we have terms which contain cosines of

numbers.It is called cosine series and its sum is denoted by C. The

series (2) contains sines of numbers.It is called sine series and its

sum is denoted by S.

Now, using Euler,s Theorem

[ ][ ] ⋅∞⋅⋅+β+α+β+α

+β+α+β+α+α+α=+)2sin(i)2cos(a

)sin(i)cos(a)sini(cosaiSC

2

10

⋅∞⋅⋅+++= β+αβ+αα )2(i2

)(i1

i0 eaeaea (3)

[ ][ ] ⋅∞⋅⋅+β+α−β+α

+β+α−β+α+α−α=−)2sin(i)2cos(a

)sin(i)cos(a)sini(cosaiSC

2

10

∞⋅⋅⋅⋅+++= β+α−β+α−α− )2(i2

)(i1

i0 eaeaea (4)

From the series (3) and (4),we use

[ ])iSC()iSC(21

C −++=

and [ ])iSC()iSC(i2

1S −−+=

to find the values of C and S respectively.



13.4.2 Series Based on Geometric or Arithmetico-

Geometric Series

Sum of n terms in G.P

1n32 ar.......ararara −+++++

−−=

r1r1

an

or

−−1r1r

an

according as 1r < or 1r > .

Sum of the infinite Geometric series:

⋅∞⋅⋅⋅⋅+++++++ − n1n32 arar.......ararara

,r1

a−

= if 1r < i.e., 1r1 <<− .

13.4.3 Sum of a Series of Sines(or Cosines) of Angles in

Arithmetical Progression

Let

{ }β−+α+⋅⋅⋅⋅+β+α+β+α+β+α+α= )1n(sin)3sin()2sin()sin(sinS

We assume that,

{ }β−+α+⋅⋅⋅⋅+β+α+β+α+β+α+α= )1n(cos)3cos()2cos()cos(cosC

So,

{ } { } ⋅⋅⋅⋅+β+α+β+α+β+α+β+α+α+α=+ )2sin(i)2cos()sin(i)cos()sini(cosiSC

{ } { }[ ]β−+α+β−+α+ )1n(sini)1n(cos{ }β−+αβ+αβ+αα +⋅⋅⋅+++= )1n(i)2(i)(ii eeee

{ }β−ββα +⋅⋅⋅+++= i)1n(i2ii eee1e

−−= β

βα

i

nii

e1e1

e

β

β

β

βα

−−

−−= i

i

i

nii

e1e1

.e1e1

e

{ }

( ) 1ee1eeee

ii

)1n(i)n(i)(ii

++−+−−= β−β

β−+αβ+αβ−αα

{ }

β−+−−=

β−+αβ+αβ−αα

cos22eeee )1n(i)n(i)(ii

( ) ( ){ } { } { } { }[ ])cos1(2

)1n(sini)1n(cos)nsin(i)ncos(sini)cos(sinicosβ−

β−+α+β−+α+β+α+β+α−β−α+β−α−α+α=



{ }[ ] ( ) { }[ ])cos1(2

)1n(sin)nsin(sinsini

cos1(2)1n(cos)ncos()cos(cos

β−β−+α+β+α−β−α−α+

β−β−+α+β+α−β−α−α=

Equating real and imaginary parts, we get

{ })cos1(2

)1n(cos)ncos()cos(cosC


{ }[ ] { }

)cos1(2)ncos()cos()1n(coscos

β−β+α+β−α−β−+α+α=

2sin4

21n

cos2

1ncos2

21n

cos2

1ncos2

2 β

β

+

β

−+α−

β

−

β

−+α

=

2sin2

21n

cos2

1ncos

21n

cos

2 β

β+−β−

β

−+α

=

2sin2

2sin

2n

sin2.2

1ncos

2 β

ββ

β−+α

=

2sin

2n

sin2

1ncos

β

β

β−+α

=

and { })cos1(2

)1n(sin)nsin()sin(sinS


{ }[ ] { })cos1(2

)nsin()sin()1n(sinsinβ−

β+α+β−α−β−+α+α=

2sin4

21n

cos2

1nsin2

21n

cos2

1nsin2

2 β

β

+

β

−−α−

β

−

β

++α

=

2sin2

21n

cos2

1ncos

21n

sin

2 β

β+−β−

β

−+α

=



2sin2

2sin

2n

sin2.2

1nsin

2 β

ββ

β−+α

=

2sin

2n

sin2

1nsin

β

β

β−+α

=

Hence,

{ }2

sin

2n

sin2

1nsin

)1n(sin)sin(sinβ

β

β−+α

=β−+α+⋅⋅⋅⋅+β+α+α (1)

