Use $\cos 5\theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$

Question

I used $\cos(3\theta + 2\theta)$ to prove the first part, but I don't know how to the $2$nd part.

Show that $\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta,$ and hence show that $$\text{the roots of }x(16x^4 - 20x^2 + 5) \text{ are: } 0,\cos\frac{\pi}{10}, \cos\frac{3\pi}{10},\cos\frac{7\pi}{10}, \cos\frac{9\pi}{10}$$

Answer 1

• The equation $$\;x(16x^4 - 20x^2 + 5) = 0\; \tag 1$$ can have $5$ real solutions (incl. multiplicity).

• From $\;\cos 5\theta = 16\cos^5 \theta - 20\cos^3 \theta + 5\cos \theta \;$ we deduce that we will look for solutions of $(1)$ in the form of $\cos \theta\;$ where $\;\cos 5\theta = 0.$

• $\cos 5\theta = 0$ holds for $10$ different values of $\;\theta \in (-\pi,\pi]\;$ obtained when solving $$\cos 5\theta =\frac \pi2 +2k\pi\quad \text{or} \quad \cos 5\theta = -\frac \pi2 + 2k\pi,\; k=0,1,\dots,4.$$

These are $$\frac{\pi}{10},\; \frac{5\pi}{10},\; \frac{9\pi}{10},\;\frac{13\pi}{10},\;\frac{18\pi}{10}\;\text{and}\;-\frac{\pi}{10},\;\frac{3\pi}{10},\;\frac{7\pi}{10},\;\frac{11\pi}{10},\;\frac{15\pi}{10},\;$$ between them two and two have equal cosines (can you check which are the pairs?).

The solutions are those listed in the question.

Answer 2

Here the value of cosine will repeat after point. Final values wil be only 5

Answer 3

An obvious root is $x=0$; besides $$16x^4-20x^2+5=\left(4x^2-\dfrac52\right)^2-\dfrac{25}{4}+5=0\iff\left(4x^2-\dfrac52\right)^2=\dfrac54$$ therefore $x=\pm\sqrt{\dfrac{5\pm\sqrt5}{8}}$. This gives the other four values of $\cos(\theta)$ in fact, we can verify that $$\cos\left(\frac{\pi}{10}\right)=\sqrt{\dfrac{5+\sqrt5}{8}}\\\cos\left(\frac{9\pi}{10}\right)=-\sqrt{\dfrac{5+\sqrt5}{8}}\\\cos\left(\frac{3\pi}{10}\right)=\sqrt{\dfrac{5-\sqrt5}{8}}\\\cos\left(\frac{7\pi}{10}\right)=-\sqrt{\dfrac{5-\sqrt5}{8}}$$