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I hope this message finds you well. I am currently working on a challenging trigonometric problem and am seeking the collective wisdom and expertise of this community to help me find an efficient and elegant solution. The problem at hand involves evaluating the following intricate trigonometric expression:

$$\cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{3\pi}{11}\right)\cdots\cos\left(\frac{10\pi}{11}\right) = -2^n $$

Find the value of $n$

Approach: Allow me to share my current approach, which involves the method of the product of cosines and subsequent algebraic manipulation. However, I am keen to explore whether there exists a more concise or elegant approach or a shortcut method to arrive at the value of $n.$

Here is my current approach:

Let's consider $ \theta = \frac{\pi}{11} $, then $ 11\theta = \pi $

$ \Rightarrow \cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{3\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\cos\left(\frac{5\pi}{11}\right)\cos\left(\frac{6\pi}{11}\right)\cos\left(\frac{7\pi}{11}\right)\cos\left(\frac{8\pi}{11}\right)\cos\left(\frac{9\pi}{11}\right)\cos\left(\frac{10\pi}{11}\right) = -2^n $

$\Rightarrow \cos\theta \cos(2\theta) \cos(3\theta) \cos(4\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(8\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$\Rightarrow \frac{1}{2\sin(\theta)} \cdot 2 \sin(\theta) \cos(\theta)\cos(2\theta) \cos(3\theta) \cos(4\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(8\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$\Rightarrow \frac{1}{2\sin(\theta)} \cdot \sin(2\theta) \cos(2\theta) \cos(3\theta) \cos(4\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(8\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{1}{2\sin(\theta)}\frac{1}{2} \cdot 2 \sin(2\theta) \cos(2\theta) \cos(3\theta) \cos(4\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(8\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{1}{2^2\sin(\theta)} \cdot \sin(4\theta)\cos(4\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(8\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$\Rightarrow \frac{1}{2^2\sin(\theta)} \cdot \frac{1}{2} \cdot 2 \sin(4\theta)\cos(4\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(8\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{1}{2^3\sin(\theta)} \cdot \sin(8\theta)\cos(8\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{1}{2^3\sin(\theta)} \cdot \frac{1}{2} \cdot 2 \sin(8\theta)\cos(8\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{1}{2^4\sin(\theta)} \cdot \sin(16\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{1}{2^4\sin(\theta)} \cdot \sin(11\theta+ 5\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$\Rightarrow \frac{1}{2^4\sin(\theta)} \cdot \sin(\pi + 5\theta) \cos(3\theta) \cos(5\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$\Rightarrow \frac{1}{2^4\sin(\theta)} -\sin( 5\theta)\cos(5\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$\Rightarrow \frac{-1}{2^4\sin(\theta)}\cdot \frac{1}{2} \cdot 2 \sin( 5\theta)\cos(5\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) \cos(10\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^5 \sin(\theta)}\cdot \sin(10\theta)\cos(10\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^5 \sin(\theta)}\cdot \frac{1}{2} \cdot 2 \sin(10\theta)\cos(10\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^6 \sin(\theta)}\cdot \sin(20\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^6 \sin(\theta)}\cdot\sin(11\theta+9\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^6 \sin(\theta)}\cdot\sin(\pi +9\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) \cos(9\theta) = -2^n $

$\Rightarrow\frac{-1}{2^6 \sin(\theta)}\cdot\ -sin(9\theta) \cos(9\theta) \cos(3\theta) \ cos(6\theta) \cos(7\theta) = -2^n $

$ \Rightarrow\frac{1}{2^6 \sin(\theta)}\cdot \frac{1}{2} \cdot 2 \sin(9\theta) \cos(9\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) = -2^n $

$ \Rightarrow\frac{1}{2^7 \sin(\theta)} \cdot \sin(18\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) = -2^n $

$ \Rightarrow\frac{1}{2^7 \sin(\theta)} \cdot \sin(11\theta+7\theta) \cos(3\theta) \cos(6\theta) \cos(7\theta) = -2^n $

