Complex Numbers and Trigonometry - Show that this is true about the roots of the polynomial

Question

I'm given $$ cos(5\theta) = 15\cos ^5(\theta ) -20\cos^3(\theta) +5\cos(\theta)$$

Find the possible values of $\theta$ within the interval [$0,\pi$] for which: $$16\cos^4(\theta) -20\cos^2(\theta) + 5=0$$

Hence show that $cos(\frac{\pi}{10}) + cos(\frac{3\pi}{10})= \frac{\sqrt{5}}{4}$

I understand that $$\frac{cos(5\theta)}{cos(\theta)}=0 \implies cos(5\theta)=0$$

So $$\theta=\frac{\pi}{10},\frac{3\pi}{10},\frac{7\pi}{10},\frac{9\pi}{10}$$

Missing out $\theta=\frac{\pi}{2}$ because you can't divide by zero.

Since the expression is equal to the 4th degree polynomial they share the same solutions. So factorising the polynomial.

$$\big(cos(\theta)-\cos(\frac{\pi}{10})\big)(\cos(\theta)-\cos(\frac{3\pi}{10}))(\cos(\theta)-\cos(\frac{7\pi}{10}))(\cos(\theta)-\cos(\frac{9\pi}{10}))=0$$

Now I don't understand how to derive the expression about the roots. I've tried using the relationships between the sum and products and product of pairs of roots etc.. Not sure what I'm doing wrong any help would be much appreciated.

Answer 1

Note that $\cos\frac{\pi}{10}+ \cos\frac{3\pi}{10}= \frac{\sqrt{5}}{4}$ is not ture. I believe that you need to show that

$$\cos\frac{\pi}{10} \cos\frac{3\pi}{10}= \frac{\sqrt{5}}{4}$$

which can be done as follows. Since $\cos\frac{\pi}{10}, \ \cos\frac{3\pi}{10}, \ \cos\frac{7\pi}{10}, \ \cos\frac{9\pi}{10}$ are the roots of,

$$16\cos^4\theta -20\cos^2\theta + 5=0$$

we have

$$\cos\frac{\pi}{10}\cos\frac{3\pi}{10}\cos\frac{7\pi}{10}\cos\frac{9\pi}{10}=\frac 5{16}$$

Recognize $\cos\frac{7\pi}{10}=-\cos\frac{3\pi}{10}$, $\cos\frac{9\pi}{10}=-\cos\frac{\pi}{10}$ to get,

$$(\cos\frac{\pi}{10}\cos\frac{3\pi}{10})^2=\frac 5{16}$$

Thus,

$$\cos\frac{\pi}{10}\cos\frac{3\pi}{10} = \frac{\sqrt5}4$$

Answer 2

Let the roots be $ a, b, c, d = \cos \frac{\pi}{10}, \cos \frac{3\pi}{10}, \cos \frac{7\pi}{10}, \cos \frac{9\pi}{10}$.

Observe that $ a = -d > 0, b = -c > 0$.

Hint: Can you find $ ab$?

Consider $ abcd = (ab)^2$.

Hint: Can you find $a^2 + b^2 $?

Consider $(a+b+c+d)^2 - 2 (ab+bc+ca+ad+ac+bd)$.

Hence, calculate $ (a+b)^2$ and thus $a + b$.

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As mentioned by Quanto, the statement should have been $ab = \frac{\sqrt{5}}{4}$. I obviously didn't bother to do the calculations, as it was enough to show that these could be done.