{ }2

sin

2n

sin2

1ncos

)1n(cos)cos(cosβ

β

β−+α

=β−+α+⋅⋅⋅⋅+β+α+α (2)

Particular case (i): Putting α=β in (1) and (2), we get

2sin

2n

sin2

1nsin

nsin2sinsinα

α

α+

=α+⋅⋅⋅⋅+α+α

and

2sin

2n

sin2

1ncos

ncos2coscosα

α

α+

=α+⋅⋅⋅⋅+α+α

ii) If n2π=β , then 0sin

2n

sin =π=β in (1) and (2),

then, 0 terms n ton4

sinn2

sinsin =⋅⋅⋅⋅+

π+α+

π+α+α

and

0 terms n ton4

cosn2

coscos =⋅⋅⋅⋅+

π+α+

π+α+α

Example 1: Sum to n terms of the series

⋅⋅⋅⋅⋅+β+α−β+α+β+α−α )3sin()2sin()sin(sinSolution: Let

terms n to)3sin()2sin()sin(sinS ⋅⋅⋅⋅⋅+β+α−β+α+β+α−α=terms n to)33sin()2sin(2)sin(sin ⋅⋅⋅+β+α+π+β+α+π+β+α+π+α=



( ) ( )

2sin

2n

sin2

1-nsin

β+π

β+π

β+π+α

=

( )( ) ( )

2cos

2n

sin2

1-nsin

β

β+π

β+π+α

=

Example 2: Sum to n terms of the series:

⋅⋅⋅⋅⋅+β+α+β+α+α )2(sin)(sinsin 222

Solution: terms n to)2(sin)(sinsinS 222 ⋅⋅⋅⋅⋅+β+α+β+α+α=

terms n to 2

)2cos2(-12

)cos2(-12

cos2-1 ⋅⋅⋅⋅+β+α+β+α+α=

[ ]terms n to)2(2cos)(2cos2cos21

terms) n to111(21 ⋅⋅⋅⋅+β+α+β+α+α−⋅⋅⋅⋅⋅+++=

β

β

β−+α

−=

22

sin

2.2n

sin.2.2

1n2cos

.21

2n

( ){ } ( )β

ββ−+α−=sin

nsin..1n2cos.

21

2n

Example 3: Sum the series:

terms n to coscos3

coscos2

coscos

1 32 ⋅⋅⋅⋅⋅+θθ+

θθ+

θθ+

Solution: ++++= θ

3cos

θ3cos

θ2cos

θ2cos

θcos

θcos1C ........ to n terms.

terms n to cossin3

cossin2

cossin

S 32 ⋅⋅⋅⋅⋅+θθ+

θθ+

θθ=

Now,

( ) ( ) ( ) terms n to

cosisin3cos3

cosisin2cos2

cosisincos

1iSC 32 ⋅⋅⋅⋅⋅+θ

θ+θ+θ

θ+θ+θ

θ+θ+=+

terms n tosecesecesece1 3i32i2i ⋅⋅⋅⋅+θ+θ+θ+= θθθ

( )1sece1sece

i

ni

−θ−θ= θ

θ

( )( )( )( )1sece1sece

1sece1seceii

inni

−θ−θ−θ−θ= θ−θ

θ−θ



1)ee(secsec1e sece sece sec

ii2

inini)1n(1n

++θ−θ+θ−θ−θ= θ−θ

θ−θθ−+

1cos2secsec1e sece sece sec

2

inini)1n(1n

+θθ−θ+θ−θ−θ=

θ−θθ−+

1sec1e sece sece sec

2

inini)1n(1n

−θ+θ−θ−θ=

θ−θθ−+

θ+θ−θ−θ=

θ−θθ−+

2

inini)1n(1n

tan1e sece sece sec

[ ] ( ) ( )

θ+θ+θθ−θ+θθ−θ−+θ−θ=

+

2

n1n

tan1sinicossecnsinincossec)1nsin(i)1ncos(sec

θθθ−θθ−θ−θ+

θ+θθ−θθ−θ−θ=

++

2

n1n

2

n1n

tancossecnsinsec)1nsin(sec

itan

1cossecncossec)1ncos(sec

Equating real parts on both sides,we get

θ+θθ−θθ−θ−θ=

+

2

n1n

tan1cossecncossec)1ncos(sec

C

θ+−θθ−θ−θ=

+

2

n1n

tan11ncossec)1ncos(sec

θθθ−θ−θ=

+

2

n1n

tanncossec)1ncos(sec

[ ]θ

θθ−θ−θ=+

2

1n

tanncoscos)1ncos(sec

θθθθ=

+

2

1n

tansinnsinsec

θθθ=

tansinsecn

CHECK YOUR PROGRESS

Q.5: Sum the series:

