question bank - answers semester: iii ma 2211 - … unit... · find the root mean square value of...
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MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
1 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI – 621213
QUESTION BANK - ANSWERS
SEMESTER: III
MA 2211 - TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS UNIT – I: FOURIER SERIES
PART-A
State Dirichlet's conditions on Fourier series
( ). ( ) is single valued ,finite and periodic function in ( , 2 )( ). ( ) is continuous or piece-wii f x c cii f x
Question :1 AUC A / M 2013, 2010 N / D 2011, N / D 2011
Answer :
se continuous with finite number of finite discontinuities in (c,c+2π)( ). ( ) has a finite number of maxima (or) minima in (c,c+2π).
What is meant by Harmonic analysis
iii f xQuestion : 2 AUC A / M 2013
Answ
0
The process of finding the Fourier series for a given function by numerical value is known as harmonic analisis. In harmonic analisis the Fourier coefficients , and of the functn na a b
er :
0
ion ( ) in (0, 2 ) given by
2 ,
cos2 ,
sin2
where 'n' is the number of terms appears in the table.
n
n
y f x
ya
n
y nxa
n
y nxb
n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
2 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
20
0 0
0
,
Find the constant term in the expansion of cos as a Fourier series in the interval ,
2 2 1 cos 2cos2
1 1 sin 21 cos 2
x
xa xdx dx
x dx x
Question : 3 : AUC A / M 2012 N / D 2010
Answer :
0
0
12
1Constant term 2 2
Define Root Mean Square value of a function ( ), over the interval ( , )
The root mean square value of ( ) over the inte
x
a
f x a b
f x
Question : 4 : AUC A / M 2012 N / D 2012
Answer :
2
rval ( , )is defined as
( )R.M.S
Give the expression for the Fourier series co-efficient for .the function ( )over the interval ( 2, 2)
2 ( )sin
b
a
n
n
a b
f x dx
b a
b f x
nb f xl
Question : 5 : AUC A / M 2011
Answer :
2 2
0 0 0
20
22 20
1
2 ( )sin ( )sin2 2 2
Without finding the values of , and , for ( ) in the interval (0, )
find the value of ( )2
l
n n
n nn
x n x n xdx f x dx f x dxl
a a b f x x
a a b
Question : 6 : AUC A / M 2011
An
22 20
10
222 20
1 0
2 22 2 20
1 0
1 1We know that ( ) ..............(1)4 2
2(1) 2 ( ) ( )2
2( )2
nn
n nn
n nn
af x dx a
a a b f x dx
a a b x dx
swer :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
3 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 5 52 2 4 50
1 0 0
2 2 2 2( ) 02 5 5 5
Find the coefficient of the Fourier series for the function sin in ( 2,2)
Since the function
n nn
n
a xa b x dx
b x x
Question : 7 : AUC N / D 2012
Answer :
3 32 2
0 0
( ) sin even 0
Find the root mean square value of the function ( ) in (0, )
( )3 3R.M.S
0 0 0 3
n
lb l
a
f x x x b
f x x l
x lf x dx x dxl
b a l l l
Question : 8 : AUC N / D 2011
Answer :
Question : 9 : AUC N / D 2010A
2
5 522 22
0 0
Find the root mean square value of the function ( ) in (0, )
( )5 5R.M.S
0 0 0 5
Obtain the first term of the
lb l
a
f x x l
x lf x dx x dxl
b a l l l
nswer :
Question :10 : AUC N / D 20092
32
00 0 0
23
Fourier series for the function ( ) ,
2 2 2Since ( ) is even, ( )3
2 203 3
f x x x
xf x a f x dx x dx
Answer :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
4 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
2
Find the Fourier series expansion for in the interval
Since ( ) is neither even nor odd the Fourier series expansionin the interval is given
x x x
f x x xx
PART B Question :1 AUC N / D 2012
Answer : .
0
1 1
0
2 32
0
2 3 2
by
( ) cos sin .....................(1)2
1 1Where ( ) , ( )cos
1 ( )sin
1 1 1Now ( ) ( )2 3
1 ( )2 3 2
n nn n
n
n
af x a nx b nx
a f x dx a f x nxdx
b f x nxdx
x xa f x dx x x dx
3
2 3 2 3 3 3
32 2
0
2
22
( )3
1 12 3 2 3 3 3
1 2 2 2 .....................(2)3 3 3
1 1( ) cos ( ) cos
1 sin cos sin( ) (1 2 ) (2)
n
a
a f x nxdx x x nxdx
nx nx nxx x xn n n
3
2
2
2
2
2
1 cos(1 2 ) since sinterms 0
1 (1 2 )cos
1 (1 2 )cos (1 2 )cos ( )
1 (1 2 )cos (1 2 )cos since cos( ) cos
1 cos 1 2 1 2
nxxn
x nxn
n nn
n nn
nn
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
5 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 2 2
2
22 3
23
4 4 4( 1) ( 1) ( 1) ............(3)
1 1( )sin ( )sin
1 cos sin cos( ) (1 2 ) (2)
1 2cos cos( )
1 2co
n n nn
n
an n n
b f x nxdx x x nxdx
nx nx nxx x xn n n
nx nxx xn n
2 23 3
2 23 3
2 2 2 2
s cos 2cos ( ) cos ( )( ) ( ( ) )
1 2cos cos 2cos cos( ) ( )
1 cos 1 cos( ) ( )
n n n nn n n n
n n n nn n n nn n
n n
2
02
1 1 1 1
2
21
)
1 ( 1) 2 22 ( 1) ( 1) ..................(4)
Use (2), (3), and (4) in (1) we get 2
4 23( ) cos sin ( 1) cos ( 1) sin2 2
( 1)4 cos3
nn n
n
n nn n
n n n n
n
n
bn n n
af x a nx b nx nx nxn n
nxn
1
0
1
00
( 1)2 sin
Find the half range Fourier cosine series expansion of ( ) sin in 0 .
