Question

The value of \[\sin \dfrac{\pi }{14}\cdot \sin \dfrac{3\pi }{14}\cdot \sin \dfrac{5\pi }{14}\cdot \sin \dfrac{7\pi }{14}\cdot \sin \dfrac{9\pi }{14}\cdot \sin \dfrac{11\pi }{14}\cdot \sin \dfrac{13\pi }{14}\] is equal to:
a. 1
b. \[\dfrac{1}{16}\]
c. \[\dfrac{1}{64}\]
d. None of These

Answer 1

Hint: Express the obtuse angles in forms of smaller acute angles of sine. Then reduce it to simplified forms by changing sine to cosine.

Complete step-by-step answer:

Let, S = \[\sin \dfrac{\pi }{14}\cdot \sin \dfrac{3\pi }{14}\cdot \sin \dfrac{5\pi }{14}\cdot \sin \dfrac{7\pi }{14}\cdot \sin \dfrac{9\pi }{14}\cdot \sin \dfrac{11\pi }{14}\cdot \sin \dfrac{13\pi }{14}\]
Now we know that \[\sin x=\sin \left( \pi -x \right)\] for all real values of x.

So, S = \[\sin \dfrac{\pi }{14}\cdot \sin \dfrac{3\pi }{14}\cdot \sin \dfrac{5\pi }{14}\cdot \sin \dfrac{7\pi }{14}\cdot \sin \left( \pi -\dfrac{9\pi }{14} \right)\cdot \sin \left( \pi -\dfrac{11\pi }{14} \right)\cdot \sin \left( \pi -\dfrac{13\pi }{14} \right)\]
\[\Rightarrow \] S = \[\sin \dfrac{\pi }{14}\cdot \sin \dfrac{3\pi }{14}\cdot \sin \dfrac{5\pi }{14}\cdot \sin \dfrac{7\pi }{14}\cdot \sin \dfrac{5\pi }{14}\cdot \sin \dfrac{3\pi }{14}\cdot \sin \dfrac{\pi }{14}\]

Now we know that \[\sin \dfrac{7\pi }{14}=\sin \dfrac{\pi }{2}=1\].
\[\Rightarrow \] S = \[{{\left( \sin \dfrac{\pi }{14}\cdot \sin \dfrac{3\pi }{14}\cdot \sin \dfrac{5\pi }{14} \right)}^{2}}\]

We know that \[\sin x=\cos \left( \dfrac{\pi }{2}-x \right)\] for all real values of x.
So, now S = \[{{\left( \sin \dfrac{\pi }{14}\cdot \cos \left( \dfrac{\pi }{2}-\dfrac{3\pi }{14} \right)\cdot \cos \left( \dfrac{\pi }{2}-\dfrac{5\pi }{14} \right) \right)}^{2}}\]
\[\Rightarrow \] S = \[{{\left( \sin \dfrac{\pi }{14}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{\pi }{7} \right)}^{2}}\]

We know from multiple angle formulas that \[2\sin x\cos x=\sin 2x\] for all real values of x.
\[\Rightarrow \] S = \[{{\left( \dfrac{1}{2\cos \dfrac{\pi }{14}}\cdot 2\sin \dfrac{\pi }{14}\cdot \cos \dfrac{\pi }{14}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{\pi }{7} \right)}^{2}}\] [Multiplying numerator & denominator by same]
\[\Rightarrow \] S = \[{{\left( \dfrac{1}{2\cos \dfrac{\pi }{14}}\cdot \sin \dfrac{\pi }{7}\cdot \cos \dfrac{2\pi }{7}\cdot \cos \dfrac{\pi }{7} \right)}^{2}}\] [Applying the above double angle formula]
\[\Rightarrow \] S = \[{{\left( \dfrac{1}{4\cos \dfrac{\pi }{14}}\cdot 2\sin \dfrac{\pi }{7}\cdot \cos \dfrac{\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)}^{2}}\] [Repeating the process]
\[\Rightarrow \] S = \[{{\left( \dfrac{1}{4\cos \dfrac{\pi }{14}}\cdot \sin \dfrac{2\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)}^{2}}\] [Applying the above double angle formula]
\[\Rightarrow \] S = \[{{\left( \dfrac{1}{8\cos \dfrac{\pi }{14}}\cdot 2\sin \dfrac{2\pi }{7}\cdot \cos \dfrac{2\pi }{7} \right)}^{2}}\] [Repeating the process]
\[\Rightarrow \] S = \[{{\left( \dfrac{1}{8\cos \dfrac{\pi }{14}}\cdot \sin \dfrac{4\pi }{7} \right)}^{2}}\] [Applying the above double angle formula]

Now \[\sin \dfrac{4\pi }{7}=\cos \left( \dfrac{\pi }{2}-\dfrac{4\pi }{7} \right)=\cos \left( -\dfrac{\pi }{14} \right)\].

However, cosine is an even function. So \[\cos (-x)=\cos x\] for all real values of x.
\[\Rightarrow \]S = \[{{\left( \dfrac{1}{8\cos \dfrac{\pi }{14}}\cdot \cos \dfrac{\pi }{14} \right)}^{2}}\] = \[{{\left( \dfrac{1}{8} \right)}^{2}}=\dfrac{1}{64}\].

Hence, we get the value of S. This is the required value for the given product.

Hence, the correct answer to the given question is option (c) \[\dfrac{1}{64}\].

Note: Expressing large multiples into smaller ones will lead the process simpler and make the calculations easier. You can also assume the common value as x and apply the same rules of multiples angles, and finally put back the value to get the desired result.