a) terms n to3sin2sinsin 222 ⋅⋅⋅⋅+α+α+α

b) terms n to)2(cos)(coscos 222 ⋅⋅⋅⋅+β+α+β+α+α

c) terms n to3cos2coscos 222 ⋅⋅⋅⋅+α+α+α

d) terms n to3sin2sinsin 333 ⋅⋅⋅⋅+α+α+α

e) terms n tocos5cos3cos ⋅⋅⋅⋅⋅+α+α+α



13.4.4 Summation of Series using Binomial Series

We should remember the following formulae :

i) When n is a positive integer and x,a are any complex number,

we have

n22n1nnn aax!2

)1n(nanxx)ax( +⋅⋅⋅⋅+−++=+ −−

nn22n1nnn a)1(ax!2

)1n(nanxx)ax( −+⋅⋅⋅⋅−−+−=− −−

n32n xx!3

)2n)(1n(nx

!2)1n(n

nx1)x1( +⋅⋅⋅⋅+−−+−++=+

nn32n x)1(x!3

)2n)(1n(nx

!2)1n(n

nx1)x1( −+⋅⋅⋅⋅+−−−−+−=−

ii) When n is any rational index and x is a complex number such

that 1x < , we have

∞⋅⋅⋅⋅+−−+−++=+ 32n x!3

)2n)(1n(nx

!2)1n(n

nx1)x1(

∞⋅⋅⋅⋅+−−−−+−=− 32n x!3

)2n)(1n(nx

!2)1n(n

nx1)x1(

∞⋅⋅⋅⋅+++++++=− − 32n x!3

)2n)(1n(nx

!2)1n(n

nx1)x1(

Also, ∞⋅⋅⋅⋅−+−+=+ 3221

x6.4.25.3.1

x4.23.1

x21

1)x1(

∞⋅⋅⋅⋅++++=−− 322

1

x6.4.25.3.1

x4.23.1

x21

1)x1(

∞⋅⋅⋅⋅−+−+=+ 3231

x9.6.35.2.1

x6.32.1

x31

1)x1(

and ∞⋅⋅⋅⋅++++=−− 323

1

x9.6.35.2.1

x6.32.1

x31

1)x1(

Example 1: Sum the Series: ∞⋅⋅⋅⋅+α+α+α to5sin4.23.1

3sin21

sin

Solution: Let ∞⋅⋅⋅⋅+α+α+α= to5sin4.23.1

3sin21

sinS

Now, ∞⋅⋅⋅⋅+α+α+α= to5cos4.23.1

3cos21

cosC

Then, using Bionomial Theorem



( ) ( ) ( ) ∞⋅⋅⋅⋅+α+α+α+α+α+α=+ to5sini5cos4.23.1

3sini3cos21

sinicosiSC

∞⋅⋅⋅⋅+++= ααα toe4.23.1

e21

e i5i3i

∞⋅⋅⋅⋅+++= ααε toe

4.23.1

e21

1e i4i2i

[ ] 21

i2i e1e−αα −=

( )[ ] 21

2 cossin2isin2sinicos−

αα−αα+α=

( ) ( )

α−ααα+α=

−− 21

)cosi(sinsin2sinicos 21

2

( ) ( )

α−π−−

α−π−α+αα=

−

221

sini22

1cossinicossin2 2

12

( ) ( )

α−π+

α−πα+αα=

−

21

4sini

21

4cossinicossin2 2

12

( )

α−π+α+

α−π+αα=

−

21

4sini

21

4cossin2 2

12

(Since ( )( ) ( ) ( )φ+θ+φ+θ=φ+φθ+θ sinicossinicossinicos )

( )

α+π+

α+πα=

−

21

4sini

21

4cossin2 2

12

Equating the imaginary parts on both sides,we get

( )

