Proving Trig Identities (Complex Numbers)

Question

Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then

<p>$$ z^n +  {1\over z^n} = 2\cos(n\theta) $$</p>

<p>Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)</p>

I learnt to prove the first part in another post linked here.

The second part is where I am confused because there is a 'hence'

so I thought of taking 2 approaches:

either $$ z^6 + \frac 1{z^6} $$

or $$ z^6 $$ and equating real parts.

I will start with my first approach

$$ z^6 + \frac 1{z^6} $$

$$ (\cos(x)+i\sin(x))^6 + \frac 1{(\cos(x)+i\sin(x))^6} $$

$$ (\cos(x)+i\sin(x))^6 + {(\cos(x)+i\sin(x))^{-6}} $$

$$ \cos(6x) + i\sin(6x) + \cos(-6x)+ i \sin(-6x) $$

$$ 2\cos(6x) $$

Which is no where near what I am suppose to prove..

So with my second approach (expanding and equation real parts)

$$ z^6 $$

$$ (\cos(x) + i \sin(x))^6 $$

Using pascals

$$ \cos^6(x) + i*6\cos^5(x)\sin(x) + i^2*15\cos^4(x)\sin^2(x) + i^3*20\cos^3(x)\sin(x) + i^4*15\cos^2(x)\sin^4(x) + i^5*6\cos(x)\sin^5(x) + i^6 * \sin^6(x) $$

Simplifying

$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) +i(6\cos^5(x)\sin(x)-20\cos^3(x)\sin(x) + 6cos(x)\sin^5(x)$$

Now considering only real

$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) $$

At this point I'm confused , am I on the right approach?

Answer 1

$n=1\implies2\cos x=z+\dfrac1z$

$$\cos^6x=\left(\dfrac{z+\dfrac1z}2\right)^6$$

$$64\cos^6x=\left(z+\dfrac1z\right)^6=z^6+\dfrac1{z^6}+\binom61\left(z^4+\dfrac1{z^4}\right)+\binom62\left(z^2+\dfrac1{z^2}\right)+\binom63$$

Can you take it from here?

Answer 2

By De Moivre's formula if $z = \cos (\theta) + i\sin(\theta)$ then $z^n = \cos (n\theta) + i\sin(n\theta)$.

So $$z^n = \cos (n\theta) + i\sin(n\theta)$$

and

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

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

Thus when $n=1$ ,

$$z+z^{-1}=2\cos (\theta) \Rightarrow \left(z+\frac{1}{z} \right)^6=2^6 \cdot \cos^6(\theta) $$

$$2^6 \cdot \cos^6(\theta)=\left(z+\frac{1}{z} \right)^6 $$

$$2^6 \cdot \cos^6(\theta)=z^6+6z^4+15z^2+20+15+\frac{1}{z^2} +6\frac{1}{z^4} +\frac{1}{z^6} $$

$$2^6 \cdot \cos^6(\theta)=\left(z^6+\frac{1}{z^6}\right)+6\left(z^4+\frac{1}{z^4}\right)+15\left(z^2+\frac{1}{z^2}\right)+20 $$

$$64 \cdot \cos^6(\theta)=\left(2\cos 6\theta \right)+6\left(2\cos 4\theta\right)+15\left(2\cos 2\theta\right)+20 $$

Therefore

$$ \cos^6(\theta)=\frac{1}{32}\{\left(\cos 6\theta \right)+6\left(\cos 4\theta\right)+15\left(\cos 2\theta\right)+10 \}$$

Answer 3

This is simply because for such a $z$, $\;\frac1{z^n}=\bar z^n=\overline{z^n}$, and $$z^n+\overline{z^n}=2\operatorname{Re}(z^n)=2\cos n\theta\qquad\quad\text{by De Moivre's formula.}$$

Answer 4

I don't think that the hint is really that useful. It would be better to do $$ \cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2} $$ that gives you \begin{align} 64\cos^6\theta &=(e^{i\theta}+e^{-i\theta})^6 \\[3px] &=e^{6i\theta}+6e^{4i\theta}+15e^{2i\theta} +20+15e^{-2i\theta}+6e^{-4i\theta}+e^{-6i\theta} \\[3px] &=(e^{6i\theta}+e^{-6i\theta})+ 6(e^{4i\theta}+e^{-4i\theta})+ 15(e^{2i\theta}+e^{-2i\theta})+ 20 \\[3px] &=2\cos6\theta+12\cos4\theta+30\cos2\theta+20 \end{align} This is no more than De Moivre’s formula in disguise, but much easier to manage.