Question

0.2 Let z = cos θ + isin θ.
Then z
n = cos(nθ) + isin(nθ) 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 =

z +
1
z

n
and (2i)
n
sinn
θ =

z −
1
z

n
. (2)
(c) Use (b) to express sin7
θ in terms of multiple angles. (6)
(d) Express cos3
θ sin4
θ in terms of multiple angles. (6)
(e) Eliminate θ from the equations 4x = cos(3θ) + 3 cos θ; 4y = 3 sin θ − sin(3θ). (5

Answer 1

a)

$z^n+z^{-n}=cos(n\theta)+isin(n\theta)+cos(n\theta)-isin(n\theta)=2cos(n\theta)$

$2cos(n\theta)=z^n+z^{-n}$

$z^n-z^{-n}=cos(n\theta)+isin(n\theta)-(cos(n\theta)-isin(n\theta))=2isin(n\theta)$

$2isin(n\theta)=z^n-z^{-n}$

b)

$2^ncos^n(\theta)=(2cos(\theta))^n=(z+\frac{1}{z})^n$

$(2i)^nsin^n(\theta)=(2isin(\theta))^n=(z-\frac{1}{z})^n$

(Here we use formulas from part (a) when n=1 )

c)

$sin^7(\theta)=\frac{1}{(2i)^7}\cdot(2i)^7\cdot sin^7(\theta)=\\-\frac{1}{128i}(2isin(\theta))^7 =-\frac{1}{128i}(z-\frac{1}{z})^7$

Using Newton's binomial formula:

$(z-\frac{1}{z})^7=z^7-7z^5+21z^3-35z+35\frac{1}{z}-21\frac{1}{z^3}+7\frac{1}{z^5}-\frac{1}{z^7}=$

$(z^7-\frac{1}{z^7})-7(z^5-\frac{1}{z^5})+21(z^3-\frac{1}{z^7}-35(z-\frac{1}{z}))=$

$2i\cdot sin(7\theta)-7\cdot 2i\cdot sin(5\theta)+21\cdot 2i\cdot sin(3\theta)-35\cdot 2i\cdot sin(\theta)=$

$2i(sin(7\theta)-7\cdot sin(5\theta)+21\cdot sin(3\theta)-35\cdot sin(\theta))$

So, $sin^7(\theta)=-\frac{1}{128i}\cdot2i(sin(7\theta)-7\cdot sin(5\theta)+21\cdot sin(3\theta)-35\cdot sin(\theta))=$

$-\frac{1}{64}(sin(7\theta)-7\cdot sin(5\theta)+21\cdot sin(3\theta)-35\cdot sin(\theta))$

d)

$cos^3(\theta)sin^4(\theta)=\frac{1}{2^3}\cdot2^3\cdot cos^3(\theta)\cdot \frac{1}{(2i)^4}\cdot (2i)^4\cdot sin^4(\theta)=$

$\frac{1}{2^7}(2cos(\theta))^3(2isin(\theta))^4=\frac{1}{2^7}(z+\frac{1}{z})^3(z-\frac{1}{z})^4=$

$\frac{1}{2^7}((z+\frac{1}{z})(z-\frac{1}{z}))^3(z-\frac{1}{z})=\frac{1}{2^7}(z^2-\frac{1}{z^2})^3(z-\frac{1}{z})=$

$\frac{1}{2^7}(z^6-3z^2+3\cdot \frac{1}{z^2}-\frac{1}{z^6})(z-\frac{1}{z})=$

$\frac{1}{2^7}((z^6-\frac{1}{z^6})-3(z^2-\frac{1}{z^2}))(z-\frac{1}{z})=$

$\frac{1}{2^7}(2isin(6\theta)-3\cdot2isin(2\theta))\cdot2isin(\theta)=$

$-\frac{1}{2^5}(sin(6\theta)-3sin(sin(2\theta))sin(\theta)$

e)

$4x=cos(3\theta)+3cos(\theta)\\ 4y=3sin(\theta)-sin(3\theta)$

Let's multiply 4y to i and sum up equations.

$4x+4iy=cos(3\theta)+3cos(\theta)+3isin(\theta)-isin(3\theta)=$

$=(cos(3\theta)-isin(3\theta))+3(cos(\theta)+isin(\theta))=$

$=z^{-3}+3z$

So, 4x is the real part and 4y is imaginary part

$4x=Re(z^{-3}+3z)$

$4y=Im(z^{-3}+3z)$