α+πα=

−

21

4sinsin2S 2

12

Example 2: Sum the series

∞⋅⋅⋅⋅+θ−θ+θ− 3cos6.4.25.3.1

2cos4.23.1

cos21

1

Solution: Let ⋅⋅⋅⋅+θ−θ+θ−= 3cos6.4.25.3.1

2cos4.23.1

cos21

1C

⋅⋅⋅⋅+θ−θ+θ−= 3sin6.4.25.3.1

2sin4.23.1

sin21

S

Then,

( ) ( ) ( ) ⋅⋅⋅⋅+θ+θ−θ+θ+θ+θ−=+ 3sini3cos6.4.25.3.1

2isn2cos4.23.1

sinicos21

1iSC

⋅⋅⋅⋅+−+−= θθθ i3i2i e6.4.25.3.1

e4.23.1

e21

1



⋅⋅⋅⋅+−+−= 32 x6.4.25.3.1

x4.23.1

x21

1 where θ= iex

( ) 21

x1 −+=

( ) 21

ie1−θ+=

( ) 21

sinicos1 −θ+θ+=

21

2

2cos

2sin2i

2cos2

−

θθ+θ=

21

21

2sini

2cos

2cos2

−−

θ+θ

θ=

θ−θ

θ=

−

4sini

4cos

2cos2

21

[By De Moivre’s Theorem]

Equating real parts, we get 4

cos2

cos2C21

θ

θ=

−

21

2cos2

4cos

θ

θ

=

2cos2

4cos

θ

θ

=

13.4.5 Summation of Series using of Exponential Series

The following series are frequently used.

i) ∞⋅⋅⋅⋅++++=!3

x!2

xx1e

32x

ii) ∞⋅⋅⋅⋅+−+−=−

!3x

!2x

x1e32

x

iii) ∞⋅⋅⋅⋅−+−=!5

x!3

xxxsin

53

iv) ∞⋅⋅⋅⋅−+−=!4

x!2

x1xcos

42



v) ∞⋅⋅⋅⋅+++=!5

x!3

xxxsinh

53

vi) ∞⋅⋅⋅⋅+++=!4

x!2

x1xcosh

42

Example 1: Sum the Series

∞⋅⋅⋅⋅+βα−βα+βα−!3

3coscos!2

2coscoscoscos1

32

Solution: Let

∞⋅⋅⋅⋅+βα−βα+βα−=!3

3coscos!2

2coscoscoscos1C

32

and ∞⋅⋅⋅⋅+βα−βα+βα−=!33sin

cos!22sin

cossincosS 32

Then,

( ) ∞⋅⋅⋅⋅+β+βα−β+βα+β+βα−=+!3

)3sini3(coscos!2

)2sini2(coscossinicoscos1iSC

32

∞⋅⋅⋅⋅+α−α+α−= βββ i33

i22

i e!3

cose

!2cos

ecos1

βα−=iecose

( )β+βα−= sinicoscoseβα−βα−= sincosicoscos e.e

( ) ( ){ }βα−βα= βα− sincoscsinisincoscose coscos

Equating real parts on both sides, we get

( )βα= βα− sincoscoseC coscos

Example 2: Find the sum of the series

∞⋅⋅⋅⋅+θ+θ+ 4cos!4

c2cos

!2c

142

Solution: Let ∞⋅⋅⋅⋅+θ+θ+= 4cos!4

c2cos

!2c

1C42

and ∞⋅⋅⋅⋅+θ+θ+= 4sin!4

c2sin

!2c

1S42

Then ( ) ( ) ∞⋅⋅⋅⋅+θ+θ+θ+θ+=+ 4sini4cos!4

c2sini2cos

!2c

1iSC42

∞⋅⋅⋅⋅+++= θθ i44

i22

e!4

ce

!2c

1

( )θ= icecosh

( ){ }θ+θ= sinicosccosh



( ){ }θ+θ= sinicosiccos

( )θ−θ= sinccosiccos

( ) ( ) ( ) ( )θθ+θθ= sincsincoscsinhisinccoscosccosh

Equating real parts, we get ( ) ( )θθ= sinccoscosccoshC

13.4.6 Summation of series using Logarithmic series and

Gregory’ s series

i) ∞⋅⋅⋅⋅+−+−=+4x

3x

2x

x)x1log(432

ii) ∞⋅⋅⋅⋅−−−−−=−4x

3x

2x

x)x1log(432

iii) ∞⋅⋅⋅⋅−+−=−

5x

3x

xxtan53

1 where 1x1 ≤≤−

Example: Sum the Series:

i) ∞⋅⋅⋅⋅−θ+θ−θ 3cos31

2cos21

cos

ii) ∞⋅⋅⋅⋅−θ+θ−θ 3sin31

2sin21

sin

Solution: Let the series (i) and (ii) be denoted by C and S

respectively.

