Question

If 14 a = π^c, prove it!
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Answer 1

Question:

If  $14\alpha=\pi^c,$  then prove that,

$\sin\alpha\cdot\sin 3\alpha\cdot\sin 5\alpha\cdot\sin 7\alpha\cdot\sin 9\alpha\cdot\sin 11\alpha\cdot\sin 13\alpha=\dfrac{1}{64}$

$[\text{Note:\ \pi^c \ represents\ \pi \ radian, but I ignore it and take only\ \pi !}]$

Solution:

$14\alpha=\pi\ \ \Longrightarrow\ \ \alpha=\dfrac{\pi}{14}$

So, take the LHS.

$\text{LHS}\ \ \Longrightarrow\ \ \sin\alpha\cdot\sin 3\alpha\cdot\sin 5\alpha\cdot\sin 7\alpha\cdot\sin 9\alpha\cdot\sin 11\alpha\cdot\sin 13\alpha$

Since  $\sin 7\alpha=\sin\left(\dfrac{7\pi}{14}\right)=\sin\left(\dfrac{\pi}{2}\right)=1,$  we can ignore it.

$\sin\alpha\cdot\sin 3\alpha\cdot\sin 5\alpha\cdot\sin 9\alpha\cdot\sin 11\alpha\cdot\sin 13\alpha$

I just rearrange the terms as,

$\sin\alpha\cdot\sin 13\alpha\cdot\sin 3\alpha\cdot\sin 11\alpha\cdot\sin 5\alpha\cdot\sin 9\alpha$

Here some terms are taken as,

$\sin\alpha\cdot\sin(14\alpha-\alpha)\cdot\sin 3\alpha\cdot\sin(14\alpha-3\alpha)\cdot\sin 5\alpha\cdot\sin(14\alpha-5\alpha)\\ \\ \Longrightarrow\ \sin\alpha\cdot\sin(\pi-\alpha)\cdot\sin 3\alpha\cdot\sin(\pi-3\alpha)\cdot\sin 5\alpha\cdot\sin(\pi-5\alpha)$

Here we remember  $\sin(\pi-\theta)=\sin\theta.$  So,

$\sin\alpha\cdot\sin(\pi-\alpha)\cdot\sin 3\alpha\cdot\sin(\pi-3\alpha)\cdot\sin 5\alpha\cdot\sin(\pi-5\alpha)\\ \\ \Longrightarrow\ \sin\alpha\cdot\sin\alpha\cdot\sin 3\alpha\cdot\sin3\alpha\cdot\sin 5\alpha\cdot\sin5\alpha\\ \\ \Longrightarrow\ \sin^2\alpha\cdot\sin^23\alpha\cdot\sin^25\alpha\\ \\ \Longrightarrow\ (\sin\alpha\cdot\sin3\alpha\cdot\sin5\alpha)^2$

Now we have to change each sin term to cos term.

But before we consider  $14\alpha=\pi\ \ \Longrightarrow\ \ 7\alpha=\dfrac{\pi}{2}.$

And  $\sin\left(\dfrac{\pi}{2}-\theta\right)=\cos\theta.$

So,

$\leadsto\ \sin\alpha=-\sin(-\alpha)=-\sin(7\alpha-8\alpha)=-\sin\left(\dfrac{\pi}{2}-8\alpha\right)=-\cos8\alpha\\ \\ \\ \leadsto\ \sin3\alpha=\sin(7\alpha-4\alpha)=\sin\left(\dfrac{\pi}{2}-4\alpha\right)=\cos4\alpha\\ \\ \\ \leadsto\ \sin5\alpha=\sin(7\alpha-2\alpha)=\sin\left(\dfrac{\pi}{2}-2\alpha\right)=\cos2\alpha$

Hence the LHS becomes,

$(-\cos8\alpha\cdot\cos4\alpha\cdot\cos2\alpha)^2$

Now let's simplify by  $2\sin\theta\cdot\cos\theta=\sin(2\theta).$

$(-\cos8\alpha\cdot\cos4\alpha\cdot\cos2\alpha)^2\\ \\ \\ \Longrightarrow\ \dfrac{(-\cos8\alpha\cdot\cos4\alpha)^2\cdot\cos^22\alpha\cdot 4\sin^22\alpha}{4\sin^22\alpha}\\ \\ \\ \Longrightarrow\ \dfrac{(-\cos8\alpha\cdot\cos4\alpha)^2\cdot(2\sin2\alpha\cdot\cos2\alpha)^2}{(2\sin2\alpha)^2}\\ \\ \\ \Longrightarrow\ \dfrac{(-\cos8\alpha\cdot\cos4\alpha)^2\cdot\sin^24\alpha}{(2\sin2\alpha)^2}$

$\Longrightarrow\ \dfrac{(-\cos8\alpha)^2\cdot\cos^24\alpha\cdot\sin^24\alpha\cdot4}{(2\sin2\alpha)^2\cdot4}\\ \\ \\ \Longrightarrow\ \ \dfrac{(-\cos8\alpha)^2\cdot(2\sin4\alpha\cdot\cos4\alpha)^2}{(4\sin2\alpha)^2}\\ \\ \\ \Longrightarrow\ \dfrac{(-\cos8\alpha)^2\cdot\sin^28\alpha}{(4\sin2\alpha)^2}\\ \\ \\ \Longrightarrow\ \dfrac{(\sin8\alpha\cdot(-\cos8\alpha))^2}{(4\sin2\alpha)^2}\\ \\ \\ \Longrightarrow\ \dfrac{(2\sin8\alpha\cdot(-\cos8\alpha))^2}{(8\sin2\alpha)^2}$

$\Longrightarrow\ \dfrac{(-\sin16\alpha)^2}{(8\sin2\alpha)^2}\\ \\ \\ \Longrightarrow\ \dfrac{(-\sin(14\alpha+2\alpha))^2}{64\sin^22\alpha}\\ \\ \\ \Longrightarrow\ \dfrac{(-\sin(\pi+2\alpha))^2}{64\sin^22\alpha}\\ \\ \\ \Longrightarrow\ \dfrac{\sin^22\alpha}{64\sin^22\alpha}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bigg[\because\ \sin(\pi+\theta)=-\sin\theta\bigg]\\ \\ \\ \Longrightarrow\ \dfrac{1}{64}\\ \\ \\ \Longrightarrow\ \text{RHS}$

Hence Proved!