Find out the answer to "I have to provesin⁡(nθ)sin⁡(θ)=2n−1∏k=1n−1(cos⁡(θ)−cos⁡(πkn))"

Jupellodseple804

Answered question

2022-02-26

I have to prove

\frac{\sin (n θ)}{\sin (θ)} = 2^{n - 1} \prod_{k = 1}^{n - 1} (\cos (θ) - \cos (\frac{π k}{n}))

shotokan0758s

Beginner2022-02-27Added 8 answers

Let

n \in N^{\cdot}

. Let's factor the polynomial

X^{2 n} - 1

:
Its zeros are

e^{i \frac{k π}{n}} k \in [0, 2 n - 1]

, thus, we can write the following :

X^{2 n} - 1 = \prod_{k = 0}^{2 n - 1} (X - e^{i \frac{k π}{n}}) = (X - 1) \prod_{k = 1}^{n - 1} (X - e^{i \frac{k π}{n}}) (X + 1) \prod_{k = n + 1}^{2 n - 1} (X - e^{i \frac{k x}{n}})

= (X^{2} - 1) \prod_{k = 1}^{\prod_{k = 1}^{n - 1}} (X - e^{i \frac{k x}{n}}) \prod_{k = 1}^{n - 1} (X - e^{i \frac{(2 n - k) π}{n}})

= (X^{2} - 1) \prod_{k = 1}^{n - 1} (X - e^{i \frac{k x}{n}}) (X - e^{- i \frac{k π}{n}})

X^{2 n - 1} = (X^{2} - 1) \prod_{k = 1}^{n - 1} (X^{2} - 2 X \cos (\frac{k π}{n}) + 1)

Thus forall

z \in \frac{C}{- 1, 1}

we have

\frac{z^{2 n} - 1}{z^{2} - 1} = \prod_{k = 1}^{n - 1} (z^{2} - 2 z \cos (\frac{k π}{n}) + 1)

Now if we put

z = e^{i θ}

for some

θ \in \frac{R}{π} Z

we have:

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