i.e ∞⋅⋅⋅⋅−θ+θ−θ= 3cos31

2cos21

cosC

and ∞⋅⋅⋅⋅−θ+θ−θ= 3sin31

2sin21

sinS

Then

( ) ( ) ( ) ∞⋅⋅⋅⋅−θ+θ+θ+θ−θ+θ=+ 3sini3cos31

2sini2cos21

sinicosiSC

∞⋅⋅⋅⋅−+−= θθθ i3i2i e31

e21

e

)e1log( iθ+= by Logarithmic series

)sinicos1log( θ+θ+=

{ }θ+

θ+θ+θ+= −

cos1sin

tanisin)cos1(log21 122

( )

θ

θθ

+θ+θ+θ+= −

2cos2

2cos

2sin2

tanisincos2cos1log21

2

122



( )

θ

θ

+θ+= −

2cos

2sin

tanicos22log21 1

( ){ }

θ+θ+= −

2tantanicos12log

21 1

θ+

θ= −

2tantani

2cos2.2log

21 12

θ+

θ= −

2tantani

2cos.2log

21 1

2

2i

2cos.2log

θ+

θ=

Equating real and imaginary parts, we get

θ=

2cos.2logC

i.e i)

θ=∞⋅⋅⋅⋅−θ+θ−θ

2cos2log3cos

31

2cos21

cos and 2

Sθ=

i.e ii)2

3sin31

2sin21

sinθ=∞⋅⋅⋅⋅−θ+θ−θ

CHECK YOUR PROGRESS

Q.6: Sum the series:

i) ( ) ( ) ∞⋅⋅⋅⋅+β+α+β+α+α 2cos!2

xcosxcos

2

ii) ∞⋅⋅⋅⋅+α+β−βαα

α 5cos5

e3cos

3e

cose53

iii) ∞⋅⋅⋅⋅+θ+θ+θ+ 3cos!3

x2cos

!2x

cosx132

iv) ∞⋅⋅⋅⋅−θ+θ−θ 3sinx31

2sinx21

sinx 32

13.4.7 Difference Method

In order to sum a series,sometimes it is convenient to split

up each term as the difference of two expressions such that one

expression of each difference occurs in succeeding with an opposite



sign. The splitting is done in such a way that when all the terms of

the series are added together, the component terms cancel in pairs.

Finally we are left with two terms, one from the first term and one

from last term.

Suppose, we have to find the sum of n321 uuuu +⋅⋅⋅⋅+++

First we write )n(f)1n(fun −+= (1)

From (1) putting n...,,3,2,1n = , we get

)1(f)2(fu1 −=

)2(f)3(fu2 −=

)3(f)4(fu3 −=

...........................

...........................

...........................

)n(f)1n(fun −+=

Adding vertically, we get )n(f)1n(fuuuu n321 −+=+⋅⋅⋅⋅+++

(since all the intermediate terms cancel in pairs)

Example1: Sum the series to n terms

)1n(n11

tan211

tan131

tan71

tan31

tan 11111

+++⋅⋅⋅⋅⋅++++ −−−−−

Solution: Here

)1n(n11

tanT 1n ++

= −

)1n(n1n1n

tan 1

++−+= −

)n(tan)1n(tan 11 −− −+=

+−=− −−−

xy1yx

tanytanxtan 111Θ

Now, 1tan2tanT 111

−− −=

2tan3tanT 112

−− −=

3tan4tanT 113

−− −=

...................................

...................................

)n(tan)1n(tanT 11n

−− −+=

Adding, we get n321n TTTTS +⋅⋅⋅⋅⋅+++= 1tan)1n(tan 11 −− −+=

Example 2: Sum the series:

terms n to3sec3sin3sec3sin3secsin 322 ⋅⋅⋅⋅+θθ+θθ+θθ



Solution: We have θθθθ=

θθ=θθ=

3coscoscossin2

.21

3cossin

3secsinT1

( )θθ

θ−θ=θθ

θ=cos3cos

3sin21

cos3cos2sin

21

θθθθ−θθ=

cos3cossin3coscos3sin

21

[ ]θ−θ= tan3tan21

Similarly, [ ]θ−θ= 3tan3tan21

T 22

[ ]θ−θ= 233 3tan3tan

21

T

[ ]θ−θ= 344 3tan3tan

21

T

......................................

......................................

[ ]θ−θ= −1nnn 3tan3tan

21

T

Adding these,we have the required sum

n321n TTTTS +⋅⋅⋅⋅+++=

[ ]θ−θ= tan3tan21 n , other terms cancelling each other.

