How do you find the exact values of cos 2pi/5?

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Answer 1 · 2016-03-16T22:03:54

$cos (2 \frac{π}{5}) = \frac{- 1 + \sqrt{5}}{4}$ $cos(2pi/5)=(-1+sqrt(5))/4$

Here the most elegant solution I found in:

$cos (4 \frac{π}{5}) = cos (2 π - 4 \frac{π}{5}) = cos (6 \frac{π}{5})$ $cos(4pi/5)=cos(2pi-4pi/5)=cos(6pi/5)$

So if $x = 2 \frac{π}{5}$ $x=2pi/5$ :

$cos (2 x) = cos (3 x)$ $cos(2x)=cos(3x)$

Replacing the cos(2x) and cos(3x) by their general formulae:

$cos (2 x) = 2 {cos}^{2} x - 1 and cos (3 x) = 4 {cos}^{3} x - 3 cos x$ $color(red)(cos(2x)=2cos^2x-1 and cos(3x)=4cos^3x-3cosx)$ ,

we get:

$2 {cos}^{2} x - 1 = 4 {cos}^{3} x - 3 cos x$ $2cos^2x-1=4cos^3x-3cosx$

Replacing $cos x$ $cosx$ by $y$ $y$ :

$4 y^{3} - 2 y^{2} - 3 y - 1 = 0$ $4y^3-2y^2-3y-1=0$

$(y - 1) (4 y^{2} + 2 y - 1) = 0$ $(y-1)(4y^2+2y-1)=0$

We know that $y \neq 1$ $y!=1$ , so we have to solve the quadratic part:

$y = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 4 \cdot (- 1)}}{2 \cdot 4}$ $y=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)$

$y = \frac{- 2 \pm \sqrt{20}}{8}$ $y=(-2+-sqrt(20))/8$

since $y > 0$ $y>0$ , $y = cos (2 \frac{π}{5}) = \frac{- 1 + \sqrt{5}}{4}$ $y=cos(2pi/5)=(-1+sqrt(5))/4$