( ) cos .......................(1)2
2 2 2( ) sin ( co
n
n
nn
nxn
f x x x x
af x a nx
a f x dx x xdx x
Question : 2 AUC N / D 2011
Answer :
00
0 0
0
s ) (1)( sin )
2 2 2cos sin cos cos 0
2 ( 1) 2, 2 .............................(2)
x x
x x x x x
a
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
6 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
0 0 0
0 0
1 20 0
1
2 2 2( )cos sin cos cos sin
2 2 1cos sin sin( 1) sin( 1)2
1 1sin( 1) sin( 1) I I ........................(*)
cosI sin( 1)
na f x nxdx x x nxdx x nx xdx
x nx xdx x n x n x dx
x n xdx x n xdx
x n xdx x
0 0
0
1
12
1 11 2
1 1
( 1) sin( 1)(1)( 1) ( 1
1 cos( 1)( 1)
1 cos( 1) 0 1( 1) ( 1)
Similarlly I 1( 1)
I I 1 1( 1) ( 1)
1 1 Since 1( 1) ( 1)
n
n
n n
n n n
n x n xn n
x n xn
nn n
n
n n
n n
1 1
1
1 12
11 22 2 2
1
1 11( 1) ( 1)
1 1 21 1( 1)( 1) 1
2 2 ( 1) 2 ( 1)( 1) ( 1) I I ...........( )1 1 1
Use (a) in (*) we get
1
n
n
n n
n nn
n
n n
n nn n n
an n n
a
2 2 2
0
10 0
0 0
2 ( 1) 2( 1) 2( 1) , 1 ....................(3)1 1 1
2 cos sin
2 2 1sin cos 2sin cos2
1 1 cos 2 sin 2sin 2 (1)2 4
n n n
n
n
a nn n n
a x nx xdx
a x x xdx x x x dx
x xx xdx x
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
7 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
0
1
0
1
01
2
2
1 1 1 1cos 2 cos 2 0 (1)2 2 2 2
1 ....................................(4)2
Use (2),(3) and (4) in (1) we get
( ) cos2
cos cos22 1 2( 1)cos cos2 2 1
nn
nn
n
n
x x
a
af x a nx
a a x a nx
x nxn
2
22
2
4
4 4 4
2
1 ( 1)1 cos 2 cos2 1
Find the Fourier series expansion for in the interval and hence prove that1 1 11 2 3 90
Since ( ) is even 0 and t
n
n
n
x nxn
x x
f x x b
Question : 3 AUC M / J 2013
Answer :
0
1
00 0
32
00 0 0
3
0
he Fourier series expansion in the interval is given by
( ) cos .......................(1)2
2 2Where ( ) , ( ) cos
2 2 2( )3
23
nn
n
xaf x a nx
a f x dx a f x nxdx
xa f x dx x dx
x
2
3
20
0
2
0
2 203 3
2 .........................(2)3
2 ( )cos
2 cos
n
a
a f x nxdx
x nxdx
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
8 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
02
22 2 0 0
0
2 2 2
2
2 sin cos sin(2 ) (2)
2 2 cos 4 cos since sinterms 0
4 4 4cos 0 ( 1) ( 1)
4 ( 1) .................................
n n
nn
nx nx nxx xn n n n n n
x nx x nxn n
nn n n
an
2
21
22 2 20
1
22
0 2
.(3)
Use (2) and (3) in (1) we get 2
43( ) ( 1) cos2
The Parseval's identity of ( ) in is given by
1 1( ) ( ).2 4 2
2 4( 1)Put , , 0 , ( ) , we get3
1
n
n
n nn
n
n n
f x nxn
f x x
af x dx a b
a a b f x xn
22
222
21
42
44
1
5 4 2
41
45
4 4 4
45 5
4
23 1 4( 1) 0)
2 4 2
41 1 16( 1)9 .
2 4 2
1 ( 1)8 .2 5 9
1 1 1 1810 9 1 2 3
1 1( ) 810 9 1
n
n
n
n
n
n
x dxn
x dxn
xn
x
4 4
5 4
4 4 4
4 4
4 4 4
1 12 3
2 1 1 1810 9 1 2 3
1 1 185 9 1 2 3
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
9 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
4 4
4 4 4
4 4 4
4 4 4 4 4 4
4
4 4 4
4
4 4 4
1 1 181 2 3 5 91 1 1 9 5 1 1 1 48 81 2 3 45 1 2 3 45
1 1 1 41 2 3 45 81 1 11 2 3 90
Find the Fourier series to represent t
Question : 4 : AUC M / J 2012
2
21
0
1
00 0
2
00
he function ( ) , and
1hence deduce (2 1) 8
Since ( ) is even, 0
( ) cos ..............(1)2
2 2( ) , ( )cos
2 2 2( )2
n
n
nn
n
f x x x
n
f x baf x a nx
a f x dx a f x nxdx
xa f x dx xdx
Answer :
0 0
0 0
2 2( ) cos cos
2 sin
na f x nxdx x nxdx
nxxn
0
2 0
2 2 2
2
21
cos(1)
2 cos
2 1 4cos cos0 ( 1) 1 , is odd
4 , is odd ..................................(3)
use (2) and (3) in (1) we get4( ) cos ...
