Answer to Question #110821 in Complex Analysis for Hetisani Sewela

Question

Use 9.2 to evaluate sin π
5
,sin 2π
5
and cos π
5
. (9)
10.2 Let z = cosθ + sin θ.
Then z
n = cos nθ + isin θ for all n ∈ N (by de Moivre) and z
−n = cos nθ − isin nθ.
(a) Show that 2 cos nθ = z
n + z
−n
and 2isin nθ = z
n − z
n
. (2)
(b) Show that 2
n
cosn
θ =

Answer 1

Let

$$\theta=\pi/5, =>5\theta=\pi.\\ 3\theta=\pi-2\theta. \\$$

Therefore,

$$\text{sin}(3\theta)=\text{sin}(\pi-2\theta)=\text{sin}(2\theta)$$

if $\theta<90.$

Next,

$$\text{sin}3\theta=\text{sin}2\theta,\\ 3\text{sin}\theta-4\text{sin}^3\theta=2\text{sin}\theta\text{cos}\theta\space||\cdot\frac{1}{\text{sin}\theta},\\ 3-4\text{sin}^2\theta=2\text{cos}\theta,\\ 3-4(1-\text{cos}^2\theta)=2\text{cos}\theta,\\ 4\text{cos}^2\theta-2\text{cos}\theta-1=0.$$

We notice that this is a quadratic equation where cos theta is like a variable:

$$\text{cos}\theta=\frac{1\pm\sqrt5}{4},$$

we take only positive root because theta cannot give negative cosine,

$$\text{cos}\theta=\text{cos}\frac{\pi}{5}=\frac{1+\sqrt5}{4},\\ \text{sin}\theta=\text{sin}\frac{\pi}{5}=\sqrt{1-\text{cos}^2\theta}=\frac{\sqrt{10-2\sqrt5}}{4}.\\ \text{sin}\frac{2\pi}{5}=\text{sin}(2\theta)=2\text{sin}\theta\text{cos}\theta=\\ =2\cdot\frac{\sqrt{10-2\sqrt5}}{4}\cdot\frac{1+\sqrt5}{4} =\frac{\sqrt{{10+2\sqrt5}}}{4}.$$

(a)

$$z^n+z^{-n}=(\text{cos}\theta+i\text{sin}\theta)^n+(\text{cos}\theta+i\text{sin}\theta)^{-n},\\$$

according to de Moivre theorem this is

$$\text{cos}(n\theta)+i\text{sin}(n\theta)+\text{cos}(-n\theta)+i\text{sin}(-n\theta)=\\ =\text{cos}(n\theta)+i\text{sin}(n\theta)+\text{cos}(-n\theta)-i\text{sin}(n\theta)=\\ =2\text{cos}(n\theta).$$