Download Further Application of de Moivre`s Theorem 1

Further Application of de Moivre's Theorem
1.
Trigonometric Functions of Multiple Angles
de Moivre’s Theorem can help to express cos n, sin n and tan n in terms of cos , sin  and/or tan
 , which is often useful in simplifying trigonometric expressions or solving trigonometric equations:
sin 2 = 2 sin  cos 
sin 3 = 3 sin  – 4 sin3 
sin 4 = 4 sin  cos  – 8 sin3  cos 
sin 5 = 5 sin  – 20 sin3  + 16 sin5 
sin 6 = 6 sin  cos  – 32 sin3  cos  + 32 sin5  cos 
cos 2 = 2 cos2  – 1
cos 3 = 4 cos3  – 3 cos 
cos 4 = 8 cos4  – 8 cos2  + 1
cos 5 = 16 cos5  – 20 cos3  + 5 cos 
cos 6 = 32 cos6  – 48 cos4  + 18 cos2  – 1
Example 1
(i)
cos 5
= Re (cos 5 + i sin 5)
= Re (cos  + i sin )5
by de Moivre’s Theorem
where c = cos , s = sin 
= Re (c + is)5
= Re [ c5 + 5c4is + 10c3(is)2 + 10 c2(is)3 + 5c(is)4 + (is)5 ]
= Re (c5 + 5c4is – 10c3s2 – 10 c2is3 + 5cs4 + is5)
since i2 = –1
5
3 2
4
= c – 10c s + 5cs
= c5 – 10 c3 (1 – c2) + 5 c(1 – c2)2
using s2 + c2 = 1
5
3
5
2
4
= c – 10 c + 10 c + 5 c(1 – 2 c + c )
= 16 cos5  – 20 cos3  + 5 cos 
(ii)
sin 5 = Im (cos 5 + i sin 5)
= Im (c5 + 5c4is – 10c3s2 – 10 c2is3 + 5cs4 + is5) from part (i)
= 5c4s – 10 c2s3 + s5
= 5(1 – s2)2 s – 10 (1 – s2) s3 + s5
using s2 + c2 = 1
2
4
3
5
5
= 5(1 – 2s + s ) s – 10s + 10 s + s
= 5 sin  – 20 sin3  + 16 sin5 
(iii)
tan 5
sin 5
=
cos 5
5cos4  sin  – 10 cos2  sin3  + sin5 
=
from parts (i) and (ii)
cos5  – 10cos3  sin2  + 5cos  sin4 
3
5
5 tan  – 10 tan  + tan 
(by dividing throughout by cos5 )
=
1 – 10 tan2  + 5 tan4 
2.
Powers of Trigonometric Functions
de Moivre’s Theorem can also help to express powers of cos  and sin  in terms of cosines and sines
of multiple angles. This is helpful for integrating powers of cos  and sin :
1
sin2  = 2 (1 – cos 2)
1
cos2  = 2 (1 + cos 2)
1
1
sin3  = 4 (3 sin  – sin 3)
1
cos3  = 4 (3 cos  + cos 3)
1
sin4  = 8 (3 – 4 cos 2 + cos 4)
1
cos4  = 8 (3 + 4 cos 2 + cos 4)
1
sin5  = 16 (10 sin  – 5 sin 3 + sin 5)
1
cos5  = 16 (10 cos  + 5 cos 3 + cos 5)
1
sin6  = 32 (10 – 15 cos 2 + 6 cos 4 – cos 6)
1
cos6  = 32 (10 + 15 cos 2 + 6 cos 4 + cos 6)
1
If z = cos  + i sin , we know that z = cos  – i sin , so that
1
Similarly, using zn = cos n + i sin n and zn
1
z + z = 2 cos 
1
z – z = 2i sin 
= (cos  + i sin )–n
= cos (–n) + i sin (–n) by de Moivre’s Theorem
= cos n – i sin n
1
zn + zn = 2 cos n
we get
1
zn – zn = 2i sin n
Example 2
(i)
(2 cos )5
25 cos5 
32 cos5 
cos5 
 cos5  d
(ii)
(2i sin )5
25 i5 sin5 
32i sin5 
sin5 
 sin5  d
1
= ( z + )5
z
1
1
1
1 1
= z5 + 5z4 + 10z3 2 + 10z2 3 + 5z 4 + 5
z
z
z
z z
1
1 1
= z5 + 5z3+ 10z + 10 z + 5 z3 + z5
1
1
1
= ( z5 + z5 ) + 5( z3 + z3 ) + 10(z + z )
= 2 cos 5 + 10 cos 3 + 20 cos 
1
5
5
= 16 cos 5 + 16 cos 3 + 8 cos 
5
5
1
=  16 cos 5 + 16 cos 3 + 8 cos  d
1
5
5
= 80 sin 5 + 48 sin 3 + 8 sin  + C
1
= ( z – z )5
1
1
1
1 1
= z5 – 5z4 + 10z3 2 – 10z2 3 + 5z 4 – 5
z
z
z
z z
1
1 1
5
3
= z – 5z + 10z – 10 + 5 3 – 5
z
z z
1
1
1
5
3
= ( z – z5 ) – 5( z – z3 ) + 10(z – z )
= 2i sin 5 – 10i sin 3 + 20i sin 
1
5
5
=
sin 5 – sin 3 + sin 
16
16
8
5
5
1
=  16 sin 5 – 16 sin 3 + 8 sin  d
1
5
5
= – 80 cos 5 + 48 cos 3 – 8 cos  + C
2