2
n
n
n
nxn n
nxn
n nn n n
a nn
f x nxn
...........(1)
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
10 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
21
21
21
21
2
21
4 1( ) cos 2 1 .............(1)2 2 1
Put 0 in (1) we getL.H.S of (1) (0) 0
4 1R.H.S of (1)2 2 1
R.H.S L.H.S implies4 1 0
2 2 1
4 122 1
12 4 82 1
n
n
n
n
n
f x n xn
xf
n
n
n
n
Ques 2
0
1 1
2 2 32
00 0 0
Find the Fourier series expansion for ( ) ( ) in (0, 2 ).
( ) cos sin .............(1)2
1 1 1 ( )( ) ( )3( 1)
n nn n
f x x
af x a nx b nx
xa f x dx x dx
tion : 5 : AUC M / J 2012
Answer :
2
23 3 3
0
3 3
3 2
20
2
02
2
0
2
1 1( ) ( 2 ) ( 0)3 31 ( )
31 22
3 32 .........................(2)3
1 ( ) cos .
1 ( ) cos .
1 sin cos( ) 2( )( 1)
n
x
a
a f x nxdx
x nxdx
nx nxx xn n n
2
0
sin(2)( 1)( 1)
nxn n n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
11 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
22 2
2 2 0 00
2 2
2
2
2
2
0
1 2( ) cos 2 ( ) cos since sin terms 0
2 2( 2 )cos 2 ( 0) cos 0
2 2
1 4 since cos 2 1
4 ..................................(3)
1 ( )sin
n
n
x nx x nxn n
nn n
n
nn
an
b f x nxdx
22
0
2
22
30
2 23 3
1 ( ) sin
1 cos sin cos( ) 2( )( 1) (2)( 1)( 1)
1 2cos cos( )
1 2cos 2 cos 2 2cos 0 cos0( 2 ) ( 0)
x nxdx
nx nx nxx xn n n n n n
nx nxxn n
n nn n n n
2
0
22
3 3
2 2
3 3
0
1 1
since sin terms 0
1 2 1 2( )
1 2 2 0
0 ........................(4)
use (2),(3), and (4) in (1) we get
( ) cos sin2
n
n nn n
n n n n
n n n n
b
af x a nx b nx
2
21 1
2
21
243 cos 0sin
214 cos
3
n n
n
nx nxn
nxn
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
12 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
0
1
32
00 0 0
3
Find the half range Fourier cosine series of ( ) ( ) in (0, )
( ) cos .............(1)2
2 2 2 ( )( ) ( )3( 1)
2 ( )3
nn
f x x
af x a nx
xa f x dx x dx
x
Question : 6 : AUC M / J 2012
Answer :
3 3
0
3
3 2
20
0
2
0
2
2 ( ) ( 0)3
2 031 22
3 32 .........................(2)3
2 ( )cos .
2 ( ) cos .
2 sin cos sin( ) 2( )( 1) (2)( 1)( 1)
n
a
a f x nxdx
x nxdx
nx nx nxx xn n n n
2
0
22 2 0 0
0
2 2 2
0
2 2( )cos 4 ( ) cos since sin terms 0
4 4 40 ( 0) cos0 ..................................(3)
Use (2) and (3) in (1) we get
( )2
n
n n
x nx x nxn n
an n n
af x a
1
22 20
10
4
44
10
5 4
41
0
cos ..............(1)
1 1( )4 24
1 1 1694 2
1 185 1 9
nn
nn
n
n
nx
af x dx a
x dxn
xn
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
13 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
4 4
41
4 4 4
41
4 4
41
189 5
1 485 9 45
1 4 145 8 90
Find the Fourier series expansion of ( ) sin in .
Since ( ) sin odd×odd even , 0 and hence the Fourie
n
n
n
n
n
n
n
f x x x x
f x x x b
Question : 7 : AUC N / D 2011
Answer :
0
1
0 00 0
0 0
0
rseries expansion is given by
( ) cos .......................(1)2
2 2 2( ) sin ( cos ) (1)( sin )
2 2 2cos sin cos cos 0
2 ( 1) 2, 2 ...................
nn
af x a nx
a f x dx x xdx x x x
x x x x x
a
0 0 0
0 0
1 20 0
1
..........(2)
2 2 2( ) cos sin cos cos sin
2 2 1cos sin sin( 1) sin( 1)2
1 1sin( 1) sin( 1) I I ........................(*)
I sin(
na f x nxdx x x nxdx x nx xdx
x nx xdx x n x n x dx
x n xdx x n xdx
x
0 0
0
1
12
cos( 1) sin( 1)1) (1)( 1) ( 1
1 cos( 1)( 1)
1 cos( 1) 0 1( 1) ( 1)
Similarlly I 1( 1)
n
n
n x n xn xdx xn n
x n xn
nn n
n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
14 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
1 11 2
1 1 1 1
1
1 12
11 22 2
I I 1 1( 1) ( 1)
1 1 Since 1 1( 1) ( 1)
1 11( 1) ( 1)
1 1 21 1( 1)( 1) 1
2 2 ( 1)( 1) ( 1) I I1 1
n n
n n n n
n
n n
nn
n n
n n
n n
n nn n n
n n
2
2 2 2
0
1
2 ( 1) ...........( )1
Use (a) in (*) we get
1 2 ( 1) 2( 1) 2( 1) , 1 ....................(3)1 1 1
Use (2),(3) and (4) in (1) we get
( ) cos2
n
n n n
n n
nn
an
a a nn n n
af x a nx
01
2
22
22
2
0
1
cos cos22 1 2( 1)cos cos2 2 1
1 ( 1)1 cos 2 cos2 1
Find the Fourier series for ( ) 2 in 0 3.
