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For any integer $m>1$ evaluate the value of $P$:

$$P=\prod_{k=1}^{m-1} \cot{\dfrac{k\pi}{2m}}=\dfrac{\prod_{k=1}^{m-1} \cos{\dfrac{k\pi}{2m}}}{\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{2m}}}=?$$

I don't know how to proceed using complex number's basic properties (like $n$ roots of unity on polygon inscribed in unit circle) without contour integration or advance algebra… I'm class 12th student… I attempted the problem as follows:

My attempt:

we know that

$$\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{m}}=\dfrac{m}{2^{m-1}}$$

$$2^{m-1}\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{2m}} × \cos\dfrac{k\pi}{2m}=\dfrac{m}{2^{m-1}}$$

$$\prod_{k=1}^{m-1} \sin{\dfrac{k\pi}{2m}} × \cos\dfrac{k\pi}{2m}=\dfrac{m}{2^{2(m-1)}} $$ but after this step I got stuck because we want ratio of product of cosines and product of sines… and I got product of sines and cosines. I don't know how to relate further… I also tried telescoping series but it didn't work either… Any help will be appreciated, thank you…

Any other method is also welcome

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  • $\begingroup$ See math.stackexchange.com/questions/346368/… $\endgroup$ – lab bhattacharjee Feb 7 '18 at 5:31
  • $\begingroup$ Also my answer in math.stackexchange.com/questions/544228/… $\endgroup$ – lab bhattacharjee Feb 7 '18 at 5:36
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    $\begingroup$ If you know the product of the sines, and you know the product of the sines times the product of the cosines, then don't you know the product of the cosines? $\endgroup$ – Gerry Myerson Feb 7 '18 at 6:02
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    $\begingroup$ I don't see the problem. If you know the product of sines, and you know the product of cosines, how can you not know the product of cotangents? Anyway, if you forget about the sines and cosines, can't you just pair off the cotangents, $\cot\theta$ with $\cot((\pi/2)-\theta)$, and get the answer that way? $\endgroup$ – Gerry Myerson Feb 7 '18 at 23:04
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    $\begingroup$ @Gerry Myerson thank you very very much ....sir...... $\endgroup$ – Faraday Pathak Feb 7 '18 at 23:31
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by hint of Gerry myerson i am able to solve this question

we know $cot\left(\dfrac{\pi}{2}-\theta\right)=tan\theta\implies cot\dfrac{(m-1)\pi}{2m}=tan\dfrac{\pi}{2m};cot\dfrac{(m-2)\pi}{2m}=tan\dfrac{2\pi}{2m};....... $

$\implies P=\prod_{k=1}^{m-1} cot{\dfrac{k\pi}{2m}}=cot\dfrac{\pi}{2m}.cot\dfrac{2\pi}{2m}.cot\dfrac{3\pi}{2m}........cot\dfrac{(m-1)\pi}{2m}$

coupling last term with first term ;

$P=\left(cot\dfrac{\pi}{2m}tan\dfrac{\pi}{2m}\right).\left(cot\dfrac{2\pi}{2m}.tan\dfrac{2\pi}{2m}.\right)....\left(cot\dfrac{\dfrac{(m-1)}{2}\pi}{2m}.tan\dfrac{{\dfrac{(m-1)}{2}}\pi}{2m}\right)=1$

$\implies P=1$

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