1. Complex numbers
Trig identities

2. FMA2: Complex numbers Starter: KUS objectives
BAT know how to apply DeMoivres theorem to trigonometric identities
Starter:

3. apply De Moivre’s theorem to trigonometric identities
Notes 1
apply De Moivre’s theorem to trigonometric identities
This involves changing expressions involving a function of θ into one without.
For example changing a cos6θ into powers of cosθ
𝑎+𝑏 𝑛
= 𝑎 𝑛 +𝑛 𝐶 1 𝑎 𝑛−1 𝑏+𝑛 𝐶 2 𝑎 𝑛−2 𝑏 2 +𝑛 𝐶 3 𝑎 𝑛−3 𝑏 3 + ………… + 𝑏 𝑛
Remember nCr is a function you can find on your calculator
The first term has the full power of n
 As you move across you slowly swap the powers over to the second term until it has the full power of n
𝑎+𝑏 4
𝑎 4
+ 4 𝐶 1 𝑎 3 𝑏
+ 4 𝐶 2 𝑎 2 𝑏 2
+ 4 𝐶 3 𝑎 𝑏 3
+ 𝑏 4
𝑎 4
+ 4 𝑎 3 𝑏
+ 6 𝑎 2 𝑏 2
+ 4𝑎 𝑏 3
+ 𝑏 4

4. 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 3 WB 12 Express cos3θ using powers of cosθ
You have to think logically and decide where to start
If we apply De Moivre’s theorem to this, we will end up with a ‘cos3θ’ term
If we apply the binomial expansion to it, we will end up with some terms with cosθ in
So this expression is a good starting point!
𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 3
Apply De Moivre’s theorem
Apply the Binomial expansion
=𝑐𝑜𝑠3𝜃+𝑖𝑠𝑖𝑛3𝜃
= 𝑐𝑜𝑠𝜃 3
+ 3 𝐶 1 𝑐𝑜𝑠𝜃 2 𝑖𝑠𝑖𝑛𝜃
+ 3 𝐶 2 𝑐𝑜𝑠𝜃 𝑖𝑠𝑖𝑛𝜃 2
+ 𝑖𝑠𝑖𝑛𝜃 3
= 𝑐𝑜𝑠 3 𝜃+ 3 𝑖𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛𝜃+ 3 𝑖 2 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃+ 𝑖 3 𝑠𝑖𝑛 3 𝜃
= 𝑐𝑜𝑠 3 𝜃+ 3 𝑖𝑐𝑜𝑠 2 𝜃𝑠𝑖𝑛𝜃 − 3𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃− 𝑖 𝑠𝑖𝑛 3 𝜃
The two expressions we have made must be equal. Therefore the real parts in each and the imaginary parts in each must be the same
Equate the real parts
𝑐𝑜𝑠3𝜃= 𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃
𝑐𝑜𝑠3𝜃= 𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃 1− 𝑐𝑜𝑠 2 𝜃
TA DA !We have successfully expressed cos3θ as posers of cosθ!
𝑐𝑜𝑠3𝜃= 4𝑐𝑜𝑠 3 𝜃−3𝑐𝑜𝑠𝜃

5. This time we have to equate the imaginary parts as this has sin6θ in
WB 13 Express the following as powers of cosθ: 𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛𝜃
we need something that will give us sin6θ using De Moivre’s theorem
𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 6
Apply De Moivre’s theorem
=𝑐𝑜𝑠6𝜃+𝑖𝑠𝑖𝑛6𝜃
Apply the Binomial expansion
= 𝑐𝑜𝑠𝜃 6
+ 6 𝐶 1 𝑐𝑜𝑠𝜃 5 (𝑖𝑠𝑖𝑛𝜃)
+ 6 𝐶 2 𝑐𝑜𝑠𝜃 4 𝑖𝑠𝑖𝑛𝜃 2
+ 6 𝐶 3 𝑐𝑜𝑠𝜃 3 𝑖𝑠𝑖𝑛𝜃 3
+ 6 𝐶 4 𝑐𝑜𝑠𝜃 2 𝑖𝑠𝑖𝑛𝜃 4
+ 6 𝐶 5 𝑐𝑜𝑠𝜃 𝑖𝑠𝑖𝑛𝜃 5
+ 𝑖𝑠𝑖𝑛𝜃 6
= 𝑐𝑜𝑠 6 𝜃+ 15 𝑖 2 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛 2 𝜃+ 20 𝑖 3 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+ 15 𝑖 4 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛 4 𝜃+ 6 𝑖 5 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃+ 𝑖 6 𝑠𝑖𝑛 6 𝜃
= 𝑐𝑜𝑠 6 𝜃+ 6𝑖 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃− 15 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛 2 𝜃− 20𝑖 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+ 15 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛 4 𝜃+ 6𝑖𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃− 𝑠𝑖𝑛 6 𝜃
This time we have to equate the imaginary parts as this has sin6θ in
𝑖𝑠𝑖𝑛6𝜃=6𝑖 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20𝑖 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑖𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃
𝑠𝑖𝑛6𝜃=6 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 2 4
𝑠𝑖𝑛6𝜃 𝑠𝑖𝑛𝜃
= 6 𝑐𝑜𝑠 5 𝜃𝑠𝑖𝑛𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 3 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 5 𝜃 𝑠𝑖𝑛𝜃
Substitute 𝑠𝑖𝑛 2 𝜃= 1− 𝑐𝑜𝑠 2 𝜃 1− 𝑐𝑜𝑠 2 𝜃
+ 6 𝑐𝑜𝑠 5 𝜃
=6 𝑐𝑜𝑠 5 𝜃−20 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛 2 𝜃+6𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 4 𝜃
Substitute 𝑠𝑖𝑛 4 𝜃= 1− 𝑐𝑜𝑠 2 𝜃 − 𝑐𝑜𝑠 2 𝜃 2
=… =32 𝑐𝑜𝑠 5 𝜃− 32 𝑐𝑜𝑠 3 𝜃+ 6𝑐𝑜𝑠𝜃

