De Moivres Theorem

Question

This is the questions I got and I'm stuck as to where to proceed after these few steps

Answer 1

For the second part since we want to solve the equation $$x^4-6x^2+1=0$$ in terms of $\tan(\frac{n\pi}{8})$ we assume that: $$\frac{n\pi}{8}=\theta$$ Now $$4\theta = \frac{n\pi}{2}$$ Therefore, $$\tan(4\theta) = \tan(\frac{n\pi}{2})$$ RHS is always either $\infty$ or $-\infty$ so whatever expansion of $\tan(4\theta)$ is in the denominator that will obviously be $0$. Now since, denominator is $0$ or $$\tan^4\theta-6\tan^2\theta+1=0$$ Now you assume that $$x=\tan\theta$$ and solve the biquadratic equation to get the answer. The equation will become: $$x^4-6x^2+1=0$$ How to solve this equation?? Put $$x^2=t$$ and rewrite the biquadratic equation as a quadratic equation in $t$ giving $$t^2-6t+1=0$$. Solve for $t$ and then put $x^2=t$ to get the roots of x. Hope this helps ......

Answer 2

For the first part of the question: Notice by DMT:

$$(\cos\theta +i\sin\theta)^4=\cos(4\theta)+i\sin(4\theta)$$ Since $\tan x=\frac{\sin x}{\cos x}$, we have that: $$\tan(4\theta)=\frac{\Im((\cos\theta+i\sin\theta)^4)}{\Re((\cos\theta+i\sin\theta)^4)}$$

Use the binomial expansion to do the rest.

Answer 3

$$\sin(4 \theta) = 2 \sin (2 \theta) cos(2 \theta)$$ $$\cos(4 \theta) = \cos^2(2 \theta)-\sin^2(2\theta)$$ $$sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ $$\cos(2\theta) = \cos^2(\theta)-\sin^2(\theta)$$

Note that all of these can be derived from de Moivre's theorem $(\cos \theta + j \sin \theta)^n = \cos(n \theta) + j\sin(n \theta)$.

For example, when $n=2$, $$(\cos \theta + j \sin \theta)^2 = cos^2(\theta) - sin^2(\theta) + j2 \sin \theta \cos \theta = \cos(2 \theta) + j\sin(2 \theta)$$ implying $$sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ $$\cos(2\theta) = \cos^2(\theta)-\sin^2(\theta)$$

You can similarly derive the result for $n=4$.

Anyway, now you can compute $$\tan(4 \theta) = {\sin(4 \theta) \over \cos(4 \theta)} = {4 \sin \theta \cos \theta \; (\cos^2 \theta - \sin^2 \theta) \over (\cos^2 \theta - \sin^2 \theta)^2 - 4\sin^2 \theta \cos^2 \theta}$$

If you divide the numerator and denominator by $\cos^4 \theta$, this simplifies to

$$\tan(4 \theta) = {4 \tan \theta - 4 \tan^3 \theta \over 1 - 6 \tan^2 \theta + \tan^4 \theta}$$

Now, note that when $4 \theta = n \pi/2$, $\tan 4 \theta = \infty$, i.e. $1 - 6 \tan^2 \theta + \tan^4 \theta = 0$. This is exactly the equation you want to solve (with $x = \tan \theta$), so you get $x = \tan (n \pi/8)$.

Answer 4

For the second part, set $\dfrac1{\tan4\theta }=0$

$\implies4\theta=(2n+1)\dfrac\pi2$ for some integer $n$

$\theta=\dfrac{(2n+1)\pi}8$ where $n\equiv0,1,2,3\pmod4$