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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

  • See https://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – lab bhattacharjee Feb 07 '18 at 05:31
  • Also my answer in https://math.stackexchange.com/questions/544228/finite-sum-sum-limits-k-1m-1-frac1-sin2-frack-pim – lab bhattacharjee Feb 07 '18 at 05:36
  • @ lab bhattarcharjee in your first link you proved "sum" of tangents such that there arguments are in A.P. and in your second link you evaluated summation of coscec squares such that there angles are in A.P. ......i'm not able to relate both formulas to find my answer ...my purpose is evaluation of "products" cotangents such that there angles are in A.P....sir.please provide solution if possible..... or give me formula for product of cosines such that there angles are in A.P....thank you sir –  Feb 07 '18 at 05:43
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    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? – Gerry Myerson Feb 07 '18 at 06:02
  • @ Gerry Myerson ...but ' angles' are different in both.....the thing is.. i know product of sines (but each of it's angle is double of the angle of whom i know product of i.e,..sines and cosines.) –  Feb 07 '18 at 06:09
  • @veereshpandey, In both cases we have a polynomial equations. We can apply https://brilliant.org/wiki/vietas-formula/ for product as we have applied for sum – lab bhattacharjee Feb 07 '18 at 06:51
  • @ lab bhattacharjee , yes sir, we do have polynomial equations but constant term is missing ....so...without knowing constant term of that equation we can't find product of roots we need to divide constant term by coefficient of highest degree of 'x'...to find product of roots....and even if we somehow know the constant term then it's sign will also change according to (-1)^n where n is odd or even ....'n '.is here degree of equation.. –  Feb 07 '18 at 08:56
  • In the product with $m$ in the denominator, replace $m$ with $2n$, split the product into $k$ from $1$ to $n-1$ and the product from $n+1$ to $2n-1$, and use the symmetry of the sine function around $\pi/2$. – Gerry Myerson Feb 07 '18 at 12:04
  • @ Gerry Myerson,yes, by doing as you said(replacing m by 2n and using formula $cos(90-\theta)=sin\theta$ i'm again ending up with product of sines and product of cosines=$\dfrac{2n}{2^{2n-1}}=\dfrac{n}{2^{2(n-1)}}$ which i was already getting just by using formula sin2x=2sinxcosx earlier .so no progress at all........i got stuck so badly...help me out of it sir.......also one thing more..can you explain that will answer depend on if 'm' is either even or odd ? –  Feb 07 '18 at 16:55
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    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? – Gerry Myerson Feb 07 '18 at 23:04
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    @Gerry Myerson thank you very very much ....sir...... –  Feb 07 '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 ;

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

$\implies P=1$