Here 2 332
( ) cos2
nn
n
nn
n
n nn
a a x a nx
x nxn
x nxn
f x x x x
l
l
a n xf x a bl
Question : 8 : AUC N / D 2011
Answer :
1
0
1 1
0
1 1
sin .
( ) cos sin .3 322 2
2 2( ) cos sin ...............(1)2 3 3
n
n nn n
n nn n
n xl
a n x n xf x a b
a n x n xf x a b
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
15 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
32 3 2 3
20
0 0 0
332
0
0
2 32
0 0
2
1 2 2( ) 2 23 3 3 2 32
2 2 27 29 (9 9) 03 3 3 3 3
0 .....................(2)
1 2 2 2( )sin (2 )sin3 3 3
2cos2 3(2 ) ( 223
3
l
l
n
x xa f x dx x x dx
xx
a
n x n xb f x dx x x dxl
n x
x x n
2 2
2sin32 )
49
n x
xn
3
3 3
03
23 3
0
3 3 3 3
3 3
2cos3( 2)
1681
2 3 2 81 2(2 ) cos 2cos3 2 3 16 3
2 3 81 813 cos 6 cos 6 0 (cos 0)3 2 8 8
2 9 813 2 8
n x
n
n x n xx xn n
n nn n n
n n
3 3
818n
2 32
0 0
22 2 3 3
3
3 ...................(3)
1 2 2 2( ) cos (2 ) cos3 3 3
2 2 2sin cos sin2 3 3 3(2 ) (2 2 ) ( 2)2 4 1633 9 81
n
l
n
n
bn
n x n xa f x dx x x dxl
n x n x n x
x x xn n n
3
0
3
3
2 2 2 20
0
2 2
2cos2 2 9 23(2 2 ) cos43 3 4 3
93 cos 2 cos 0 0
20 ...................(4)n
n xn xx
n n
nna
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
16 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
0
1 1
1
Use (2), (3), and (4) in (1) we get
( ) cos sin2
3 sin
, 0 1Find the Fourier series expansion of ( ) . Hence show that
2 , 1 2
1
n nn n
n
af x a n x b n x
n xn
x xf x
x x
Question : 9 : AUC N / D 2012
2
2 2 2
0
1 1
0
1 1
2 1 2
00 0 1
11 2 2
0 1 0
1 1 ...... .1 3 5 8
Here 2 2 1
( ) cos sin2
( ) cos sin2
1 1( ) ( ) ( )1̀
(2 ) 22
n nn n
n nn n
l
l la n x n xf x a b
l laf x a n x b n x
a f x dx f x dx f x dxl
xxdx x dx x
Answer :
22
1
2
0
2 1 2
0 0 1
1 2
1
10
2
1 1 4 1 1 10 2(2) 2 2 2 02 2 2 2 2
0 .................(2)
1 1( ) cos cos ( ) cos 1
I I .
siI cos
l
n
x
l
a
n xa f x dx x n xdx l x n xdxl l
x n xdx x
n n xn
1
2 20
1
2 20
cos(1)
1 cos
n xn
n xn
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
17 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 2
2 2 2 2
2
21
1 cos cos 0
1 1( 1) 1 1 ( 1)
sinI (2 ) cos (2 )
n n
nn
n n
n xx n xdx xn
2
2 21
22 2 1
2 2
2 2 2 2
1 2 2 2
2 2
cos( 1)
1 cos
1 cos 2 cos
1 11 ( 1) 1 ( 1)
1 2I I 1 ( 1)
4 , if is odd.
0 , if is even
n n
nn
n xn
n xn
n nn
n n
al n
nn
n
2 1 2
0 0 1
1 2
0 1
1 2
1
1 2 20
.....................(3)
1 ( )sin ( )sin ( )sin
sin (2 )sin
I I
cos sinI sin (1)
l
nn xb f x dx f x n xdx f x n xdx
l l
x n xdx x n xdx
n x n xx n xdx xn n
1
0
1
0
2
2 2 21
1 1cos cos 0
( 1)
cos sinI (2 )sin (2 ) ( 1)
n
x n x nn n
n
n x n xx n xdx xn n
2
12
2
1
1 1(2 ) cos cos 2 0 lx n x nn n n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
18 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
1 2
0
1 1
2 2odd od
( 1) 1I I
1 11 ( 1) 1 ( 1)
2 , if is odd....................(4)
0, if is even
( ) cos sin2
0 4 2cos sin2
n
n
n n
n nn n
n n
bn n
n n
nn
naf x a n x b n x
n xn n
d
2 21 1
2 21
2
4 1 2 1( ) cos(2 1) sin(2 1) ............(*)(2 1) (2 1)
Put 0 in the above series we get. . of * (0) 2
4 1R.H.S of (*) cos 0(2 1)
R.H.S of (*) . . of *4 1
n n
n
n x
f x n x n xn n
xL H S f
nL H S
21
2
21
0
1
00
cos 0 2(2 1)1
(2 1) 8
Obtain the half range Fourier cosine series for ( ) in (0, )
( ) cos ..........................(1)2
2 2( )
n
n
nn
n
n
f x x
af x a nx
a f x dx xd
Question :10 : AUC N / D 2012, 2010
Answer :
22
00 0
0 0
20
2 2 20
2 1 0 ......................(2)2
2 2( ) cos cos
2 sin cos(1)
2 2 2cos cos cos 0 ( 1) 1
n
n
xx a
a f x nxdx x nxdx
nx nxxn n
nx nn n n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
19 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
2
0
1
2odd
21
2 1 ( 1)
4 if is odd...................(3)
=0,if is evenUse (2) and (3) in (1) we get
( ) cos2
4 cos2
4 1 cos(2 1)2 (2 1)
4 1Hence ( )2 (2
n
nn
n
n
n
nn
n
af x a nx
nxn
n xn
f xn
21
2
2 2 2
0
1 1
cos(2 1)1)
, 0Find the Fourier series expansion for ( ) .Hence show that
2 , 2
1 1 1 ...... .1 3 5 8
( ) cos sin .............2
n
n nn n
n x
x xf x
x x
af x a nx b nx
Question :11: AUC N / D 2010
Answer :
2 2
00 0
2
0
22 2
0
2 2 2
2 2 22 2
(1)
1 1( ) ( ) ( )
1 (2 )
1 22 2
1 (2 )0 2 (2 ) 2 ( )2 2 2
1 44 22 2 2
a f x dx f x dx f x dx
xdx x dx
x xx
2
0
1
......................(2)a
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
20 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
0
2
0
2
1 20
10 0
2 2 00
1 ( ) cos
1 ( )cos ( ) cos
1 1cos (2 ) cos I I ..........(*)
sin cosI cos (1)
cos 1 cos
na f x nxdx
f x nxdx f x nxdx
x nxdx x nxdx
nx nxx nxdx xn n n
nx nxn n
2
2 2
2
2
2
2
22 2
2
1 cos cos 0
1 1( 1) 1 1 ( 1) ...........................( )
I (2 )cos
sin cos(2 ) ( 1)
1 1cos cos 2 cos
1 1 ( 1) ............
