n-2-n-1-sin-pi-n-sin-2pi-n-sin-n-1-n-pi-prove-

Question Number 124302 by Dwaipayan Shikari last updated on 02/Dec/20

n = 2^{n - 1} s i n (\frac{π}{n}) s i n (\frac{2 π}{n}) \dots . s i n (\frac{n - 1}{n} π) p r o v e

Answered by mnjuly1970 last updated on 02/Dec/20

s o l u t i o n

z^{n} - 1 = 0 [{r e}^{i n θ} = e^{i 2 k π}, k \in Z]

[r = 1, θ_{k} = \frac{2 k π}{n}, k = 1, \dots, n - 1]

(z - 1) (z - z_{1}) (z - z_{2}) \dots (z - z_{n - 1}) = 0

w h e r e z_{1}, \dots, z_{n - 1} \in C a n d

(z - z_{1}) \dots (z - z_{n - 1}) = z^{n - 1} + z^{n - 2} + \dots + z + 1 (*)

z = 1 b o t h s i d e s o f (*)

(1 - z_{1}) \dots (z - z_{n - 1}) = (1 + \dots + 1) + 1 = n

C = (1 - e^{\frac{i 2 π}{n}}) (1 - e^{\frac{i 4 π}{n}}) \dots (1 - e^{\frac{i (2 n - 2) π}{n}}) = n

∣ C ∣= 1 \Rightarrow∣ 2 {s i n}^{2} (\frac{π}{n}) - 2 i s i n (\frac{π}{n}) c o s (\frac{π}{n}) ∣ \dots ∣ 2 {s i n}^{2} (\frac{(n - 1) π}{n}) - 2 i s i n (\frac{(n - 1) π}{n}) c o s (\frac{(n - 1) π}{n}) ∣= n

\Rightarrow 2^{n - 1} s i n (\frac{π}{n}) s i n (\frac{2 π}{n}) \dots s i n (\frac{(n - 1) π}{n}) = n

\dots \dots \dots ✓

c o r a l l a r y :

\int_{0}^{\frac{π}{2}} l o g (s i n (x)) d x = - \frac{π}{2} l o g (2)

Commented by Dwaipayan Shikari last updated on 02/Dec/20

G r e a t! t h a n k i n g y o u

Commented by mnjuly1970 last updated on 02/Dec/20

g r a t e f u l \dots