De Moivres Trig application

Report
Using de Moivres Theorem to find cos nq and sin nq
Find
i) cos(5q) in terms of cosq
ii) sin(5q) in terms of sinq
iii) tan(5q) in terms of tanq
cos(5q) + isin(5q) = (cosq + isinq)5
Separating
Using
De
Moivres
Real
and
Pascals
Triangle
Theorem
Imaginary
Parts
line 5
= cos5q + 5icos4qsinq + 10i2cos3qsin2q+ 10i3cos2qsin3q+ 5i4cosqsin4q+ i5sin5q
= cos5q – 10cos3q sin2q + 5cosq sin4q + i(5cos4qsinq – 10cos2qsin3q + sin5q)
cos5q = cos5q – 10cos3qsin2q + 5cosqsin4q
sin5q = 5cos4qsinq – 10cos2qsin3q + sin5q
Equating real parts
Equating imaginary parts
Replace sin2q by 1– cos2q and sin4q by (1–cos2q)2 in cos5q formula
cos5q = cos5q – 10cos3q(1-cos2q) + 5cosq(1-cos2q)2 Expand and group terms
Replace cos2q by 1 – sin2q and cos4q by (1 – sin2q)2 in sin5q formula
sin5q = 5(1–sin2q)2sinq – 10(1–sin2q)sin3q + sin5q
Expand and group terms
sin5θ
tan5q =
cos5θ
5cos 4qsinq – 10cos 2qsin3q + sin5q

cos5q – 10cos3qsin2q + 5cosqsin4q
Divide top and bottom terms by cos5q
5cos 4qsinq
10cos2qsin3q
sin5q
–
+
5
5
5
cos
q
cos
q
cos
q

5
3
2
4
cos q
10cos qsin q
5cosqsin q
–
+
cos5q
cos5q
cos5q
5 tan q – 10 tan3 q + tan5 q

1– 10 tan2 q + 5 tan4 q
Cancel cos terms
Using de Moivres Theorem to express
cosnq and sinnq in terms of cosq and sinq
z = cosq + isinq 
1
= (cosq + isinq)-1 = cosq - isinq  e–iq
z
eiq
As cos(–q) = cosq and sin(–q) = –sinq
1
–inq
n = cosnq – isin(nq)  e
z
zn = cos(nq) + isin(nq)  einq
z+
1
 2cosθ
z
1
z + n  2cos(nq)
z
n
z-
1
 2isinθ
z
1
z - n  2isin(nθ)
z
n
Express cos2q in terms of cosq
1
z +  2cosθ
z
2
2
1
2

2
z
+

2cosθ

4cos
θ
(
)


z

2
1
1
1  1

2
2

z
+
+2
z
+

z
+
2z

+
2




z
z
z z

= 2cos(2q) + 2
4cos2q = 2cos(2q) + 2
cos2q =
(cos(2q) + 1)
n
as z +
1
 2cos(nq)
n
z
5
sin
 qd q
1) Express sin5q in terms of sinq
2) Integrate the answer
1
z -  2isinθ
z
zn -
1
 2isin(nθ)
n
z
5
1
5

5
z

2i
sinθ

32i
sin
q
(
)


z


z 
5
Pascals Triangle line 5
2
3
4
1
1
 1  1
5
4
3  1
2  1

z
5
z

+
10
z
10
z
+
5
z
- 







z
z
z
z
z z
5
3
 1
 1
5
3
 z -   - 5z + 5  
 z 5
 z 3
 3  1
 1
5
= z -   - 5 z -  
z
z

 1
+ 10z - 10  
z


 1 
 + 10  z -   
 z 


= 2isin5q – 52isin3q + 102isinq
= 2isin5q – 10isin3q + 20isinq
5
Grouping
powers
5
1
5

5
z

2i
sinθ

32i
sin
q
(
)


z

32isin5q = 2isin5q – 10isin3q + 20isinq
sin5q =
Hence
1
5
10
sin5q - sin3q +
sin q
16
16
16
5
sin
 qdq  
-
1
5
10
sin5q - sin3q + sin qdq
16
16
16
1
5
10
cos5q +
cos3q - cos q + K
80
48
16

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