# Evaluate sin(2pi/5)

## Homework Statement

Show that sin(2pi/5) = ((sqrt(10) + (2sqrt(5)))/4)

## The Attempt at a Solution

i showed that cos(2pi/5) + isin(2pi/5) was a primitive 5th root of unity, and that cos(2pi/5) = (-1 + sqrt(5))/4 but i cannot figure out how to do this one any ideas?

i also can use the fact that x^5 - 1 = 0 and i did a substitution to prove that cos(2pi/5) = (-1 + sqrt(5))/4 using the fact that x^5 - 1 + ( x - 1)(x^4 + x^3 + x^2 + x + 1) and x-1 is nonzero so the second equation on rhs has to be zero and then i used a variable substitution but i don't know how to go about proving the sin(2pi/5) part

dextercioby
Homework Helper
Well

$$\sin^2 \frac{2\pi}{5} + \cos^2 \frac{2\pi}{5} =1$$

, obviously.

touche

dextercioby
Homework Helper
How did you get cos(2pi/5) ? I couldn't factor any polynomial.

you divide x^4 + x^3 + x^2 + x + 1 by x^2 because it is non zero this is valid. so we get x^2 + x + 1 + 1/x + 1/x^2. then say y = x + 1/x and plugging this back in you get y^2 + y -1 = 0
then use the quadratic formula to get (-1 + sqrt(5))/2. back to y we know that x = cos(2pi/5) + isin(2pi/5) so add that to x^-1 and you end up getting 2cos(2pi/5) = (-1 + sqrt(5))/2

dextercioby
Homework Helper
Thanks for the solution. Really nice one.

uart
Hi bigubau I used a much more mundane method.

Denoting $\frac{2 \pi}{5} = \theta$ we have,

$$Re\{(\cos \theta + i \sin \theta)^5\} = 1$$

$$Im\{(\cos \theta + i \sin \theta)^5\} = 0$$

Expanding the second of these two equations and denoting $x = \cos \theta$ (and using $\sin^2 \theta = 1 - x^2$ where appropriate) gives,

$$i \sin \theta \, \left[ 5x^4 - 10 x^2 (1-x^2) + (1-x^2)^2 \right] = 0$$

which reduces to a quadratic in $x^2$.

$$16x^4 - 12 x^2 + 1 = 0$$

Solving gives,

$$x^2 = \frac{6 \pm 2 \sqrt{5}} {16} = \frac{(\sqrt{5} \pm 1)^2}{4^2}$$

$$x = \frac{\sqrt{5} - 1}{4}$$

BTW. Something a bit weak here, I selected the plus or minus by comparing the floating point approx of the surd with that of the cosine. I couldn't think of a better way but if someone else can then please let me know. :)

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