$\Rightarrow \frac{1}{2^7 \sin(\theta)} \cdot -sin(7\theta) \cos(7\theta) \cos(3\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^7 \sin(\theta)}\cdot \frac{1}{2} \cdot 2 sin(7\theta) \cos(7\theta) \cos(3\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^8 \sin(\theta)} \cdot sin(14\theta) \cos(3\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^8 \sin(\theta)} \cdot sin(11\theta+3\theta) \cos(3\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^8 \sin(\theta)} \cdot -sin(3\theta) \cos(3\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{1}{2^8 \sin(\theta)} \cdot \frac{1}{2} \cdot 2 sin(3\theta) \cos(3\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{1}{2^9 \sin(\theta)} \cdot sin(6\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{1}{2^9 \sin(\theta)} \cdot sin(6\theta) \cos(6\theta) = -2^n $

$\Rightarrow \frac{1}{2^9 \sin(\theta)}\cdot\frac{1}{2} \cdot 2 \cdot sin(6\theta) \cos(6\theta) = -2^n $

$ \Rightarrow\frac{1}{2^{10} \sin(\theta)}\cdot 2 \cdot sin(12\theta) = -2^n $

$ \Rightarrow\frac{1}{2^{10} \sin(\theta)}\cdot sin(11\theta+\theta) = -2^n $

$ \Rightarrow\frac{1}{2^{10} \sin(\theta)}\cdot -sin(\theta) = -2^n $

$ \Rightarrow\frac{-1}{2^{10} }\cdot 1 = -2^n $

$\Rightarrow-2^{-10} = -2^n $

$\Rightarrow n = -10 $

While my current method is functional but too long, I believe that leveraging the collective expertise of this community might uncover a more efficient solution or a clever mathematical technique. Your insights, shortcuts, or alternative approaches to simplify this expression would benefit me and assist others who encounter similar trigonometric challenges.

Thank you in advance for your time and expertise in helping me tackle this trigonometric conundrum. Your contributions are greatly appreciated.

Blue
  • 75,673
Raihan Sarker
  • 522

2 Answers2

5

Let $\theta=\frac\pi n$, there is an equality $$\prod_{k=1}^{n}\cos k\theta=\begin{cases}-(\frac12)^{n-1}&n=4m+1\\(\frac12)^{n-1}&n=4m+3\end{cases}$$ see this link for a perfect geometric interpretation.

In your case, $n=11$, so$$\prod_{k=1}^{11}\cos(\frac{k\pi}{11})=\frac{1}{2^{10}}$$ $$\prod_{k=1}^{10}\cos(\frac{k\pi}{11})=-\frac{1}{2^{10}}$$

Ricky
  • 3,148
4

Let $C=\prod_{k=1}^{10}\cos(k\pi /11)$ and $S=\prod_{k=1}^{10}\sin(k\pi /11)$. Then $$ CS=\prod_{k=1}^{10}\sin(k\pi /11)\cos(k\pi /11)=\frac{1}{2^{10}}\prod_{k=1}^{10}\sin(2k\pi / 11)=-\frac{1}{2^{10}}S, $$ which together with $S \neq 0$ implies that $C=-2^{-10}$. To justify the last identity, note that $$ \begin{eqnarray} \prod_{k=1}^{10}\sin\frac{2k\pi}{11} &=& \left(\sin\frac{12\pi}{11} \right) \left(\sin\frac{14\pi}{11} \right)\left( \sin\frac{16\pi}{11} \right)\left( \sin\frac{18\pi}{11} \right)\left( \sin\frac{20\pi}{11} \right) \prod_{k=1}^{5}\sin\frac{2k\pi}{11} \\ &=& \left(-\sin\frac{\pi}{11}\right)\left(-\sin\frac{3\pi}{11}\right)\left(-\sin\frac{5\pi}{11}\right)\left(-\sin\frac{7\pi}{11}\right)\left(-\sin\frac{9\pi}{11}\right) \prod_{k=1}^{5}\sin\frac{2k\pi}{11}\\ &=& -\prod_{k=1}^{10}\sin\frac{k\pi}{11} \\ &=& -S, \end{eqnarray} $$ by repeatedly applying $\sin(x+\pi)=-\sin x$.

mjqxxxx
  • 41,358