# Show that the roots of the equation 16x^4 - 20x^2 + 5= 0 are cos((kx)/10) for k = 1,3,7  and  9 And Deduce that cos^2(pi/10)cos^2((3pi)/10) = 5/16?

Apr 8, 2015

Antoine, I think that your roots are without $x$ but with $\pi$ instead:
To solve your equation substitute ${x}^{2} = t$ to reduce the degree but remember at the end to go back to x: Your roots corresponds to $\cos \left(k \frac{\pi}{10}\right)$ with $k = 1 , 3 , 7 , 9$

and if you have: Apr 8, 2015

$f \left(x\right) = 16 {x}^{4} - 20 {x}^{2} + 5$

Has real zeroes, because
$- b \pm \sqrt{{b}^{2} - 4 a c} = - \left(- 20\right) \pm \sqrt{{\left(20\right)}^{2} - 4 \left(16\right) \left(5\right)}$
$= 20 \pm \sqrt{{20}^{2} - 16 \cdot 20} = 20 \pm \sqrt{4 \cdot 20}$ is positive.

For the full solution, see Gio's answer.)

$f$ is an even function, so its four (real) zeroes occur in opposite pairs: ${p}_{1} , {p}_{2}$ (the positive zeros) and $- {p}_{1} , - {p}_{2}$.

The factor theorem gives us:
$f \left(x\right) = 16 \left(x - {p}_{1}\right) \left(x + {p}_{1}\right) \left(x - {p}_{2}\right) \left(x + {p}_{2}\right)$

The constant for this product is clearly $16 {\left({p}_{1}\right)}^{2} {\left({p}_{2}\right)}^{2}$ which must be equal to $5$.

Therefore, ${\left({p}_{1}\right)}^{2} {\left({p}_{2}\right)}^{2} = \frac{5}{16}$.

For the first part

Expand $\cos \left(5 t\right)$ or use the multiple angle formula to get:

$\cos \left(5 t\right) = 16 {\cos}^{4} t - 20 {\cos}^{2} t + 5$

Then with $t = \frac{\pi}{10}$ we get:

$16 {\cos}^{4} \left(\frac{\pi}{10}\right) - 20 {\cos}^{2} \left(\frac{\pi}{10}\right) + 5 = \cos \left(5 \left(\frac{\pi}{10}\right)\right) = \cos \left(\frac{\pi}{2}\right) = 0$.

In fact, with $t = \frac{k \pi}{10}$ we get:
$16 {\cos}^{4} \left(\frac{k \pi}{10}\right) - 20 {\cos}^{2} \left(\frac{k \pi}{10}\right) + 5 = \cos \left(5 \left(\frac{k \pi}{10}\right)\right) = \cos \left(\frac{k \pi}{2}\right) = 0$ whenever $k$ is odd.