Question

Find the value of the following trigonometric expression:
$\sin {{12}^{\circ }}\sin {{24}^{\circ }}\sin {{48}^{\circ }}\sin {{84}^{\circ }}$
(a) $\dfrac{2}{5}$
(b) $\dfrac{1}{8}$
(c) $\dfrac{3}{5}$
(d) $\dfrac{1}{16}$

Answer 1

In the above expression, if we look at the terms $\sin {{12}^{\circ }}\sin {{48}^{\circ }}$ and $\sin {{24}^{\circ }}\sin {{84}^{\circ }}$ then you will see that the first expression in one of the two forms is of the form $\sin A\sin \left( {{60}^{\circ }}-A \right)$ and the other expression is of the form $\sin B\sin \left( {{60}^{\circ }}+B \right)$ where $A\And B$ is equal to ${{12}^{\circ }}\And {{24}^{\circ }}$ respectively. Then use the identity of the sine which is equal to $\sin A\sin \left( {{60}^{\circ }}-A \right)\sin \left( {{60}^{\circ }}+A \right)=\dfrac{1}{4}\sin 3A$. After that solve the expression and get the value of the given expression.

Complete step by step answer:
We have given the following trigonometric expression:
$\sin {{12}^{\circ }}\sin {{24}^{\circ }}\sin {{48}^{\circ }}\sin {{84}^{\circ }}$
Rearranging the above equation we get,
$\sin {{12}^{\circ }}\sin {{48}^{\circ }}\sin {{24}^{\circ }}\sin {{84}^{\circ }}$
As you can see the in the above expression $\sin {{12}^{\circ }}\sin {{48}^{\circ }}$ and $\sin {{24}^{\circ }}\sin {{84}^{\circ }}$ then you will see that the first expression in one of the two forms is of the form $\sin A\sin \left( {{60}^{\circ }}-A \right)$ and the other expression is of the form $\sin B\sin \left( {{60}^{\circ }}+B \right)$ where $A\And B$ is equal to ${{12}^{\circ }}\And {{24}^{\circ }}$ respectively.
Now, writing the above expression in the way that we shown above we get,
$\sin {{12}^{\circ }}\sin \left( {{60}^{\circ }}-{{12}^{\circ }} \right)\sin {{24}^{\circ }}\sin \left( {{60}^{\circ }}+{{24}^{\circ }} \right)$……..Eq. (1)
There is an identity of the sine as follows:
$\sin A\sin \left( {{60}^{\circ }}-A \right)\sin \left( {{60}^{\circ }}+A \right)=\dfrac{1}{4}\sin 3A$
We can rewrite the above expression as:
$\sin A\sin \left( {{60}^{\circ }}-A \right)=\dfrac{1}{4}\dfrac{\sin 3A}{\sin \left( {{60}^{\circ }}+A \right)}$
$\sin A\sin \left( {{60}^{\circ }}+A \right)=\dfrac{1}{4}\dfrac{\sin 3A}{\sin \left( {{60}^{\circ }}-A \right)}$
Now, using the above relations in the eq. (1) we get,
$\dfrac{1}{4}\dfrac{\sin 3\left( {{12}^{\circ }} \right)}{\sin \left( {{60}^{\circ }}+{{12}^{\circ }} \right)}\dfrac{1}{4}\dfrac{\sin 3\left( {{24}^{\circ }} \right)}{\sin \left( {{60}^{\circ }}-{{24}^{\circ }} \right)}$
Solving the above expression we get,
$\dfrac{1}{4}\dfrac{\sin {{36}^{\circ }}}{\sin \left( {{72}^{\circ }} \right)}\dfrac{1}{4}\dfrac{\sin {{72}^{\circ }}}{\sin \left( {{36}^{\circ }} \right)}$
Now, as you can see that $\sin {{36}^{\circ }}\And \sin {{72}^{\circ }}$ which is written in the numerator and denominator of the above expression is cancelling out and the remaining expression is:
$\dfrac{1}{4}\left( \dfrac{1}{4} \right)$
Multiplying the above two fractions we get,
$\dfrac{1}{16}$
In the above solution, we have simplified the given trigonometric expression into $\dfrac{1}{16}$.

So, the correct answer is “Option D”.

Note: A thought that might pop in your mind is that how one should know, which property of the trigonometric function should use like in this problem we have used the following property of the sine function:
$\sin A\sin \left( {{60}^{\circ }}-A \right)\sin \left( {{60}^{\circ }}+A \right)=\dfrac{1}{4}\sin 3A$
The answer to this question is that you should first see if any two pairs of sine angles are complementary or supplementary so basically you should see the sum or difference of any two angles like in the above question, the sum or difference of the any two pairs is ${{60}^{\circ }}$ then think which property of the sine is satisfying this relation.