## find 5th roots of unity solving x^5 -1=0 and use the result for sin18 and cos18

1. The problem statement, all variables and given/known data

Find 5th roots of unity solving algebraically x^5-1=0. Using the result, find sin18 and cos18

3. The attempt at a solution

$x^5 = 1\\ x = \sqrt[5]{1}$

since we have 5 roots:
$x_k, k = 0,1,2,3,4 \\ \\ x_k = e^{i\frac{2k\pi}{n}}, n=5 \\ x_0 = e^{i0} = 1\\ x_1 = e^{i\frac{2\pi}{5}} = cos\frac{2\pi}{5} + isin\frac{2\pi}{5} \\ x_2 = e^{i\frac{4\pi}{5}} = cos\frac{4\pi}{5} + isin\frac{4\pi}{5} \\ x_3 = e^{i\frac{6\pi}{5}} = cos\frac{6\pi}{5} + isin\frac{6\pi}{5} \\ x_5 = e^{i\frac{8\pi}{5}} = cos\frac{8\pi}{5} + isin\frac{8\pi}{5}$

now how do I find sin18 and cos18??
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 Perhaps you could use de Moivre's formula
 Hi tonit! You have $x_1 = e^{72i} = cos72 + isin72$, where the angle is in degrees. Can you express these in terms of cosine and sine 18? Then use the binomial expansion to the index 5 of the root you get(in terms of cos and sin 18) and equate the imaginary and real coefficients. You will get two equations, one of which you can easily solve for their values.

## find 5th roots of unity solving x^5 -1=0 and use the result for sin18 and cos18

$e^{i\frac{2\pi }{5}} = sin(18) + icos18$ right?

 Quote by tonit $e^{i\frac{2\pi }{5}} = sin(18) + icos18$ right?
Yes!
 I got $sin^5θ - 10sin^3θ cos^2θ + 5sinθ cos^4θ = 1$ and $i5sin^4θcosθ - i10sin^2θcos^3θ + icos^5θ = 0$ where $θ = \frac{\pi}{10} = 18^{\circ}$ is this ok?

 Quote by tonit I got $sin^5θ - 10sin^3θ cos^2θ + 5sinθ cos^4θ = 1$ and $i5sin^4θcosθ - i10sin^2θcos^3θ + icos^5θ = 0$ where $θ = \frac{2\pi}{5} = 18^{\circ}$ is this ok?
Yes!

Now try solving one of these equations, as a single trigonometric ratio. Which one would be easier for you to solve?
 I guess the first one would be easier, and I got $16sin^5\theta - 20sin^3\theta + 5sin\theta = 1$ I'm stuck again.....:@

 Quote by tonit I guess the first one would be easier, and I got $16sin^5\theta - 20sin^3\theta + 5sin\theta = 1$ I'm stuck again.....:@
Actually...the second one would be easier. You have a zero in the RHS, so you can divide the equation by icosθ and get rid of a worry
 alright so after simplifying I get $16sin^4\theta - 12sin^2\theta + 1 = 0$ now it's pretty obvious to solve. thanks :D

 Quote by tonit 1. The problem statement, all variables and given/known data Find 5th roots of unity solving algebraically x^5-1=0. Using the result, find sin18 and cos18 3. The attempt at a solution $x^5 = 1\\ x = \sqrt[5]{1}$ since we have 5 roots: $x_k, k = 0,1,2,3,4 \\ \\ x_k = e^{i\frac{2k\pi}{n}}, n=5 \\ x_0 = e^{i0} = 1\\ x_1 = e^{i\frac{2\pi}{5}} = cos\frac{2\pi}{5} + isin\frac{2\pi}{5} \\ x_2 = e^{i\frac{4\pi}{5}} = cos\frac{4\pi}{5} + isin\frac{4\pi}{5} \\ x_3 = e^{i\frac{6\pi}{5}} = cos\frac{6\pi}{5} + isin\frac{6\pi}{5} \\ x_5 = e^{i\frac{8\pi}{5}} = cos\frac{8\pi}{5} + isin\frac{8\pi}{5}$ now how do I find sin18 and cos18??
I would use the multiple-angle identity.

 Quote by dimension10 I would use the multiple-angle identity.
Hi dimension10!
Could you please explain how, keeping in mind the OP was asked to derive the result using the roots of unity?

 Quote by Infinitum Hi dimension10! Could you please explain how, keeping in mind the OP was asked to derive the result using the roots of unity?
Oh, I was thinking that he wanted help on finding the 5th roots of unity using the cosine of pi/5 and sine of pi/5.

 Quote by tonit alright so after simplifying I get $16sin^4\theta - 12sin^2\theta + 1 = 0$ now it's pretty obvious to solve. thanks :D
Yes, but just remember you will need to reject one of the values you get as,

$$0 \leq sin^2\theta \leq 1$$
 yeah, I'll keep that in my mind

 Tags complex algebra, roots of unity

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