CHECK YOUR PROGRESS

Q.7: Sum the series to n terms by Difference

Method.

a) 21111

)1n(2

tan162

tan92

tan42

tan+

+⋅⋅⋅⋅+++ −−−−

b) ⋅⋅⋅⋅+αα+αα+αα 22 23

sin2

sin2

3sin

2sinsin3sin

c) ⋅⋅⋅⋅+θθ+θθ+θθ 32

22

222

21

tan21

tan221

tan21

tan221

tantan



13.5 LET US SUM UP

l If θ lies within the closed interval

ππ−

4,

4 , i.e., if 44π≤θ≤π− , then

the series ⋅∞⋅⋅⋅⋅+θ−θ+θ−θ=θ 753 tan71

tan51

tan31

tan is called

Gregory’s series.We also derive important results from Gergory’s

series.

l We discuss important methods for summing up trigonometric series

which may be finite or infinite.There are two important methods for

summation.These are (a) iSC + method (b) The difference method.

13.6 ANSWERS TO CHECK YOUR PROGRESS

Ans. to Q. No. 1: We have ∞⋅⋅⋅⋅+++11.91

7.51

3.11

∞⋅⋅⋅⋅+++=

11.92

7.52

3.12

21

∞⋅⋅⋅⋅+−+−+−=

11.9711

7.557

3.113

21

∞⋅⋅⋅⋅+

−+

−+

−=

11.99

11.911

7.55

7.57

3.11

3.13

21

∞⋅⋅⋅⋅+−+−+−=

111

91

71

51

31

121

1tan21 1−=

π=π=81

41

.21

Ans. to Q. No. 2: From R.H.S,

∞⋅⋅⋅⋅−

++

+−

+= 5533 7

132

51

71

32

31

71

32

⋅⋅⋅−+−+

⋅⋅⋅⋅−+−= 5353 7

1.

51

71

.31

71

71

31

.51

31

.31

31

2



71

tan31

tan2 11 −− += (By Gregory’s series )

( )71

tan

91

1

32

tan 11 −− +−

=

71

tan43

tan 11 −− +=

41tan

71

.43

1

71

43

tan 11 π==

−

+= −−

= The L.H.S

Ans. to Q. No. 3:

R.H.S

.141

because series sGregory,By 41

tan4

ad.inf4.51

4.31

41

4

ad.inf4.51

3.41

-1L.H.S

1

53

42

=

<=

⋅⋅⋅−+−=

⋅⋅⋅−+=

−

Ans to Q No 4 : We have

2

tantan

2cos2

2sin2

tancos1cos1

tan 21

2

2

11-

θ=

θ

θ

=

θ+θ− −− (1)

2 and 0 between lies Sinceπθ

12

tan that so,4

and between lies 21

2 <θπθθ∴ .

2

tantan

2cos2

2sin2

tancos1cos1

tan 21

2

2

11-

θ=

θ

θ

=

θ+θ− −−

2 and 0 between lies Sinceπθ

.12

tan that so,4

and between lies 21

2 <θπθθ∴



Therefore,

θ−

2tantan 21 can be expanded by Gregory’s series.

⋅⋅⋅−

θ+

θ−θ=

θ−

52

32221

2tan

51

2tan

31

2tan

2tantan

⋅⋅⋅θ+θθ= - 2

tan51

2

tan 31

-2

tan 1062 (2)

Hence, from (1) and (2), we have

⋅⋅⋅θ+θθ=

θ+θ−

- 2

tan51

2

tan 31

-2

tan cos1cos1

tan 10621- Hence proved.

Ans. to Q. No. 5: a) α+⋅⋅⋅⋅+α+α+α= nsin3sin2sinsinS Let 2222n

( )

