Question

Value of $\sin {{12}^{\circ }}\sin {{24}^{\circ }}\sin {{48}^{\circ }}\sin {{84}^{\circ }}$ is equal to
A . $\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$
B . $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}$
C . $\dfrac{3}{15}$
D . None of these

Answer 1

Hint: Given expression is $\sin {{12}^{\circ }}\sin {{24}^{\circ }}\sin {{48}^{\circ }}\sin {{84}^{\circ }}$. We see from the expression that it in the form of $\sin A\sin ({{60}^{\circ }}-A)$ and $\sin B\sin ({{60}^{\circ }}+B)$. Then by using the identities and simplifying the expression we get the value of the given expression. Then we solve the options to match the value with the value of given expression and choose the correct option.

Formula Used:
We use the following trigonometric identity to solve the question:
$\sin A\sin ({{60}^{\circ }}+A)=\dfrac{1}{4}\dfrac{\sin 3A}{\sin ({{60}^{\circ }}-A)}$
And $\cos \theta \cos 2\theta \cos {{2}^{n-1}}\theta =\dfrac{\sin {{2}^{n}}\theta }{{{2}^{n}}\sin \theta }$

Complete Step- by- step Solution:
 We have given that $\sin {{12}^{\circ }}\sin {{24}^{\circ }}\sin {{48}^{\circ }}\sin {{84}^{\circ }}$
Rearranging the above equation as follow:-
$\sin {{12}^{\circ }}\sin {{48}^{\circ }}\sin {{24}^{\circ }}\sin {{84}^{\circ }}$
Now we will see that the first expression in the form of $\sin A\sin ({{60}^{\circ }}-A)$ and the other is in the form of $\sin B\sin ({{60}^{\circ }}+B)$
Where $A={{12}^{\circ }}$and $B={{24}^{\circ }}$
Now we write the above expression as
$\sin {{12}^{\circ }}\sin ({{60}^{\circ }}-{{12}^{\circ }})\sin {{24}^{\circ }}\sin ({{60}^{\circ }}+{{24}^{\circ }})$………………………………. (1)
We know the identity $\sin A\sin ({{60}^{\circ }}-A)\sin ({{60}^{\circ }}+A)=\dfrac{1}{4}\sin 3A$
We can write it as $\sin A\sin ({{60}^{\circ }}+A)=\dfrac{1}{4}\dfrac{\sin 3A}{\sin ({{60}^{\circ }}-A)}$
Now we will use the above identity in equation (1), we get
$\sin {{12}^{\circ }}\sin ({{60}^{\circ }}-{{12}^{\circ }})\sin {{24}^{\circ }}\sin ({{60}^{\circ }}+{{24}^{\circ }})$ = $\dfrac{1}{4}\dfrac{\sin 3({{12}^{\circ }})}{\sin ({{60}^{\circ }}+{{12}^{\circ }})}\dfrac{1}{4}\dfrac{\sin 3({{24}^{\circ }})}{\sin ({{60}^{\circ }}-{{24}^{\circ }})}$
Now we solve the above equation, we get
$\sin {{12}^{\circ }}\sin ({{60}^{\circ }}-{{12}^{\circ }})\sin {{24}^{\circ }}\sin ({{60}^{\circ }}+{{24}^{\circ }})$= $\dfrac{1}{4}\dfrac{\sin ({{36}^{\circ }})}{\sin ({{72}^{\circ }})}\dfrac{1}{4}\dfrac{\sin ({{72}^{\circ }})}{\sin ({{36}^{\circ }})}$
On solving the above equation, we get
$\sin {{12}^{\circ }}\sin ({{60}^{\circ }}-{{12}^{\circ }})\sin {{24}^{\circ }}\sin ({{60}^{\circ }}+{{24}^{\circ }})$ = $\dfrac{1}{4}\times \dfrac{1}{4}$= $\dfrac{1}{16}$
Now we will solve the options to find out which value matches the value of our equation:-
$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$
$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{2}(\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{80}^{\circ }})$
Let ${{20}^{\circ }}=\theta $, then ${{40}^{\circ }}=2\theta \,\And \,{{80}^{\circ }}=4\theta $
Then $\dfrac{1}{2}(\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{80}^{\circ }})$= $\dfrac{1}{2}(\cos \theta \cos 2\theta \cos 4\theta )$
We know $\cos \theta \cos 2\theta \cos {{2}^{n-1}}\theta =\dfrac{\sin {{2}^{n}}\theta }{{{2}^{n}}\sin \theta }$
Hence $\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{2}\left( \dfrac{\sin {{2}^{3}}\theta }{{{2}^{3}}\sin \theta } \right)$
$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{2}\left( \dfrac{\sin 8\theta }{8\sin \theta } \right)$
$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{16}\left( \dfrac{\sin {{160}^{\circ }}}{\sin {{20}^{\circ }}} \right)$
$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{16}\left( \dfrac{\sin ({{180}^{\circ }}-{{20}^{\circ }})}{\sin {{20}^{\circ }}} \right)$
$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{16}\left( \dfrac{\sin ({{20}^{\circ }})}{\sin {{20}^{\circ }}} \right)$
Hence $\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}$= $\dfrac{1}{16}$
 Hence we see that value of option(1) matches the value of given equation.

Thus, Option (A) is correct.

Note: We know many trigonometric identities are related to many trigonometric equations. Remember that sine and tangent are odd functions as both functions are symmetric about the origin. Cosine is an even function since the function is symmetric about the y- axis.