n n
n
nn
an n
x nxdx
nx nxxn n
nx n nn n
n
2 2
2
2
2
0
........( )
Use (a) and (b) in * we get 1 1 11 ( 1) 1 ( 1)
1 2 1 ( 1)
4 if is odd................................(3)
0 if is even
1 ( )sin
1 (
n nn
n
n
b
an n
n
nn
n
b f x nxdx
f
2
0
2
0
)sin ( )sin
1 sin (2 )sin
x nxdx f x nxdx
x nxdx x nxdx
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
21 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
1 2
10
0 0
200
0
1 I I ..............................(A)
I sin .
cos sinsin (1)
cos sinsince 0
1 1cos cos 0
1 ( 1)n
x nxdx
nx nxx nxdx xn n n
nx nxxn n
x nx nn n
n
1
2
2
2
2
2
( 1)I .............................( )
I (2 )sin
cos sin(2 ) ( 1)
1 (2 ) cos
1 (2 2 ) cos 2 (2 ) cos
1 cos
( 1) ...................
n
n
cn
x nxdx
nx nxxn n
x nxn
n nn
nn
n
1 2
0
1
................( )
use (c) and (d) in (A) we get
1 1 ( 1) ( 1)I I
0 ....................................(4)
Use (2) ,(3) and (4) in (1) we get
( ) cos2
n n
n
n
nn
d
bn n
b
af x a nx
1
2 21,3,5 1 1
21
sin
4 4 1cos 0sin cos(2 1)2 2 (2 1)
4 1( ) cos(2 1)2 (2 1)
nn
n n n
n
b nx
nx nx n xn n
f x n xn
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
22 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
21
21
21
21
2
4 1( ) cos(2 1) .............(*)2 (2 1)
Put 0 we get L.H.S of (*) 04 1R.H.S of (*) cos0
2 (2 1)R.H.S of (*) L.H.S of (*)
4 1 02 (2 1)
4 1(2 1) 21
(2 1)
n
n
n
n
f x n xn
x
n
n
n
n
2
1
2
2 2 2
2
2 4 8
1 1 1Hence ......1 3 5 8
Find the Fourier series expansion for ( ) 1 in the interval ( 1,1)
Here 1Since ( ) is even the Fourier series expa
n
f x x
lf x
Question :12 : AUC N / D 2010
Answer :
0
1
00
11 32
0 00 0
1
0 0
nsion is
given by ( ) cos .............(1) , 2
2where ( ) ,
1 4 42 1 2 2 1 .....................(2)3 3 3 3
2 ( ) cos 2 ( ) cos 2
nn
l
l
n
af x a n x
a f x dxl
xa x dx x a
n xa f x dx f x n xdxl l
12
0
2
1 cos
sin2 1
x n xdx
n xxn
2 2 3 3
cos sin( 2 ) ( 2)n x n xxn n
1
0
12 2 2 2 2 20
2 2
4 4 4cos cos 0 ( 1)
4Hence ( 1) ...............(3)
n
nn
x n x nn n n
an
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
23 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
0
1
2 21 1
2
Use (2) and (3) in (1) we get
( ) cos .............(1)2
4 2 43 cos ( 1) cos2 3
Find the Fourier series expansion for ( ) in the
nn
nn
n n
af x a n x
a n x n xn
f x x x
Question :13 : AUC N / D 2009
0
1 1
0
0
interval
Since ( ) is neither even nor odd , the Fourier series expansion is
given by ( ) cos sin ..............(1) , 2
1where ( ) ,
1 ( )
n nn n
l
l
l x l
f xa n x n xf x a b
l l
a f x dxl
a f x dxl
Answer :
2 32
2 3 2 3 2 3 2 3 3
3
0
2
2
1 1( )2 3
1 1 22 3 2 3 2 3 2 3 3
2 .....................(2)3
1 1( )cos ( ) cos .
sin1 ( )
ll l
l l l
l l
nl l
x xx x dxl l
l l l l l l l l ll l
la
n x n xa f x dx x x dxl l l l
n xlx x nl
l
2 2 3 3
2 3
cos sin(1 2 ) ( 2)
n x n xl lx
n nl l
2
2 2
2 2
2
2 2 2 2
2
2 2
1 (1 2 ) cos
(1 2 ) cos (1 2 ) cos since cos( ) cos
( 1) 4 ( 1)1 2 1 2
4 ( 1) .................(3)
l
ll
l
n n
n
n
l n xxl n l
l l n l nnl ll ln n
lan
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
24 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
2
0
2 2
20
1 1( )sin ( )sin .
