Find a polynomial in terms of cos $θ$

Question

The question gives a hint; first, 'find the general solution for $cos$ $5θ = cos$ $4θ$'

So far my progress with this hint is;

$ 5θ = 2πk + 4θ $ or $ 5θ = 2πn - 4θ $

$ θ = 2πk$ or $ 9θ = 2πn $ , where $n , k $ are integers

I'm stumped after this and have to idea how to attack the question. Any hints would be appreciated

Answer 1

From what you've got and using the hint, we see that $\cos\frac{2\pi}{9},\cos\frac{4\pi}{9},\cos\frac{8\pi}{9}$ satisfy $$\cos(5\theta)=\cos(4\theta),$$ i.e. $$\cos(2\theta+3\theta)=2\cos^2(2\theta)-1,$$ i.e. $$\cos(2\theta)\cos(3\theta)-\sin(2\theta)\sin(3\theta)=2(2c^2-1)^2-1$$ where $c:=\cos\theta$.

$$(2c^2-1)(4c^3-3c)-2sc(3s-4s^3)=2(2c^2-1)^2-1$$ where $s:=\sin\theta$.

$$(2c^2-1)(4c^3-3c)-2c(3s^2-4s^4)=2(2c^2-1)^2-1,$$ i.e. $$(2c^2-1)(4c^3-3c)-2c(3(1-c^2)-4(1-c^2)^2)=2(2c^2-1)^2-1,$$ i.e. $$16 c^5 - 8 c^4 - 20 c^3 + 8 c^2 + 5 c - 1=0$$ Since $c=1=\cos(2n\pi)$ and $c=-\frac 12=\cos(\frac{6\pi}{9})$ are the solutions, we get $$(2 c + 1) (c - 1) (8 c^3 - 6 c + 1)=0$$

It follows that the answer is$$8\cos^3\theta - 6\cos\theta + 1=0$$

(Note that we have only five distinct values in $\cos(2k\pi),\cos(\frac{2n\pi}{9})$.)

Answer 2

$$\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}+\cos\frac{8\pi}{9}=2\cos\frac{3\pi}{9}\cos\frac{\pi}{9}+\cos\frac{8\pi}{9}=\cos\frac{\pi}{9}+\cos\frac{8\pi}{9}=0;$$ $$\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}+\cos\frac{2\pi}{9}\cos\frac{8\pi}{9}+\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}=$$ $$=\frac{1}{2}\left(\cos\frac{2\pi}{3}+\cos\frac{2\pi}{9}+\cos\frac{2\pi}{3}+\cos\frac{8\pi}{9}+\cos\frac{4\pi}{9}+\cos\frac{2\pi}{3}\right)=\frac{1}{2}\cdot\left(-\frac{3}{2}\right)=-\frac{3}{4}$$ and $$\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}=\frac{8\sin\frac{2\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}}{8\sin\frac{2\pi}{9}}=\frac{\sin\frac{16\pi}{9}}{8\sin\frac{2\pi}{9}}=-\frac{1}{8}.$$ Id est, we got the following polynomial. $$x^3-\frac{3}{4}x+\frac{1}{8}.$$ For $x=\cos\theta$ we obtain: $$\cos^3\theta-\frac{3}{4}\cos\theta+\frac{1}{8}.$$

Answer 3

If $9t=2n\pi,t=\dfrac{2n\pi}9$ where $n=0,\pm1,\pm2,\pm3,\pm4$

$$\implies\sin5t=\sin(2n\pi-4t)=-\sin4t\ \ \ \ (1)$$

Using Prosthaphaeresis Formulas

$$\sin5t-\sin t=2\sin2t\cos3t=4\sin t\cos t\cos3t$$

By $(1),$ $$\sin t(4\cos t\cos3t+1)=-4\sin t\cos t\cos2t$$

If $\sin\ne t=0,n\ne0$

$$4\cos t\cos3t+1=-4\cos t\cos2t$$

$$\iff4\cos t(4\cos^3t-3\cos t)+1=-4\cos t(2\cos^2t-1)$$ which is a bi-quadratic equation on rearrangement whose roots are $\cos\dfrac{2n\pi}9$ where $n=1,2,3,4$

As for $n=3,\cos\dfrac{2n\pi}9=\dfrac12,$ divide the bi-quadratic equation by $(2\cos t-1)$