3

If $\displaystyle A= \prod^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)$ and $\displaystyle B= \sum^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)$ then show that $\displaystyle A\cdot B= 2^{-6}\bigg(\csc \frac{\pi}{22}-1\bigg)$

Attempt: $\displaystyle B= \sum^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)=\cos \frac{\pi}{11}+\cos \frac{2\pi}{11}+\cos \frac{3\pi}{11}+\cos \frac{4\pi}{11}+\cos \frac{5\pi}{11}$

$\displaystyle =\frac{\sin \frac{5\pi}{22}}{\sin \frac{\pi}{22}}\bigg[\cos \bigg(\frac{\pi}{11}+4\frac{\pi}{22}\bigg)\bigg] =\frac{\sin \frac{5\pi}{22}}{\sin \frac{\pi}{22}}\bigg[\cos \frac{6\pi}{22}\bigg]=\frac{1}{2}\frac{-\sin(\frac{11\pi}{22})+\sin\frac{\pi}{11}}{\sin \frac{\pi}{22}}=\frac{1}{2}\bigg[-\csc \frac{\pi}{22}+1\bigg]$

could some help me how to solve it, thanks

Raffaele
  • 26,371
DXT
  • 11,241
  • 1
    see here http://mathworld.wolfram.com/TrigonometryAnglesPi11.html i hope this will help you – Dr. Sonnhard Graubner Nov 24 '17 at 18:56
  • 1
    It might be worth pointing out that your sum for $B$ isn't evaluated correctly. – B. Mehta Nov 24 '17 at 19:08
  • $$\prod^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)=\frac{1}{32}$$ – Raffaele Nov 24 '17 at 19:08
  • 1
    Do you mean a product by the dot between A and B in $A.B$? If so, you could use a LaTeX symbol \cdot in A\cdot B, which renders as $A\cdot B$. – CiaPan Nov 24 '17 at 19:12
  • And with Raffaele's hint, at this stage it looks like you've solved it yourself! – B. Mehta Nov 24 '17 at 19:14
  • https://math.stackexchange.com/questions/1105236/prove-sum-of-cos-pi-11-cos3-pi-11-cos9-pi-11-1-2-using-eulers-fo and https://math.stackexchange.com/questions/1605366/proving-that-sec-frac-pi11-sec-frac2-pi11-sec-frac3-pi11-sec-frac – lab bhattacharjee Nov 25 '17 at 06:55

1 Answers1

3

COMMENT.- This is not properly a hint but a (nice) remark that I can not avoid mentioning. Given the value $x=\dfrac{\pi}{11}$ there are triangles $\triangle{ABC}$ like that of the attached figure intimately linked for which there must be a counterpart of the proposed problem in angles with the sides and bisectors $BB'$and $CC'$ of the triangle (in particular $\triangle{ACC'}$ is isosceles and $\triangle{ABB'}\sim \triangle{ABC}$).

One has all the five angles, $x,2x,3x,4x,5x$ of the problem in $\triangle{ABC}$ because of $5x+4x+2x=\pi$.

enter image description here

Piquito
  • 29,594