1 sin sin
1 sin
2 sin
cos sin2 (1)
l l
nl l
l l
l l
l
l
l
l
n x n xb f x dx x x dxl l l l
n x n xx dx x dxl l l
n xx dxl l
n xx dxl l
n x n xl lx n nl
l l
0
0
1 1
2 2
2 2
2 cos
2 2cos 0 ( 1)
2 ( 1) ..........................(4)
Use (2) , (3) and (4) in (1) we get
( ) cos sin2
4 ( 1) cos3
l
n
n
n
n nn n
n
n
l n xxl n l
ll nn n
lbn
a n x n xf x a bl l
l l n xn l
1 1
2 2
2 21 1
2
2 ( 1) sin
4 ( 1) 2 ( 1)cos sin3
Find the half range Fourier cosine series expansion for in the interval 0 and hence
deduce
n
n
n n
n n
l n xn l
l l n x l n xn l n l
x x
Question :14 : AUC M / J 2013
4
4 4
0
1
00 0
1 1that .....1 2 90
( ) cos .......................(1)2
2 2Where ( ) , ( ) cos
nn
n
af x a nx
a f x dx a f x nxdx
Answer :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
25 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
32
00 0 0
233
0
20
0
2
0
2
2 2 2( )3
2 2 203 3 3
2 .........................(2)3
2 ( ) cos
2 cos
2 sin cos sin(2 ) (2)
n
xa f x dx x dx
x
a
a f x nxdx
x nxdx
nx nx nxx xn n n n n n
02
22 2 0 0
0
2 2 2
2
2
21
2 2 cos 4 cos since sinterms 0
4 4 4cos 0 ( 1) ( 1)
4 ( 1) ..................................(3)
Use (2) and (3) in (1) we get 2
43( ) ( 1) cos2
n n
nn
n
n
x nx x nxn n
nn n n
an
f x nxn
2
21
2
21
22 20
10
22
0 2
22
222
0
( 1)4 cos3
( 1)( ) 4 cos3
By parseval's identity we have
1 1( ) ( ).4 2
2 4( 1)Put , , ( ) , we get3
231 1 4( 1)4 2
n
n
n
n
nn
n
n
n
nxn
f x nxn
af x dx a
a a f x xn
x dxn
2
1
0)n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
26 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
42
44
10
5 4 2
410
45
4 4 40
45
4 4 4
4 4
4 4 4
4 4
4 4 4
41 1 16( 1)9 .
4 2
1 ( 1)8 .5 9
1 1 1 185 9 1 2 31 1 1 10 8
5 9 1 2 31 1 18
5 9 1 2 31 1 18
5 9 1 2 3
n
n
n
n
x dxn
xn
x
4 4
4 4 4
4 4 4
4 4 4 4 4 4
4
4 4 4
4
4 4 4
1 1 181 2 3 5 91 1 1 9 5 1 1 1 48 81 2 3 45 1 2 3 45
1 1 1 41 2 3 45 81 1 11 2 3 90
Find the half range Fourier co
Question :15 : AUC M / J 20132
2
2 2
0
1
0
1
00
sine series expansion for ( 1) in the interval 0 1 and hence1 1deduce that .....1 2 6
Here 1
( ) cos .........(1)2
( ) cos .........(1)2
2= ( ) 2
nn
nn
l
x x
la n xf x a
laf x a n x
a f x dx xl
Answer :
131
2
0 0
0
11 2
3
1 2 22 0 .................(2)3 3 3
xdx
a
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
27 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
12
0 0
2
2 ( ) cos 2 1 cos
sin2 1
l
nn xa f x dx x n xdx
l l
n xxn
2 2 3 3
cos sin2( 1) 2n x n xxn n
1
0
12 2 2 2 2 20
2 2
2 21,3,5....
