# finding the exact trigonometric ratios of an equation

• Nov 15th 2009, 08:25 PM
differentiate
finding the exact trigonometric ratios of an equation
Find the exact trigonometric roots of $\displaystyle 16x^5 - 20x^3 + 5x + 1 = 0$
given that$\displaystyle cos5\theta = 16cos^5\theta - 20cos^3\theta + 5cos\theta$

heres how I did it

let x= cos@
cos5@ = 16x^5 - 20x^3 + 5x + 1
cos5@ = 0, 16x^5 -20x^3 + 5x + 1 = 0

5@ = 2(pi)k +/- pi/2 where k is any integer
= (2(pi)k +/- pi)/10

x = cos@
= cos {2(pi)k +/- pi]/10}
where k is any integer

but how do I proceed to find the five roots of the equation. and is the above correct
• Nov 16th 2009, 05:06 AM
Hello differentiate
Quote:

Originally Posted by differentiate
Find the exact trigonometric roots of $\displaystyle 16x^5 - 20x^3 + 5x + 1 = 0$
given that$\displaystyle cos5\theta = 16cos^5\theta - 20cos^3\theta + 5cos\theta$

heres how I did it

let x= cos@
cos5@ = 16x^5 - 20x^3 + 5x + 1
cos5@ = 0, 16x^5 -20x^3 + 5x + 1 = 0

5@ = 2(pi)k +/- pi/2 where k is any integer
= (2(pi)k +/- pi)/10

x = cos@
= cos {2(pi)k +/- pi]/10}
where k is any integer

but how do I proceed to find the five roots of the equation. and is the above correct

You have more-or-less the right idea, but are wrong in one or two details.

Let $\displaystyle x = \cos\theta$

Then $\displaystyle \cos5\theta = 16x^5 -20x^3+5x$ (not, as you said, $\displaystyle 16x^5 -20x^3+5x+1$)

So $\displaystyle 16x^5 -20x^3+5x+1=0$

$\displaystyle \Rightarrow 16x^5 -20x^3+5x=-1$

$\displaystyle \Rightarrow \cos5\theta = -1$

$\displaystyle \Rightarrow 5\theta = (2n+1)\pi, n \in \mathbb{Z}$

$\displaystyle \Rightarrow \theta = \frac{(2n+1)\pi}{5}$

So the $\displaystyle 5$ roots are $\displaystyle \cos\left(\frac{(2n+1)\pi}{5}\right), n = 0,1, 2, 3, 4$

(Note that these are not all distinct. The equation has two pairs of equal roots, because $\displaystyle \cos\left( \frac{\pi}{5}\right)=\cos\left( \frac{9\pi}{5}\right)$ and $\displaystyle \cos\left( \frac{3\pi}{5}\right)=\cos\left( \frac{7\pi}{5}\right)$.)