The value of $\sin 1 º \sin 2 º \sin 3 º . ... \sin 90 º$ equals

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$\frac{3 \sqrt{5}}{2^{89}}$

$\frac{3 \sqrt{5}}{2^{90}}$

$\frac{3 \sqrt{10}}{2^{90}}$

$\frac{3 \sqrt{10}}{2^{89}}$

answer is A.

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Detailed Solution

$L e t 1, α_{1}, α_{2}, . . . . . α_{n} b e n t h r o o t s o f u n i t y$

$x^{n} - 1 = (x - 1) (x - α_{1}) (x - α_{2}) . . . . . . (x - α_{n}) \Rightarrow \frac{x^{n} - 1}{(x - 1)} = (x - α_{1}) (x - α_{2}) . . . . . . (x - α_{n - 1}) \Rightarrow 1 + x + x^{2} + . . . . + x^{n - 1} = (x - α_{2}) . . . . . . (x - α_{n - 1}) ∴ x = 1 \Rightarrow \prod_{i = 1}^{n - 1} (1 - α_{i}) = n {\Rightarrow \prod}_{i = 1}^{n - 1} ((1 - \cos \frac{2 k π}{n}) - i \sin {\frac{2 k π}{n}}_{i}) = n$

${\Rightarrow \prod}_{i = 1}^{n - 1} ((2 \sin^{2} \frac{kπ}{n}) - i 2 \sin {\frac{π}{n}}_{} \cos \frac{kπ}{n}) = n \Rightarrow \prod_{i = 1}^{n - 1} ((2 \sin \frac{kπ}{n}) (\sin \frac{kπ}{n} - i_{} \cos \frac{kπ}{n})) = n t a k i n g m o d u l u s 2^{n - 1} \sin \frac{π}{n} \sin \frac{2 π}{n} \sin \frac{3 π}{n} . . . . . \sin \frac{(n - 1) π}{n} = n l e t n = 180 \sin 1 ° \sin 2 ° . . . . . . \sin 179 ° = \frac{180}{2^{179}} \Rightarrow \sin 1 ° \sin 2 ° . . . \sin 89 ° = \sqrt{\frac{180}{2^{179}}} = \frac{3 \sqrt{10}}{2^{89}}$

$\sqrt{\sin º \sin 2 º . .... \sin 179 º}$
$= \sqrt{\frac{1}{2^{179}} | (1 - cis 2 º) (1 - cis 4 º) . ... (1 - cis 358 º) |}$
$= \sqrt{\frac{1}{2^{179}} | (1 - cis \frac{2 π}{180}) (1 - cis \frac{4 π}{180}) . ... (1 - cis \frac{358 π}{180}) |} (sinθ = | \frac{1 - cis 2 θ}{2} |)$
$= \sqrt{\frac{1}{2^{179}} | \frac{z^{180} - 1}{z - 1}}$
$= \sqrt{\frac{1}{2^{179}} | 1 + Z + Z^{2} + . .. + Z^{179} |}, at z = 1 = \sqrt{\frac{180}{2^{179}}} = \frac{3 \sqrt{10}}{2^{89}}$