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Geometry Level 5

A = sin ⁡ ( π 55 ) sin ⁡ ( 2 π 55 ) sin ⁡ ( 3 π 55 ) ⋯ sin ⁡ ( 54 π 55 ) B = ( 55 0 ) + ( 55 1 ) + ( 55 2 ) + ⋯ + ( 55 27 ) A=\sin\left(\frac{\pi}{55}\right)\sin\left(\frac{2\pi}{55}\right)\sin\left(\frac{3\pi}{55}\right)\cdots\sin\left(\frac{54\pi}{55}\right)\\B=\binom{55}{0}+\binom{55}{1}+\binom{55}{2}+\cdots+\binom{55}{27} Find 2 A B 2AB .


The answer is 110.

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Rohit Ner by Rohit Ner , 22, India

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1 solution

General expression of A A

n / 2 n − 1 n/2^{n-1}

General expression for B B

2 n − 1 2^{n-1}

Therefore 2 A B 2AB = 2n

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General expression wont help.. Please provide detailed approach.

Rohit Ner - 5 years, 1 month ago

https://math.stackexchange.com/questions/159214/evaluation-of-a-product-of-sines

Bhaskar Pandey - 3 years, 8 months ago

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