6. Notes 2 Let: 𝑧=𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 1 𝑧 = 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 −1 1 𝑧 =cos⁡(−𝜃)+𝑖𝑠𝑖𝑛(−𝜃)
You also need to be able to work this type of question in a different way:
For example, you might have a power or cos or sin and need to express it using several linear terms instead
Eg) Changing sin6θ to sinaθ + sinbθ
where a and b are integers
To do this we need to know some other patterns first!
Let:
𝑧=𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃
1 𝑧 = 𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃 −1
1 𝑧 =cos⁡(−𝜃)+𝑖𝑠𝑖𝑛(−𝜃)
1 𝑧 = cos 𝜃 −𝑖𝑠𝑖𝑛𝜃
We can add our two results together:
We could also subtract our two results:
𝑧+ 1 𝑧 =
𝑧− 1 𝑧 =
𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃
+ 𝑐𝑜𝑠𝜃−𝑖𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃+𝑖𝑠𝑖𝑛𝜃
− 𝑐𝑜𝑠𝜃−𝑖𝑠𝑖𝑛𝜃
𝑧+ 1 𝑧 =
𝑧− 1 𝑧 =
2𝑐𝑜𝑠𝜃
2𝑖𝑠𝑖𝑛𝜃

7. Notes 3 Taking this further
Let:
𝑧 𝑛 =𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃
1 𝑧 𝑛 = 𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃 −1
1 𝑧 𝑛 =cos⁡(−𝑛𝜃)+𝑖𝑠𝑖𝑛(−𝑛𝜃)
1 𝑧 𝑛 = cos 𝑛𝜃 −𝑖𝑠𝑖𝑛𝑛𝜃
We could add our two results together:
We could also subtract our two results:
𝑧 𝑛 + 1 𝑧 𝑛 =
𝑧 𝑛 − 1 𝑧 𝑛 =
𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃
+ 𝑐𝑜𝑠𝑛𝜃−𝑖𝑠𝑖𝑛𝑛𝜃
𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃
− 𝑐𝑜𝑠𝑛𝜃−𝑖𝑠𝑖𝑛𝑛𝜃
𝑧 𝑛 + 1 𝑧 𝑛 =
𝑧 𝑛 − 1 𝑧 𝑛 =
2𝑐𝑜𝑠𝑛𝜃
2𝑖𝑠𝑖𝑛𝑛𝜃

8. WB 14 Express cos5θ in the form acos5θ + bcos3θ + ccosθ
Where a, b and c are constants to be found
𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃
𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃
𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃
𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃
Creating a cos5θ term
𝑧+ 1 𝑧 5
= 2𝑐𝑜𝑠𝜃 5
=32 𝑐𝑜𝑠 5 𝜃
Creating the other cos terms – use the Binomial expansion!
𝑧+ 1 𝑧 = 𝑧 𝑧 𝑧 𝑧 𝑧 𝑧 𝑧 𝑧 1 𝑧 𝑧 5
= 𝑧 𝑧 𝑧 𝑧 𝑧 𝑧 5
= 𝑧 𝑧 𝑧 𝑧 𝑧+ 1 𝑧
=2𝑐𝑜𝑠5𝜃+ 5 2𝑐𝑜𝑠3𝜃 + 10(2𝑐𝑜𝑠𝜃)
=2𝑐𝑜𝑠5𝜃 𝑐𝑜𝑠3𝜃+ 20𝑐𝑜𝑠𝜃
These two expressions must be equal to each other
32 𝑐𝑜𝑠 5 𝜃=2𝑐𝑜𝑠5𝜃+10𝑐𝑜𝑠3𝜃+ 20𝑐𝑜𝑠𝜃
𝑐𝑜𝑠 5 𝜃= 1 16 𝑐𝑜𝑠5𝜃 𝑐𝑜𝑠3𝜃 𝑐𝑜𝑠𝜃
TA DA! we have written cos5θ using cos5θ, cos3θ and cosθ

