Prove-it-sin-pi-2m-1-sin-2pi-2m-1-sin-mpi-2m-1-2m-1-2-m-

Question Number 127206 by MathSh last updated on 27/Dec/20

P r o v e i t :

s i n \frac{π}{2 m + 1} \cdot s i n \frac{2 π}{2 m + 1} \cdot \dots \cdot s i n \frac{m π}{2 m + 1} = \frac{\sqrt{2 m + 1}}{2^{m}}

Answered by Olaf last updated on 28/Dec/20

Ω = \underset{k = 1}{\prod^{k = m}} \sin (\frac{k π}{2 m + 1})

Solve {(z + 1)}^{2 m + 1} = 1 (1)

z_{k} = e^{\frac{2 i k π}{2 m + 1}} - 1, k = 0, 1, 2, \dots, 2 m

z_{k} = 2 {i e}^{\frac{i k π}{2 m + 1}} \sin (\frac{k π}{2 m + 1})

Let P (z) = \underset{k = 0}{\sum^{2 m}} {(z + 1)}^{k} (2)

\Rightarrow P (0) = 2 m + 1 (3)

P (z) = 0 \Leftrightarrow \underset{k = 0}{\sum^{2 m}} {(z + 1)}^{k} = 0

\Leftrightarrow \frac{{(z + 1)}^{2 m + 1} - 1}{(z + 1) - 1} = 0, z \neq 0

\Leftrightarrow z = z_{k}, k = 1, 2, \dots, 2 m (4)

(2) and (4) :

P (z) = \underset{k = 1}{\prod^{2 m}} (z - z_{k})

\Rightarrow P (0) = \underset{k = 1}{\prod^{2 m}} (- z_{k}) = \underset{k = 1}{\prod^{2 m}} z_{k}

P (0) = \underset{k = 1}{\prod^{2 m}} [2 {i e}^{\frac{i k π}{2 m + 1}} \sin (\frac{k π}{2 m + 1})]

P (0) = 2^{2 m} i^{2 m} e^{\frac{2 m (2 m + 1)}{2} . \frac{i π}{2 m + 1}} \underset{k = 1}{\prod^{2 m}} \sin (\frac{k π}{2 m + 1})

P (0) = 2^{2 m} {(- 1)}^{m} e^{i m π} \underset{k = 1}{\prod^{2 m}} \sin (\frac{k π}{2 m + 1})

P (0) = 2^{2 m} \underset{k = 1}{\prod^{2 m}} \sin (\frac{k π}{2 m + 1}) (5)

(3) and (5) :

P (0) = 2 m + 1 = 2^{2 m} \underset{k = 1}{\prod^{2 m}} \sin (\frac{k π}{2 m + 1})

\Rightarrow \underset{k = 1}{\prod^{2 m}} \sin (\frac{k π}{2 m + 1}) = \frac{2 m + 1}{2^{2 m}} (6)

But \sin (\frac{k π}{2 m + 1}) = \sin (π - \frac{k π}{2 m + 1})

= \sin (\frac{(2 m + 1 - k) π}{2 m + 1})

For example :

\sin (\frac{π}{2 m + 1}) = \sin (\frac{2 m π}{2 m + 1})

\Rightarrow \underset{k = 1}{\prod^{2 m}} \sin (\frac{k π}{2 m + 1}) = \underset{k = 1}{\prod^{m}} \sin^{2} (\frac{k π}{2 m + 1})

= {(\underset{k = 1}{\prod^{m}} \sin (\frac{k π}{2 m + 1}))}^{2} = Ω^{2} (7)

(6) and (7) : Ω = \frac{\sqrt{2 m + 1}}{2^{m}}

Commented by MathSh last updated on 28/Dec/20

T h a n k s s i r

Answered by mnjuly1970 last updated on 28/Dec/20

z^{2 m + 1} - 1 = 0, z \in C (r o o t s ::= 1, z_{1}, \dots, z_{2 m})

z = {r e}^{i θ}

r^{2 m + 1} e^{i (2 m + 1) θ} = 1 = 1 . e^{i 2 k π}

θ_{k} = \frac{2 k π}{2 m + 1} \Rightarrow z_{k} = e^{\frac{i (2 k π)}{2 m + 1}} (k \in {0, 1, 2, \dots, 2 m})

(z - 1) (z^{2 m} + z^{2 m - 1} + \dots + z + 1) = 0

P (z) = z^{2 m} + z^{2 m - 1} + \dots + z + 1

P (z) = (z - z_{1}) (z - z_{2}) \dots (z - z_{2 m})

P (1) = (1 - z_{1}) (1 - z_{2}) \dots (1 - z_{2 m}) = 2 m + 1

∣ 1 - z_{1} ∣∣ z - z_{2} ∣ \dots . ∣ z - z_{2 m} ∣= 2 m + 1

∣ 1 - e^{\frac{i 2 π}{2 m + 1}} ∣∣ 1 - e^{\frac{i 4 π}{2 m + 1}} ∣ \dots ∣ 1 - e^{\frac{i (2 (2 m π))}{2 m + 1}} ∣= 2 m + 1

2^{2 m} s i n (\frac{π}{2 m + 1}) s i n (\frac{2 π}{2 m + 1}) \dots s i n (\frac{m π}{2 m + 1}) s i n (\frac{(m + 1) π}{2 m + 1}) \dots s i n (\frac{2 m π}{2 m + 1}) = 2 m + 1 (*)

n o t e :∣ s i n (θ) - i c o s (θ) ∣= 1

n o t e : 1 - c o s (2 θ) = 2 {s i n}^{2} (θ)

n o t e : s i n (2 θ) = 2 s i n (θ) c o s (θ)

n o t e : s i n (\frac{2 m π}{2 m + 1}) = s i n (π - \frac{2 m π}{2 m + 1}) = s i n (\frac{π}{2 m + 1})

⋮

s i n (\frac{(m + 1) π}{2 m + 1}) = s i n (π - \frac{(m + 1) π}{2 m + 1}) = s i n (\frac{m π}{2 m + 1})

e x a m p l e :∣ 1 - e^{\frac{i 2 π}{2 m + 1}} ∣=∣ 1 - c o s (\frac{2 π}{2 m + 1}) - i s i n (\frac{2 π}{2 m + 1}) ∣

=∣ 2 {s i n}^{2} (\frac{π}{2 m + 1}) - 2 i s i n (\frac{π}{2 m + 1}) c o s (\frac{π}{2 m + 1}) ∣

= 2 s i n (\frac{π}{2 m + 1}) ∣ s i n (\frac{π}{2 m + 1}) - i c o s (\frac{π}{2 m + 1}) ∣= 2 s i n (\frac{π}{2 m + 1})

f r o m (*) a n d (n o t e s) =::

2^{2 m} {(s i n (\frac{π}{2 m + 1}) \dots s i n (\frac{m π}{2 m + 1}))}^{2} = 2 m + 1

s i n (\frac{π}{2 m + 1}) \dots s i n (\frac{m π}{2 m + 1}) = \frac{\sqrt{2 m + 1}}{2^{m}} ✓

Commented by MathSh last updated on 28/Dec/20

T h a n k s s i r

Commented by mnjuly1970 last updated on 28/Dec/20

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