An Easy Trigo Product

Geometry Level 4

If the value of ∏ k = 1 17 sin ⁡ k π 18 \prod _{ k=1 }^{ 17 }{ \sin { \frac { k\pi }{ 18 } } } can be expressed in the form 3 m 2 n \frac { { 3 }^{ m } }{ { 2 }^{ n } } find the sum ( m + n ) (m+n)


The answer is 18.

Ninad Akolekar by Ninad Akolekar , 23, India

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1 solution

Aman Sharma
Nov 26, 2014

Consider 18th roots of unity:-

α k = cos ⁡ ( 2 k π 18 ) + i sin ⁡ ( 2 k π 18 ) \alpha_k=\cos\left(\frac{2k\pi}{18}\right)+i\sin\left(\frac{2k\pi}{18}\right) Now using the property:- ∏ k = 1 17 ∣ 1 − α k ∣ = 18 \prod_{k=1}^{17} |1-\alpha_k|=18 ∏ k = 1 17 ∣ 1 − cos ⁡ ( 2 k π 18 ) + i sin ⁡ ( 2 k π 18 ) ∣ = 18 \prod_{k=1}^{17} \left|1-\cos\left(\frac{2k\pi}{18}\right)+i\sin\left(\frac{2k\pi}{18}\right)\right|=18 ∏ k = 1 17 2 sin ⁡ ( k π 18 ) ∣ sin ⁡ ( k π 18 ) + i cos ⁡ ( k π 18 ) ∣ = 18 \prod_{k=1}^{17} 2\sin\left(\frac{k\pi}{18}\right)\left|\sin\left(\frac{k\pi}{18}\right)+i\cos\left(\frac{k\pi}{18}\right)\right|=18 ∏ k = 1 17 2 sin ⁡ ( k π 18 ) = 18 \prod_{k=1}^{17} 2\sin\left(\frac{k\pi}{18}\right)=18 2 17 ∏ k = 1 17 sin ⁡ ( k π 18 ) = 18 2^{17}\prod_{k=1}^{17}\sin\left(\frac{k\pi}{18}\right)=18 ∏ k = 1 17 sin ⁡ ( k π 18 ) = 3 2 2 16 \prod_{k=1}^{17}\sin\left(\frac{k\pi}{18}\right)=\frac{3^2}{2^{16}}

6 Helpful 0 Interesting 1 Brilliant 0 Confused

Minor mistake in the first line. Edit it! By the way, can you cite the source from where you got the property that you used in the second line? I'm quite unfamiliar with using roots of unity to solve problems like this one. I solved this problem merely by expanding the product sequence and then using basic trigonometric identities (sum and product formulas)!

Prasun Biswas - 6 years, 6 months ago

Sory, but ∣ 1 − α k ∣ = 18 | 1 - \alpha_{k}| = 18 is a mistake and your solution is wrong

Guillermo Templado - 5 years, 4 months ago

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