Prove-the-equality-sin-pi-2n-sin-2pi-2n-sin-n-1-pi-2n-n-2-n-1-

Question Number 19292 by Tinkutara last updated on 08/Aug/17

Prove the equality

\sin \frac{π}{2 n} \sin \frac{2 π}{2 n} \dots \sin \frac{(n - 1) π}{2 n} = \frac{\sqrt{n}}{2^{n - 1}} .

Commented by prakash jain last updated on 09/Aug/17

you can try with the following

substitution

\sin \frac{π}{2 n} = \frac{e^{\frac{i π}{2 n}} - e^{- \frac{i π}{2 n}}}{2 i}

Answered by Tinkutara last updated on 23/Aug/17

x^{2 n} - 1 = (x - 1) (x - α) (x - α^{2}) \dots . (x - α^{2 n - 1}) where α = e^{\frac{2 π}{2 n} i} = e^{\frac{π}{n} i}

On differentiating both sides,

2 {n x}^{2 n - 1} = (x - α) (x - α^{2}) \dots . (x - α^{2 n - 1})

+ (x - 1) (x - α^{2}) (x - α^{3}) \dots . (x - α^{2 n - 1})

+ (x - 1) (x - α) (x - α^{3}) \dots . + \dots .

Put x = 1

2 n = (1 - α) (1 - α^{2}) (1 - α^{3}) \dots . (1 - α^{2 n - 1})

Since 1 - e^{i θ} = 2 \sin \frac{θ}{2} e^{(\frac{θ}{2} - \frac{π}{2}) i}

2 n = (1 - e^{\frac{π}{n} i}) (1 - e^{\frac{2 π}{n} i}) (1 - e^{\frac{3 π}{n} i}) \dots . (1 - e^{\frac{π}{n} (2 n - 1) i})

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

= 2^{2 n - 1} \sin \frac{π}{2 n} \sin \frac{2 π}{2 n} \dots . \sin \frac{(2 n - 1) π}{2 n} e^{i [\frac{π}{2 n} (\frac{2 n (2 n - 1)}{2}) - (2 n - 1) \frac{π}{2}]}

2 n = 2^{2 n - 1} \sin \frac{π}{2 n} \sin \frac{2 π}{2 n} \dots . \sin \frac{(2 n - 1) π}{2 n} e^{0}

\frac{2 n}{2^{2 n - 1}} = \sin \frac{π}{2 n} \sin \frac{2 π}{2 n} \dots . \sin \frac{(n - 1) π}{2 n} \sin \frac{n π}{2 n} \dots . \sin \frac{π}{2 n}

\frac{n}{2^{2 n - 2}} = \sin^{2} \frac{π}{2 n} \sin^{2} \frac{2 π}{2 n} \dots . \sin^{2} \frac{(n - 1) π}{2 n}

\frac{\sqrt{n}}{2^{n - 1}} = \sin \frac{π}{2 n} \sin \frac{2 π}{2 n} \dots . \sin \frac{(n - 1) π}{2 n}