2 21
4 4 4( 1) cos 0 cos 0
4 ................(3)
Use (2) and (3) in (1) we get2 43( ) cos2
1 4 1( ) cos ..............(*)3
Put 0 in
n
n
n
x n xn n n
an
f x n xn
f x n xn
x
2 21
2 2
2 21
2 21
2 2
21
(*) we get1 4 1R.H.S of (*) cos 03
(0) (1) (0 1) (1 1) 1L.H.S of (*) (0)2 2 2
R.H.S of (*) L.H.S of (*)1 4 1 13 24 1 1 1 3 2 1
2 3 6 61 1
6 4 24
n
n
n
n
nf ff
n
n
n
Question :16 : AUC M / J 2
0
1
2
00 0
2
, 02Find the half range Fourier cosine series for ( )
( ),2
( ) sin ................(1)2
2 2( ) ( )
nn
ll l
l
lkx xf x
lk l x x l
a n xf x al
a f x dx kx dx k l x dxl l
013
Answer :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
28 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 22 2
0 022
2 2 2 2 2 2 2 22
2
0
0
2
0
2 2( )2 2
2 28 2 2 8 8 2 2 8
2 28 2 2
2 ( )cos
2 ( ) cos
l l ll
ll
l
n
l
k k x xx dx l x dx lxl l
k l l l l k l l l lll l
k l kl klal
n xa f x dxl l
f xl
2
2
1 20
2
2
2
1 2 20
20
2
2 2
( ) cos
2 2cos ( ) cos I I
sin cosI cos (1)
cos sin
l
l
ll
l
l
l
n x n xdx f x dxl l
n x n x kkx dx k l x dxl l l l
n x n xn x l lx dx x n nl
l l
l n x l nxn l n
2
0
2 2 2
2 2 2 2
2 2 2
2 2 2 2
2
2
2 2
2
2
cos sin cos 0 02 2 2
cos sin2 2 2
I ( ) cos
sin cos( ) ( 1)
l
l
l
l
xl
l n l n ln n n
l n l n ln n n
n xl x dxl
n x n xl ll x n n
l l
l
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
29 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
2 2
2
2 2
2 2 2 2
2 2 2
2 2 2 2
2 2
1 2 2 2
( )sin cos
0 cos sin sin2 2 2
cos sin cos2 2 2
I I cos sin2 2 2
l
l
l n x l n xl xn l n l
l l l n l nnn n n
l l n l nnn n n
l n l nn n
2 2 2
2 2 2 2 cos sin2 2
l l l nnn n n
2
2 2
2 2 2 2
2 2 2 2 2 2 2 2
2 2
2 2 2 2
2
2 2
2
2 2 2 2
cos2
2 2cos 1 cos cos 1 ( 1)2 2
2 2cos if is even, 0 otherwise2
2 cos 12
2 2 4cos 1 cos 12 2
n
n
l nn
l n l l n lnn n n n
l n l nn n
l nn
k l n kl nal n n
2 2
.
4 cos 1 ..................(3)2
Use (2) and (3) in (1) we get the result
nkl na
n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
30 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 2
2 20
0 0
0
0
sin cos
cos2 2 2sin sin
2 2cos cos cos 0
2 2( 1) 1 1 ( 1)
4 , if is odd
0 if is even
l
l
l l
l
n n
n
l n x l n xxn l n l
n xn x k n x k lk dx dx nl l l l l
l
k l n x k nl n l nk k
n nk n
b nn
,
1
odd 1
..................(2)
Use (2) in (1) we get
( ) sin
4 4 1 (2 1)sin sin(2 1)
, 02Obtain the sine series for ( )
( ),2
nn
n n
n xf x bl
k n x k n xn l n l
lx xf x
ll x x l
Question :17 : AUC M / J 2011
1
0
2
02
2
1 20
2
( ) sin
2 ( )sin
2 ( )sin ( )sin
2 2sin ( )sin I I
nn
l
n
ll
l
ll
l
n xf x bl
n xb f x dxl l
n x n xf x dx f x dxl l l
n x n xx dx l x dxl l l l
Answer :
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
31 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2
2
1 2 20
20
2 2
2 20
2 2
2 20
2 2 2
2 2 2 2
cos sinI sin (1)
sin cos
sin cos
sin cos sin 02 2 2
l
l
l
l
n x n xn x l lx dx x n nl
l l
l n x l n xxn l n l
l n x l n xxn l n l
l n l n ln n n
2 2
2 2
2
2
2 2
2
2
2
2 2
2
0
sin cos2 2 2
I ( )sin
cos sin( ) ( 1)
( ) cos sin
0 0 cos2 2
l
l
l
l
l
l
l n l nn n
n xl x dxl
n x n xl ll x n n
l l
l n x l n xl xn l n l
l l n ln
2
2 2
2 2
2 2
2 2 2 2
1 2 2 2 2 2
2
2 2
2
1 2 2 2
2 2 2 2
sin2
cos sin2 2 2
I I sin cos cos sin2 2 2 2 2 2
2 sin .2
2 2 2I I sin2
4 4sin sin ........(2)2 2
Use (2) i
n
n
nn
l n l nn n
l n l n l n l nn n n n
l nn
l nbl l n
l n l nbn n
n (1) we get
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
32 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 21 1
2 21
2
0
1 1
0
4( ) sin sin sin2
4 1 sin sin2
Find the Fourier series for ( ) 2 in 0 2.
Here 2 21
( ) cos sin .2
nn n
n
n nn n
n x l n n xf x bl n l
l n n xn l
f x x x x
ll
a n x n xf x a bl l
a
Question :18 : AUC M / J 2010
Answer :
1 122 2 2 3
20
0 0 023
2
0
0
2 22
0 0
2
cos sin ..............(1)2
1 ( ) 2 22 3
8 12 8 443 3 3 3
4 .....................(2)3
1 ( )sin 2 sin
c2
n nn n
l
l
n
a n x b n x
x xa f x dx x x dxl
xx
a
n xb f x dx x x n xdxl l
x x
2
2 2 3 30
22
3 30
3 3 3 3
2
os sin cos(2 2 ) ( 2)
cos cos2 2
2cos 2 2cos 00 0
0 ...................(3)
(2 )cos
n
n
n x n x n xxn n n
n x n xx xn n
nn n
b
a x x n x
2
0
22
2 2 3 30
2
2 20
2 2 2 2
sin cos sin(2 ) (2 2 ) ( 2)
cos(2 2 )
cos 2 cos 0(2 4) (2 0)
dx
n x n x n xx x xn n n
n xxn
nn n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
33 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2 2 2 2 2 2 2 2
0
1 1
2 2 2 21 1 1
2 2 4 4 ...................(4)
Use (2), (3), and (4) in (1) we get
( ) cos sin2
2 4 2 4 1cos 0sin cos3 3
n
n nn n
n n n
an n n n
af x a n x b n x
n x n x n xn n
Question :19 : AUC M /
0
1
2 32
00 0 0
3 3 3 3 3
Find the half range cosine series of ( ) ( ) , when 0
( ) cos .............(1),2
2 2 2( )2 3
2 2 3 2 22 3 6 6
nn
f x x x x
af x a nx
x xa f x dx x x dx
J 2010
Answer :
2
2
0
0
2
0 0
2
3
Hence ..............(2)3
2 a ( ) cos
2 2( ) cos ( )cos
2 sin( )
n
a
f x nxdx
x x nxdx x x nxdx
nxx xn
2 3
cos sin( 2 ) ( 2)nx nxxn n
0
2 20
2 2
2
2 2( 2 ) cos ( 2 ) cos ( 0) cos0
2 2( 1) 1 ( 1)
4 , if is even..............(3)
0 , if is oddUse (2) and (3) in (1) we get the r
n n
x nx nn n
n n
nn
n
esult
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
34 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
0
The process of finding the Fourier series for a given function by numerical value is known as harmonic analisis. In harmonic analisis the Fourier coefficients a
Definition :HARMONIC ANALYSIS
0
1 1
, and of the function ( ) in (0, 2 ) given by
cos sin2 , 2 , 2
where 'n' is the number of terms appears in the table.