ααα+−=

α

αα+α

−=

α+⋅⋅⋅⋅+α+α+α−=

α−+⋅⋅⋅⋅+α−+α−+α−=

eccosnsin)1ncos(21

2n

22

sin

2)2(n

sin2

n22cos

21

2n

n2cos6cos4cos2cos21

2n

2n2cos1

26cos1

24cos1

22cos1

b) Let

( )β−+α+⋅⋅⋅⋅+β+α+β+α+α= )1n(cos)2(cos)(coscosS 2222n

( ) ( ){ } { })42cos(121

22cos(121

2cos121 β+α++β+α++α+=

{ }α−+α++⋅⋅⋅⋅+ )1n(22cos(121

[ ]α−+α+⋅⋅⋅⋅+β+α+β+α+α+= )1n(22cos()42cos()22cos(2cos21

2n

22

sin

2)2(n

sin2

)1n(222cos

21

2n

β

ββ−+α+α

+=

βββ−+α+= eccosnsin))1n(2cos(21

2n

c) ⋅⋅⋅⋅+α+α+α= 3cos2coscosS Let 222n

( ) ( ) ( ) ( )α++⋅⋅⋅⋅+α++α++α+= n2cos121

6cos121

4cos121

2cos121



[ ]α+⋅⋅⋅⋅+α+α+α+= n2cos6cos4cos2cos21

2n

22

sin

2)2(n

sin2

n22cos

21

2n

α

αα+α

+=

αα+α+=

sinnsin)1n(cos

21

2n

d) Try yourself

α

αα+

−α

αα+

=

⋅⋅⋅⋅+α+α+α=

23

sin

2n3

sin2

)1n(3sin

2sin

2n

sin2

)1n(sin3

41

3sin2sinsinS Let 333n

e) Try yourself

αα=

⋅⋅⋅⋅+α+α+α=

sinn2sin

21

5cos3coscosS Let n

Ans. to Q. No. 6: i) Let ( ) ( ) ∞⋅⋅⋅⋅+β+α+β+α+α= 2cos!2

xcosxcosC

2

and ( ) ( ) ∞⋅⋅⋅⋅+β+α+β+α+α= 2sin!2

xsinxsinS

2

then

( ) ( ) ( )[ ] ( ) ( )[ ] ∞⋅⋅⋅⋅+β+α+β+α+β+α+β+α+α+α=+ 2sini2cos!2

xsinicosxsinicosiSC

2

∞⋅⋅⋅⋅+++= β+αβ+αα )2(i2

)(ii e!2

xxee

∞⋅⋅⋅⋅+++= ββα i2

2ii e

!2x

xe1e

( )βα=ie.xi ee this is an exponential series

)sini(cosxi e.e β+βα=)sinx(icosx e.e β+αβ=

[ ])sinxsin(i)sinxcos(e cosx β+α+β+α= β

Equating real parts on both sides,

we get )sinxcos(.eC cosx β+α= β



ii) Let ∞⋅⋅⋅⋅+α+β−β=αα

α 5cos5

e3cos

3e

coseC53

∞⋅⋅⋅⋅+α+β−β=αα

α 5sin5

e3sin

3e

sineS53

( ) ( ) ( ) ∞⋅⋅⋅⋅+β+β+β+β−β+β=+αα

α 5sini5cos5

e3sini3cos

3e

sinicoseiSC53

∞⋅⋅⋅⋅++−= αα

βα

βα 5i5

3i3

i e5

ee

3e

ee

∞⋅⋅⋅⋅++−=β+αβ+α

β+α

5e

3e

e)i(5)i(3

i

iii) Let( ) ( ) ∞⋅⋅⋅⋅−β+α+β+α−α=

!44sin

!22sin

sinS

( ) ( ) ∞⋅⋅⋅⋅−β+α+β+α−α=!4

4cos!2

2coscosC

( ) ( )[ ] ( )[ ] ∞⋅⋅⋅⋅−β+α+β+α+β+α+β+α−α+α=+!4

)4sin(i4cos!2

)2sin(i2cossinicosiSC

∞⋅⋅⋅⋅−+−=β+αβ+α

α

!4e

!2e

e)4(i)2(i

i

∞⋅⋅⋅⋅−+−=

ββα

!4e

!2e

1e4i2i

i

)ecos(e ii βα= By cosine series

( ) )sinicos(cossinicos β+βα+α=

( )[ ])sinisin()sin(cos)sinicos()cos(cossinicos ββ−ββα+α=

( )[ ])sinh(sini)sin(cos)cosh(sin)cos(cossinicos ββ−ββα+α=

( )[ ])sinh(sin)sin(cosi)cosh(sin)cos(cossinicos ββ−ββα+α=

+ββα−ββα= )sinh(sin)sin(coscosi)cosh(sin)cos(coscos

+ββα−ββα= )sinh(sin)sin(coscosi)cosh(sin)cos(coscos

)cosh(sin)cos(cossini ββα )sinh(sin)sin(cossin ββα+

[ ]+ββα+ββα )sinh(sin)sin(cossin)cosh(sin)cos(coscos

Equating imaginary parts, we get

)sinh(sin)sin(coscos)cosh(sin)cos(cossinS ββα−ββα=

iii) Let ∞⋅⋅⋅⋅+θ+θ+θ+= 3cos!3

x2cos

!2x

cosx1C32

and ∞⋅⋅⋅⋅+θ+θ+θ= 3sin!3

x2sin

!2x

sinxS32



Then ( ) ( ) ( ) ∞⋅⋅⋅⋅+θ+θ+θ+θ+θ+θ+=+ 3sini3cos!3

x2sini2cos

!2x

sinicosx1iSC32

∞⋅⋅⋅⋅++++= θθθ i33

2i2

i e!3

xe

!2x

