Product and cos

Geometry Level 3

∏ k = 1 14 cos ⁡ ( k π 15 ) = ? \large \displaystyle\prod \limits^{14}_{k=1}\cos \left( \frac{k \pi }{15} \right) =\ ?

Image Credit: Wikimedia László Németh .
0 14 1 2 \frac {1}{2} 2 3 \frac {2}{3} ( 1 2 ) 14 \bigg(\frac {1}{2}\bigg)^{14} − ( 1 2 ) 14 -\bigg(\frac {1}{2}\bigg)^{14} 1

View wiki
Refaat M. Sayed by Refaat M. Sayed , 24, Egypt

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Refaat M. Sayed
Aug 4, 2015

Assume that x = ∏ k = 1 14 cos ⁡ ( k π 15 ) x = \displaystyle \prod \limits^{14}_{k=1}\cos \left( \frac{k\pi }{15} \right) y = ∏ k = 1 14 sin ⁡ ( k π 15 ) y = \displaystyle\prod \limits^{14}_{k=1}\sin\left( \frac{k\pi }{15} \right) x . y = ∏ k = 1 14 sin ⁡ ( k π 15 ) cos ⁡ ( k π 15 ) x. y =\displaystyle\prod \limits^{14}_{k=1}\sin\left( \frac{k\pi }{15} \right) \cos \left( \frac{k\pi }{15} \right) x . y = ∏ k = 1 14 1 2 sin ⁡ ( 2 k π 15 ) = ( 1 2 ) 14 sin ⁡ ( 2 π 15 ) sin ⁡ ( 4 π 15 ) ⋯ sin ⁡ ( 14 π 15 ) sin ⁡ ( 16 π 15 ) sin ⁡ ( 18 π 15 ) ⋯ sin ⁡ ( 28 π 15 ) x. y = \displaystyle\prod_{k=1}^{14} \frac{1}{2} \sin\left( \frac{2k\pi}{15} \right) = \bigg(\frac{1}{2}\bigg)^{14}\sin \left( \frac{2\pi }{15} \right) \sin \left( \frac{4\pi }{15} \right) \cdots \sin \left( \frac{14\pi }{15} \right) \sin \left( \frac{16\pi }{15} \right) \sin \left( \frac{18\pi }{15} \right) \cdots \sin \left( \frac{28\pi }{15} \right) Now : sin ⁡ ( 16 π 15 ) = ( − 1 ) sin ⁡ ( π 15 ) & sin ⁡ ( 18 π 15 ) = ( − 1 ) sin ⁡ ( 3 π 15 ) & ⋯ & sin ⁡ ( 28 π 15 ) = ( − 1 ) sin ⁡ ( 13 π 15 ) \text {Now} :\sin \left( \frac{16\pi }{15} \right) = \left( -1\right) \sin \left( \frac{\pi }{15} \right) \& \sin \left( \frac{18\pi }{15} \right) = \left( -1\right) \sin \left( \frac{3\pi }{15} \right)\& \cdots \&\sin \left( \frac{28\pi }{15} \right) = \left( -1\right) \sin \left( \frac{13\pi }{15} \right) x . y = ( 1 2 ) 14 ( − 1 ) 7 [ sin ⁡ ( 2 π 15 ) sin ⁡ ( 4 π 15 ) ⋯ sin ⁡ ( 14 π 15 ) sin ⁡ ( π 15 ) sin ⁡ ( 3 π 15 ) ⋯ sin ⁡ ( 13 π 15 ) ] x. y = \left( \frac{1}{2} \right) ^{14}\left( -1\right) ^{7} \bigg[ \sin \left( \frac{2\pi }{15} \right) \sin \left( \frac{4\pi }{15} \right) \cdots \sin \left( \frac{14\pi }{15} \right) \sin \left( \frac{\pi }{15} \right) \sin \left( \frac{3\pi }{15} \right) \cdots \sin \left( \frac{13\pi }{15} \right) \bigg] x . y = ( 1 2 ) 14 ( − 1 ) 7 . y x. y =\left( \frac{1}{2} \right) ^{14}\left( -1\right) ^{7}. y so : x = − ( 1 2 ) 14 \text{so} : x =-\left( \frac{1}{2} \right) ^{14}

13 Helpful 1 Interesting 1 Brilliant 0 Confused

Moderator note:

There's a much simpler approach. Are you familiar with Chebyshev Polynomials?

Another simple and direct solution by by using this identity ∏ k = 1 n − 1 cos ⁡ ( k π n ) = sin ⁡ ( π n / 2 ) 2 n − 1 \large \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}

Refaat M. Sayed - 5 years, 10 months ago

Log in to reply

Wow. Nice identity. How to prove this?

Pi Han Goh - 5 years, 10 months ago

Log in to reply

Enjoy Mr. Pi Han Goh....... If n n is even, then cos ⁡ ( π 2 ) = 0 \cos(\frac{\pi}{2})=0 appears in the product w h e n k = n / 2 when k=n/2 and sin ⁡ ( n π 2 ) = 0 \sin(\frac{n\pi}{2})=0

If n n is odd, then combining lim ⁡ z = 1 z n + 1 z + 1 = 1 (2a) \lim_{z=1}\frac{z^n+1}{z+1}=1\tag{2a} z n + 1 z + 1 = ∏ k = 1 n − 1 ( z + e 2 π i k / n ) (2b) \frac{z^n+1}{z+1}=\prod_{k=1}^{n-1}(z+e^{2\pi ik/n})\tag{2b} 1 + e i 2 k π / n = 2 cos ⁡ ( k π / n ) e i k π / n (2c) 1+e^{i2k\pi/n}=2\cos(k\pi/n)e^{ik\pi/n}\tag{2c} and noting that ∑ k = 1 n − 1 k = n ( n − 1 ) 2 \displaystyle\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2} so that ∏ k = 1 n − 1 e i k π / n = ( − 1 ) ( n − 1 ) / 2 \displaystyle\prod_{k=1}^{n-1}e^{ik\pi/n}=(-1)^{(n-1)/2} which matches the sign of sin ⁡ ( π n / 2 ) \sin(\pi n/2) , yields 2 n − 1 ∏ k = 1 n − 1 cos ⁡ ( k π / n ) = ( − 1 ) ( n − 1 ) / 2 = sin ⁡ ( π n / 2 ) 2^{n-1}\prod_{k=1}^{n-1}\cos(k\pi/n)=(-1)^{(n-1)/2}=\sin(\pi n/2)

Refaat M. Sayed - 5 years, 10 months ago

Log in to reply

@Refaat M. Sayed – OH MY GOD YOU ROCK!!

You should post this here !

Pi Han Goh - 5 years, 10 months ago

Log in to reply

@Pi Han Goh – I will do... by the way are you challenge Master???

Refaat M. Sayed - 5 years, 10 months ago

Log in to reply

@Refaat M. Sayed – ahah no. But @Calvin Lin is.

Pi Han Goh - 5 years, 10 months ago

Log in to reply

@Pi Han Goh – OH.... Calvin Lin that is great

Refaat M. Sayed - 5 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...