Trigonometry! #140

Geometry Level 4

Find the value of ∏ n = 1 7 sin ⁡ ( ( 2 n − 1 ) π 14 ) \displaystyle\prod_{n=1}^{7}\sin\left(\frac{(2n-1)\pi}{14}\right)

Enter the reciprocal of your answer.

This problem is part of the set Trigonometry .


The answer is 64.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Mateus Gomes
Feb 13, 2016

∏ n = 1 7 sin ⁡ ( ( 2 n − 1 ) π 14 ) = ∏ n = 1 3 sin ⁡ 2 ( ( 2 n − 1 ) π 14 ) = ( S 1 ) 2 \prod_{n=1}^{7}\sin \left(\frac{(2n-1)\pi}{14}\right)=\prod_{n=1}^{3}\sin^2 \left(\frac{(2n-1)\pi}{14}\right)=({S_1})^2 S 1 = sin ⁡ ( π 14 ) × sin ⁡ ( 3 π 14 ) × sin ⁡ ( 5 π 14 ) S_1=\sin\left(\frac{\pi}{14}\right)\times\sin\left(\frac{3\pi}{14}\right)\times\sin\left(\frac{5\pi}{14}\right) = sin ⁡ ( π 14 ) × cos ⁡ ( π 14 ) × sin ⁡ ( 3 π 14 ) × sin ⁡ ( 5 π 14 ) cos ⁡ ( π 14 ) =\frac{\sin\left(\frac{\pi}{14}\right)\times \cos\left(\frac{\pi}{14}\right)\times\sin\left(\frac{3\pi}{14}\right)\times\sin\left(\frac{5\pi}{14}\right)}{\cos\left(\frac{\pi}{14}\right)} = 1 2 sin ⁡ ( π 7 ) × sin ⁡ ( 3 π 14 ) × sin ⁡ ( 5 π 14 ) cos ⁡ ( π 14 ) =\frac{\frac{1}{2}\sin\left(\frac{\pi}{7}\right)\times\sin\left(\frac{3\pi}{14}\right)\times\sin\left(\frac{5\pi}{14}\right)}{\cos\left(\frac{\pi}{14}\right)} = 1 2 sin ⁡ ( π 7 ) × sin ⁡ ( 3 π 14 ) × cos ⁡ ( π 7 ) sin ⁡ ( 6 π 14 ) =\frac{\frac{1}{2}\sin\left(\frac{\pi}{7}\right)\times\sin\left(\frac{3\pi}{14}\right)\times\cos\left(\frac{\pi}{7}\right)}{\sin\left(\frac{6\pi}{14}\right)}

= 1 4 sin ⁡ ( 2 π 7 ) × sin ⁡ ( 3 π 14 ) 2 sin ⁡ ( 3 π 14 ) cos ⁡ ( 3 π 14 ) =\frac{\frac{1}{4}\sin\left(\frac{2\pi}{7}\right)\times\sin\left(\frac{3\pi}{14}\right)}{2\sin\left(\frac{3\pi}{14}\right)\cos\left(\frac{3\pi}{14}\right)}

= 1 4 sin ⁡ ( 2 π 7 ) 2 cos ⁡ ( 3 π 14 ) =\frac{\frac{1}{4}\sin\left(\frac{2\pi}{7}\right)}{2\cos\left(\frac{3\pi}{14}\right)} S 1 = 1 8 S_1=\frac{1}{8} ( S 1 ) 2 = 1 64 \large\color{#3D99F6}{\boxed{{(S_1)}^{2}=\frac{1}{64}}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I think there is a mistake in your big pi notation. You use k as variable in big pi but n as variable in your expression.

谦艺 伍 - 4 years, 9 months ago

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thanks I edit my solution

Mateus Gomes - 4 years, 9 months ago

@Mateus Gomes ∏ n = 1 7 sin ⁡ ( ( 2 n − 1 ) π 14 ) \displaystyle\prod_{n=1}^{7}\sin\left(\frac{(2n-1)\pi}{14}\right) , where is the term ( sin ^2 (7 pie / 14 )=1) ?

sanyam goel - 3 years, 10 months ago

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