9. WB 15 Show that: 𝑠𝑖𝑛 3 𝜃=− 1 4 𝑠𝑖𝑛3𝜃+ 3 4 𝑠𝑖𝑛𝜃
𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃
𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃
𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃
𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃
Creating a sin3θ term
𝑧− 1 𝑧 3
= 2𝑖𝑠𝑖𝑛𝜃 3
=8 𝑖 3 𝑠𝑖𝑛 3 𝜃
=−8𝑖 𝑠𝑖𝑛 3 𝜃
Creating the other sin terms – use the Binomial expansion!
𝑧− 1 𝑧 = 𝑧 𝑧 2 − 1 𝑧 + 3𝑧 − 1 𝑧 − 1 𝑧 3
= 𝑧 3 − 3𝑧 𝑧 − 1 𝑧 3
= 𝑧 3 − 1 𝑧 3 − 3 𝑧− 1 𝑧
=2𝑖𝑠𝑖𝑛3𝜃 − 3(2𝑖𝑠𝑖𝑛𝜃)
=2𝑖𝑠𝑖𝑛3𝜃 − 6𝑖𝑠𝑖𝑛𝜃
−8𝑖 𝑠𝑖𝑛 3 𝜃=2𝑖𝑠𝑖𝑛3𝜃−6𝑖𝑠𝑖𝑛𝜃
−8 𝑠𝑖𝑛 3 𝜃=2𝑠𝑖𝑛3𝜃−6𝑠𝑖𝑛𝜃
𝑠𝑖𝑛 3 𝜃=− 1 4 𝑠𝑖𝑛3𝜃+ 3 4 𝑠𝑖𝑛𝜃
TA DA!

10. Express sin4θ in the form: 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓
WB 16
Express sin4θ in the form: 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓
Where d, e and f are constants to be found.
b) Hence, find the exact value of the following integral: 0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃
𝑧+ 1 𝑧 =2𝑐𝑜𝑠𝜃
𝑧− 1 𝑧 =2𝑖𝑠𝑖𝑛𝜃
𝑧 𝑛 + 1 𝑧 𝑛 =2𝑐𝑜𝑠𝑛𝜃
𝑧 𝑛 − 1 𝑧 𝑛 =2𝑖𝑠𝑖𝑛𝑛𝜃
Creating a sin4θ term
𝑧− 1 𝑧 4
= 2𝑖𝑠𝑖𝑛𝜃 4
=16 𝑖 4 𝑠𝑖𝑛 4 𝜃
=16 𝑠𝑖𝑛 4 𝜃
Creating the cos terms – use the Binomial expansion!
𝑧− 1 𝑧 4 = 𝑧 𝑧 3 − 1 𝑧 𝑧 2 − 1 𝑧 𝑧 − 1 𝑧 − 1 𝑧 4
= 𝑧 4 − 4 𝑧 − 𝑧 𝑧 4
0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃 = 0 𝜋 𝑐𝑜𝑠4𝜃− 1 2 𝑐𝑜𝑠2𝜃 𝑑𝜃
= 𝑧 𝑧 4 − 4 𝑧 𝑧
=2𝑐𝑜𝑠4𝜃− 8𝑐𝑜𝑠2𝜃 +6
= 𝑠𝑖𝑛4𝜃− 1 4 𝑠𝑖𝑛2𝜃 𝜃 0 𝜋 2
16 𝑠𝑖𝑛 4 𝜃=2𝑐𝑜𝑠4𝜃−8𝑐𝑜𝑠2𝜃+6
𝑠𝑖𝑛 4 𝜃= 1 8 𝑐𝑜𝑠4𝜃− 1 2 𝑐𝑜𝑠2𝜃+ 3 8
TA DA!

11. Express sin4θ in the form: 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓
WB 17
Express sin4θ in the form: 𝑑𝑐𝑜𝑠4𝜃+𝑒𝑐𝑜𝑠2𝜃+𝑓
Where d, e and f are constants to be found.
b) Hence, find the exact value of the following integral: 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃
𝑐𝑜𝑠4𝜃 = 1 4 𝑠𝑖𝑛4𝜃
0 𝜋 2 𝑠𝑖𝑛 4 𝜃 𝑑𝜃 = 0 𝜋 𝑐𝑜𝑠4𝜃− 1 2 𝑐𝑜𝑠2𝜃 𝑑𝜃
𝑐𝑜𝑠2𝜃 = 1 2 𝑠𝑖𝑛2𝜃
= 𝑠𝑖𝑛4𝜃− 1 4 𝑠𝑖𝑛2𝜃 𝜃 0 𝜋 2
= 𝑠𝑖𝑛4 𝜋 2 − 1 4 𝑠𝑖𝑛2 𝜋 𝜋 2 − 𝑠𝑖𝑛4 0 − 1 4 𝑠𝑖𝑛
= 3 16 𝜋

12. One thing to improve is –
KUS objectives BAT know how to apply DeMoivres theorem to trigonometric identities
self-assess
One thing learned is –
One thing to improve is –

13. END