Question

The value of $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}$ is:
a). $\dfrac{1}{16}$
b). $\dfrac{1}{8}$
c). $\dfrac{1}{2}$
d). $1$

Answer 1

Hint: The above question is of trigonometry. In this question we have to find the value of $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}$ but since we do not know the value of any of them so we will use trigonometric properties to solve the above question. We will first convert then into cosine function by using the property $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $ and then we will multiply numerator and denominator both with $\sin \dfrac{\pi }{7}$ and then by using the property $\sin 2A=2\sin A\cos A$ we will simplify it to get the required option.

Complete step by step answer:
We can see that the above question is of trigonometry in which we have to find the value of $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}$ but since we do not know the value of any of then so we will use the trigonometric properties to simplify and solve it.
We will first convert all the sine in $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}$ into cosine.
Since, we know that $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $ so, we can say that:
$\cos \left( \dfrac{\pi }{2}-\dfrac{\pi }{14} \right)=\sin \dfrac{\pi }{14}$
So, $\sin \dfrac{\pi }{14}=\cos \dfrac{3\pi }{7}$
Similarly, we can say that $\cos \left( \dfrac{\pi }{2}-\dfrac{3\pi }{14} \right)=\sin \dfrac{3\pi }{14}$
$\Rightarrow \sin \dfrac{3\pi }{14}=\cos \left( \dfrac{2\pi }{7} \right)$
Similarly, $\Rightarrow \sin \dfrac{5\pi }{14}=\cos \left( \dfrac{\pi }{2}-\dfrac{5\pi }{14} \right)=\cos \left( \dfrac{\pi }{14} \right)$
So, we can write $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}$ as $\cos \left( \dfrac{3\pi }{7} \right)\cos \left( \dfrac{2\pi }{7} \right)\cos \left( \dfrac{\pi }{7} \right)$
Now, we will multiply numerator and denominator with $\sin \dfrac{\pi }{7}$ we will get:
 $=\dfrac{\cos \left( \dfrac{3\pi }{7} \right)\cos \left( \dfrac{2\pi }{7} \right)\cos \left( \dfrac{\pi }{7} \right)\sin \left( \dfrac{\pi }{7} \right)}{\sin \left( \dfrac{\pi }{7} \right)}............(1)$
Since, we know that $\sin 2A=2\sin A\cos A$, so $\sin \left( \dfrac{\pi }{7} \right)\cos \left( \dfrac{\pi }{7} \right)=\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{2}$
So, after putting $\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{2}$, in place of $\sin \left( \dfrac{\pi }{7} \right)\cos \left( \dfrac{\pi }{7} \right)$, we will get:
\[=\dfrac{\cos \left( \dfrac{3\pi }{7} \right)\cos \left( \dfrac{2\pi }{7} \right)\sin \left( \dfrac{2\pi }{7} \right)}{2\sin \left( \dfrac{\pi }{7} \right)}\]
Similarly, we can also write $\sin \left( \dfrac{2\pi }{7} \right)\cos \left( \dfrac{2\pi }{7} \right)$ as $\dfrac{\sin \left( \dfrac{4\pi }{7} \right)}{2}$. So, after replacing we will get:
\[=\dfrac{\cos \left( \dfrac{3\pi }{7} \right)\sin \left( \dfrac{4\pi }{7} \right)}{4\sin \left( \dfrac{\pi }{7} \right)}\]
Now, we will convert $\sin \left( \dfrac{4\pi }{7} \right)$ into $\sin \left( \dfrac{3\pi }{7} \right)$ by using the trigonometric property $\sin \left( \pi -\theta \right)=\sin \theta $
So, we can rewrite the above function as:
\[=\dfrac{\cos \left( \dfrac{3\pi }{7} \right)\sin \left( \dfrac{3\pi }{7} \right)}{4\sin \left( \dfrac{\pi }{7} \right)}\]
Now, again we will use the property $\sin 2A=2\sin A\cos A$, then we will say that $\cos \dfrac{3\pi }{7}\sin \dfrac{3\pi }{7}$ is equal to $\dfrac{\sin \dfrac{6\pi }{7}}{2}$ .
So, after putting $\dfrac{\sin \dfrac{6\pi }{7}}{2}$ in place of $\cos \dfrac{3\pi }{7}\sin \dfrac{3\pi }{7}$ we will get:
\[=\dfrac{\sin \left( \dfrac{6\pi }{7} \right)}{8\sin \left( \dfrac{\pi }{7} \right)}\]
Now, we know that $\sin \left( \pi -\theta \right)=\sin \theta $ .
So, we can write $\sin \left( \pi -\dfrac{6\pi }{7} \right)=\sin \dfrac{6\pi }{7}$ .
$\Rightarrow \sin \left( \dfrac{\pi }{7} \right)=\sin \dfrac{6\pi }{7}$
So, we can write \[\dfrac{\sin \left( \dfrac{6\pi }{7} \right)}{8\sin \left( \dfrac{\pi }{7} \right)}\] as \[\dfrac{\sin \left( \dfrac{\pi }{7} \right)}{8\sin \left( \dfrac{\pi }{7} \right)}\]
Hence, value of $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}$= \[\dfrac{\sin \left( \dfrac{\pi }{7} \right)}{8\sin \left( \dfrac{\pi }{7} \right)}=\dfrac{1}{8}\]

So, the correct answer is “Option b”.

Note: Students are required to memorize all the trigonometric properties and use them efficiently in such a way that they should be able to simplify them and get a simpler form of it. When we are given a sine function then we generally convert them first into cosine and then divide both numerator and denominator by the same sine function to simplify it.