"Prove:cos⁡(2π5)+cos⁡(4π5)+cos⁡(6π5)+cos⁡(8π5)=−1" was answered by our community

haciendodedorcp

Answered question

2022-02-24

Prove:

\cos (\frac{2 π}{5}) + \cos (\frac{4 π}{5}) + \cos (\frac{6 π}{5}) + \cos (\frac{8 π}{5}) = - 1

vazen2bl

Beginner2022-02-25Added 9 answers

z^{5} - 1 = 0

, then

(z^{4} + z^{3} + z^{2} + z + 1) (z - 1) = 0

because

z^{5} - 1 = (z^{4} + z^{3} + z^{2} + z + 1) (z - 1)

. This implies that

z^{4} + z^{3} + z^{2} + z + 1 = 0

z - 1 = 0

. Since

z = \cos \frac{2 π}{5} + i \sin \frac{2 π}{5} = e^{i 2 \frac{π}{5}} \neq 1

is a root of

z^{5} - 1 = 0

and

z^{k} = \cos \frac{2 k π}{5} + i \sin \frac{2 k π}{5} = e^{i 2 k \frac{π}{5}}

,
for

k \in [1, 2, 3, 4]

we have

0 = z^{4} + z^{3} + z^{2} + z + 1 = (\cos \frac{8 π}{5} + i \sin \frac{8 π}{5}) + (\cos \frac{6 π}{5} + i \sin \frac{6 π}{5}) + \dots + 1

= (\cos \frac{8 π}{5} + \cos \frac{6 π}{5} + \dots + 1) + i (\sin \frac{8 π}{5} + \cos \frac{6 π}{5} + \dots + \sin \frac{2 π}{5})

\cos \frac{8 π}{5} + \cos \frac{6 π}{5} + \dots + 1 = 0

litoshypinaylh4

Beginner2022-02-26Added 6 answers

\sum_{n = 1}^{4} \cos (\frac{2 n π}{5}) = R \sum_{n = 1}^{4} e^{2 n π \frac{i}{5}} = R \sum_{n = 1}^{4} {(e^{2 π \frac{i}{5}})}^{n}

= R [\frac{e^{2 π \frac{i}{5}} [{(e^{2 π \frac{i}{5}})}^{4} - 1]}{e^{2 π \frac{i}{5}} - 1}]

= R [\frac{e^{2 \frac{π}{5}} (e^{8 π \frac{i}{5}} - 1} {e^{2 π \frac{i}{5}} - 1}}{=} \frac{- \sin (4 \frac{π}{5})}{\sin (\frac{π}{5})}

= \frac{- \sin (π - \frac{π}{5})}{\sin (\frac{π}{5})} = \frac{- \sin (\frac{π}{5})}{\sin (\frac{π}{5})} = - 1

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