Product of three sines - does this approach go anywhere?

Question

THIS IS NOT A HOMEWORK QUESTION

Context
I am trying to simplify the expression $\sin{\frac{\pi}{14}}\times\sin{\frac{3\pi}{14}}\times\sin{\frac{5\pi}{14}}$.

I know from numerical approximations that it is equal to $\frac{1}{8}$ and I can prove this using complex numbers and Euler's formula.

However, when trying to use the product-to-sum identities, things get very messy (working shown below). Can anyone see where I have missed something important or where this can be re-directed?

Approach using product-to-sum identities
$\sin{\frac{\pi}{14}}\times\sin{\frac{3\pi}{14}}\times{\frac{5\pi}{14}}=\frac{1}{2}[\cos{\frac{2\pi}{14}}-\cos{\frac{4\pi}{14}}]\times\sin{\frac{5\pi}{14}}$

$=\frac{1}{2}[\sin{\frac{7\pi}{14}}+\sin{\frac{3\pi}{14}}-\sin{\frac{9\pi}{14}}-\sin{\frac{\pi}{14}}]$

Which seems to be going nowhere.

Grouping the terms and multiplying in a different order seems to lead to the same result.

Because I can see that the product has an exact value of $\frac{1}{8}$, and can be shown using Euler's formula, I would assume that it was possible to show this using product-to-sum and sum-to-product identities, so I am guessing that I have missed an obvious step.

If anyone can identify that obvious step that I have missed it is greatly appreciated.

Answer 1

Since $\sin A=\cos(\frac{\pi}{2}-A)$, we have $$\begin{align}&\sin{\frac{\pi}{14}}\sin{\frac{3\pi}{14}}\sin{\frac{5\pi}{14}} \\\\&=\cos\bigg(\frac{\pi}{2}-\frac{\pi}{14}\bigg)\cos\bigg(\frac{\pi}{2}-\frac{3\pi}{14}\bigg)\cos\bigg(\frac{\pi}{2}-\frac{5\pi}{14}\bigg) \\\\&=\cos\frac{3\pi}{7}\cos\frac{2\pi}{7}\cos\frac{\pi}{7}\tag1\end{align}$$

Multiplying $(1)$ by $\dfrac{8\sin(\pi/7)}{8\sin(\pi/7)}(=1)$ and using $2\sin A\cos A=\sin(2A)$ twice, we have

$$\begin{align}(1)&=\frac{8\sin\frac{\pi}{7}\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}}{8\sin\frac{\pi}{7}} \\\\&=\frac{4\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}}{8\sin\frac{\pi}{7}} \\\\&=\frac{2\sin\frac{4\pi}{7}\cos\frac{3\pi}{7}}{8\sin\frac{\pi}{7}}\tag2\end{align}$$

Since $2\sin A\cos B=\sin(A+B)+\sin(A-B)$, we finally get $$(2)=\frac{\sin\pi+\sin\frac{\pi}{7}}{8\sin\frac{\pi}{7}}=\color{red}{\frac 18}$$

Answer 2

The main ideas here are product to sum trig identities and telescoping sums.

Define for brevity $\,s_n := \sin(n\frac{\pi}{14}),\, c_n := \cos(n\frac{\pi}{14})\,$ and $\,S := s_1 \, s_3 \, s_5.\,$

Use product to sum trig identities and $\,s_7 = 1,\, s_{-n} = -s_n,\, s_{14-n} = s_n \,$ to get

$$ S = \frac14(s_7 + s_3 - s_9 - s_1) = \frac14 - \frac14(s_{-3} + s_1 + s_5). \tag1 $$

Use product to difference trig identities and $\,s_{n-7} = -c_n\,$ to get

$$ 2\,s_2\,s_n = -c_{n+2} + c_{n-2} = s_{n-5} - s_{n-9}. \tag2 $$

Apply this identity to equation $(1)$ and telescope the sum to get

$$ S = \frac14 - \frac14\frac{s_0-s_{-12}}{2\,s_2} = \frac14 - \frac14\frac{s_2}{2\,s_2} = \frac18. \tag3 $$

Answer 3

Trying to derive how the $\dfrac\pi{14}$ was chosen.

If $\sin8x=\sin x,\sin x(8\cos x\cos2x\cos4x-1)=0$

and $\sin x\ne0\implies x\ne n\pi \ \ \ \ (1)$ where $n$ is any integer

$8\cos x\cos2x\cos4x=1$

Case$\#1:$, $8x=2m\pi+x\implies x=\dfrac{2m\pi}7,$ by $(1), m\ne0, m\in[\pm1,\pm2,\pm3]$

$m=\pm1\implies\cos x\cos2x\cos4x=\cos\dfrac{2\pi}7\cos\dfrac{4\pi}7\cos\dfrac{8\pi}7=\cos\dfrac{2\pi}7\left(-\cos\dfrac{3\pi}7\right)\left(-\cos\dfrac\pi7\right)$

$m=\pm2\implies\cos x\cos2x\cos4x=\cdots=\left(-\cos\dfrac{3\pi}7\right)\left(-\cos\dfrac\pi7\right)\left(\cos\dfrac{2\pi}7\right)$

$m=\pm3\implies\cos x\cos2x\cos4x=\cdots=\left(-\cos\dfrac\pi7\right)\left(\cos\dfrac{2\pi}7\right)\left(-\cos\dfrac{3\pi}7\right)$

So, all the non-zero values of $m$ give us the same identity $8\cos\dfrac\pi7\cos\dfrac{2\pi}7\cos\dfrac{3\pi}7=1$

Case$\#2:$, $8x=(2m+1)\pi-x\implies x=\dfrac{(2m+1)\pi}9, 0\le m\le8$

By $(1), 9\nmid(2m+1)$

If $2m+1=9q\iff2(m-4)=9(q-1)\iff9\mid(m-4), m-4=9t$(say) for any integer $t$

So in $[0,8],m\ne4$

Observe that if $\dfrac{(2m_1+1)\pi}9+\dfrac{(2m_2+1)\pi}9=2r\pi\iff m_1+m_2=9r-1$ where $r$ is any integer, $\cos\dfrac{(2m_1+1)\pi}9=\cos\dfrac{(2m_2+1)\pi}9$

Setting $r=1, m_1+m_2=8$

$\implies m=0,8\implies\cos x\cos2x\cos4x=\cdots=\cos\dfrac{\pi}9\cos\dfrac{2\pi}9\cos\dfrac{4\pi}9$

$\implies m=3,5\implies\cos x\cos2x\cos4x=\cdots=\left(-\cos\dfrac{2\pi}9\right)\left(-\cos\dfrac{4\pi}9\right)\cos\dfrac{\pi}9$

$\implies m=2,6\implies\cos x\cos2x\cos4x=\cdots=\left(-\cos\dfrac{4\pi}9\right)\left(-\cos\dfrac{\pi}9\right)\cos\dfrac{2\pi}9$

If $2m+1=3p\iff2(m-1)=3(p-1)\implies 3$ divides $m-1, m-1=3p$(say)

$\cos\dfrac{3p\pi}9=\cos\dfrac{p\pi}3$

More trivially if $p=1\iff m=1,\cos\dfrac{(2m+1)\pi}9=?$

For $m=1,\cos x\cos2x\cos4x=?$

So, $m=0,8,2,6,3,5$ all lead to the Morrie's law