xe1

( )θ

=ixee By expoential series

( )θ+θ= sinicosxeθθ= sinixcosx e.e

( ) ( )[ ]θ+θ= θ sinxsinisinxcose cosx

( ) ( )θ+θ= θθ sinxsiniesinxcose cosxcosx

Equating the real parts on both sides,we get

( )θ= θ sinxcoseC cosx

iv) Let ∞⋅⋅⋅⋅−θ+θ−θ= 3sinx31

2sinx21

sinxS 32

and ∞⋅⋅⋅⋅−θ+θ−θ= 3cosx31

2cosx21

cosxC 32

Then ( ) ( ) ( ) ∞⋅⋅⋅⋅−θ+θ+θ+θ−θ+θ=+ 3sini3cosx31

2sini2cosx21

sinicosxiSC 32

∞⋅⋅⋅⋅−+−= θθθ i33i22i ex31

ex21

xe

( )θ+= ixe1log By Logarithmic series

( )θ+θ+= sinixcosx1log

( ) ( )θ+

θ+θ+θ+= −

cosx1sinx

tanisinxcosx1log 122

(Separating real and imaginary parts)

Equating imaginary parts, we get

θ+θ= −

cosx1sinx

tanS 1 (except when 1cosx −=θ )

Ans. to Q. No. 7: a) Here 1nn

2tan

)1n(2

tanT 21

21

n ++=

+= −−

)2n(n1n2n

tan)2n(n1

2tan 11

++−+=

++= −−

ntan)2n(tan 11 −− −+=

Putting n...,3,2,1,n = , we have

1tan3tanT 111

−− −=

2tan4tanT 112

−− −=



3tan5tanT 113

−− −=

.............................

.............................

ntan)2n(tanT 11n

−− −+=

Adding, we get the required sum to n terms

n321n TTTTS +⋅⋅⋅⋅+++=

1tan2tan)1n(tan)2n(tan 1111 −−−− −−+++=

b) Here, αα=αα= sin3sin2.21

sin3sinT1 [ ]α−α= 4cos2cos21

23

sin2

sin2.21

23

sin2

sinT2

αα=αα= [ ]α−α= 2coscos21

α−α= cos

21

cos21

T3

......................................

.....................................

α−

α= −− 3n1nn 2

cos2

cos21

T

Adding up the above n relations, we the required sum

n321n TTTTS +⋅⋅⋅⋅+++=

α−

α= − 4cos2

cos21

1n

d) Try yourself .

θ−θ= n

n

2tan2tan sum quiredRe

13.7 FURTHER READING

1) Durell, C. V. and Robson, A; Advanced Trigonometry.

2) J. Smith, Karl; Essentials of Trigonometry.

3) N. Aufmann, Richard, C. Barkar, Vernon & D. Nation, Richard;

College Algebra.



13.8 MODEL QUESTIONS

Q.1: Find the sum of the following series:

a) ( ) ∞⋅⋅⋅⋅+β+α+β+α+α )2sin(!2

xsinxsin

2

b) ∞⋅⋅⋅⋅+αα+αα−αα 3cossin31

2cossin21

cos.sin 32

c) ∞⋅⋅⋅⋅+α+α+α 5sin31

3sin31

sin 2

d) ∞⋅⋅⋅⋅+θ+θ+θ+ 3cosx2cosxcosx1 32

e) ∞⋅⋅⋅⋅++++ x3cos272

x2cos91

xcos31

1

f) terms n to334

tan92

tan31

tan 11 1- ⋅⋅⋅⋅+++ −−

g)3n4

4tan

394

tan194

tan74

tan 2111 1-

++⋅⋅⋅⋅+++ −−−

*** ***** ***



REFERENCES

1) Bali, N. P.; Trigonometry, for B.A/B.Sc. Classes (1st Edition), New

Delhi: Laxmi Publication (P) Ltd.

2) Kishan, Hari (2005); Trigonometry; (1st Edition), New Delhi: Atlantic

Publishers.

3) Lipschntz, Seymour; Linear Algebra: Schum Solved Problems

Series; Tata McGraw Hill.

4) Mapa, S. K.; Higher Algebra (Classical).

5) Ray, M. & Sharma, H. S.; A Text Book of Higher Algebra.

6) Vasistha, A.R. & Sharma, S. K.; Trigonometry.

7) Vasistha, A. R.; Matrices; Meerut: Krishna Prakashan Mandir.

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