The term ( cos sin ) is called t
n n
n n
a b y f x
y y nx y nxa a b
n n n
a x b x
Note :1
2 2
he fundamental or first harmonic,the term ( cos 2 sin 2 ) is called the second harmonic and so on.
Find the Fourier series expansion of period 2 for the function ( )which is defined in
a x b x
y f x
Problem : 20 :
(0, 2 ) by means of the table of values given below.Find the series upto second harmonic.
x 0 3
23
4
3
53
2
y 1.0 1.4 1.9 1.7 1.5 1.2 1.0
0
1 2
Since the last value of is repitition of the firstTherefore only the first six values will be used, ( . ) 6
8.72 2 2.96
cos 1.1 0.32 2 0.37 2 06 6
yi e n
ya
n
y xa a
n
Solution :
1 2
01 1 2 2
.1
0.5196 0.17322 0.17 2 0.066 6
( ) cos sin cos 2 sin 2 .2
1.45 0.37cos 0.17sin 0.1cos 2 0.06sin 2
b b
ay f x a x b x a x b x
x x x x
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
35 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
The following table gives the variations of a periodic functionover a period T.
Problem : 21
x 0 6T
3T
2T
23T
56T
T
( )f x 1.98 1.3 1.05 1.03 0.88 2.5 1.98
x y cos x
cos 2x
sin x sin 2x cosy x cos 2y x siny x sin 2y x
0 1.0 1 1 0 0 1 1 0 0
3
1.4 0.5 0.5 0.866 0.866 0.7 0.7 1.212 1.212
23
1.9 0.5 0.5 0.866 0.866
0.95 0.95 1.65 1.645
1.7 1 1 0 0 1.7 1.7 0 0
43
1.5 0.5 0.5 0.866
0.866 0.75 0.75 1.299 1.299
53
1.2 0.5 0.5 0.866
0.866
0.6 0.6 1.039 1.039
8.7 1.1 0.3 0.5196 0.1732
y
cosy x
cos 2y x
siny x
sin 2y x
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
36 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
2Show that ( ) 0.75 0.37cos 1.004sin , where = .
Here the last value is the mere repetition of the first therefore we omit that value and consider the remaining 6 values.(i.e) 6.
Therefore
xf xT
n
Solution :
01 1
0
1
2 5 when takes the values 0, , , , , takes the 6 3 2 3 6
2 4 5values 0, , , , , .3 3 3 3
Let the Fourier series be of the form
( ) cos sin 0.75 0.37cos 1.004sin2
2 1.5
cos2 0.3
T T T T Tx
af x a b
ya
n
ya
n
1
7
sin2 1.00456
yb
n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
37 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
y cos sin cosy siny
00 1.98 1.0 0 1.98 0
3
1.30 0.5 0.866 0.65 1.1258
23
1.05 0.5 0.866 0.525 0.9093
1.30 1 0 1.3 0
43
0.88 0.5 0.866 0.44 0.762
53
0.25 0.5 0.866 0.125 0.2165
4.6 1.12 3.013
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
38 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
Find the Fourier series as far as the second harmonic to representthe function given in the following data.
: 0 1 2 3 4 5( ) : 9 18 24 28 26 20
Solution:Here the length of the interval is 6( . ) , 2 6 (or)
xf x
i e l
Problem: 22
2 20
1 1
01 2 1 2
0
1
2
3
( ) cos sin .2
2 2cos cos sin sin (1)2 3 3 3 3
125Now , 2 2 41.666
cos cos ( 25)3 32 2 2 8.336 6
co2
n nn n
la n x n xf x a b
l la x x x xa a b b
ya
n
x xy ya
n
ya
1
2
2 2s cos ( 19)3 32 2 6.336 6
sin sin ( 3.4)3 32 2 2 1.136 6
2 2sin sin (20.8)3 32 2 2 6.96 6
x xy
n
x xy yb
n
x xy yb
n
MAHALAKSHMI ENGINEERING COLLEGE, TRICHY
39 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC
x
3x
2
3x
y
cos3xy
sin3xy
2cos
3xy
2sin
3xy
0 0 0 9 9 0 9
0
1 3
23
18 9 15.7 -9 15.6
2 23
43
24 -12 20.9 -24 0
3 2 28 -28 0 28 0
4
43
83
26 -13 -22.6 -13 22.6
5 53
10
3
20 10 -17.4 -10 -17.4
-25 -3.4 -19 20.8