Question

If $ \sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ } = \dfrac{1}{{2a}} $ . Find $ a $

Answer 1

Hint: To find $a$, first we should solve the left-hand side and then equating the value obtained on the left-hand side, we will be able to find the value of $a$. To simplify the left-hand side, we will be using a cofunction identity: $\sin x = \cos (90 - x)$ and a double angle formula: $\sin 2x = 2\sin x\cos x$.

Complete step by step solution:
The given question is $ \sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ } = \dfrac{1}{{2a}} $ . We have to find the value of $ a $ .
First, let us only consider Left-hand side,
That is, $ \sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ } $
To this let us multiply and divide by $ 2\cos {6^\circ } $
 $ \Rightarrow \dfrac{{2\cos {6^\circ }\sin {6^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}} $
We will be clubbing the first 3 terms and rearranging them in the numerator,
 $ \Rightarrow \dfrac{{(2\sin {6^\circ }\cos {6^ \circ })\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}} $
This was done so that we could apply double angle formula, that is $ \sin 2x = 2\sin x\cos x $
So, the above equation becomes,
 $ \Rightarrow \dfrac{{(\sin 2({6^ \circ }))\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}} $
 $ \Rightarrow \dfrac{{\sin {{12}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}} $
From cofunction identity, we know that, $ \sin x = \cos (90 - x) $
So, $ \sin {78^\circ } $ can be written as:
 \[
  \sin {78^\circ } = \cos {(90 - 78)^\circ } \\
   \Rightarrow \sin {78^\circ } = \cos {12^\circ } \;
 \]
Substituting this in $ \dfrac{{\sin {{12}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}} $ , we get:
 $ \Rightarrow \dfrac{{\sin {{12}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }\cos {{12}^\circ }}}{{2\cos {6^\circ }}} $
Again, rearranging the terms, we get
 $ \Rightarrow \dfrac{{\sin {{12}^ \circ }\cos {{12}^\circ }\sin {{42}^\circ }\sin {{66}^\circ }}}{{2\cos {6^\circ }}} $
Let us again multiply and divide by $ 2 $ ,
 $ \Rightarrow \dfrac{{2\sin {{12}^ \circ }\cos {{12}^\circ }\sin {{42}^\circ }\sin {{66}^\circ }}}{{2 \times 2\cos {6^\circ }}} $
We shall again make use of double angle formula $ \sin 2x = 2\sin x\cos x $
 $
   \Rightarrow \dfrac{{\sin 2 \times{{12}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }}}{{4\cos {6^\circ }}} \\
   \Rightarrow \dfrac{{\sin {{24}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }}}{{4\cos {6^\circ }}} \;
 $
Following the same procedure as earlier, we may write $ \sin 66^\circ $ as
 $
  \sin 66^\circ = \cos (90 - 66) \\
   \Rightarrow \sin 66^\circ = \cos {24^\circ } \;
 $
Using this, $ \dfrac{{\sin {{24}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }}}{{4\cos {6^\circ }}} $ becomes:
 $ \Rightarrow \dfrac{{\sin {{24}^ \circ }\sin {{42}^\circ }\cos {{24}^\circ }}}{{4\cos {6^\circ }}} $
Again, multiplying and dividing by $ 2 $ and rearranging the terms, we write:
\[ \Rightarrow \dfrac{{2\sin {{24}^ \circ }\cos {{24}^\circ }\sin {{42}^\circ }}}{{2 \times 4\cos {6^\circ }}}\]
We will use double angle formula $ \sin 2x = 2\sin x\cos x $
\[
   \Rightarrow \dfrac{{\sin 2 \times{{24}^ \circ }\sin {{42}^\circ }}}{{8\cos {6^\circ }}} \\
   \Rightarrow \dfrac{{\sin {{48}^ \circ }\sin {{42}^\circ }}}{{8\cos {6^\circ }}} \;
 \]
 $ \sin {48^\circ } $ can again be written as:
 $
  \sin {48^\circ } = \cos {(90 - 48)^\circ } \\
   \Rightarrow \sin {48^\circ } = \cos {42^\circ } \;
 $
On substituting this, we get
\[\dfrac{{\cos {{42}^ \circ }\sin {{42}^\circ }}}{{8\cos {6^\circ }}}\]
Multiplying and dividing by $ 2 $ ,
\[\dfrac{{2\sin {{42}^\circ }\cos {{42}^ \circ }}}{{2 \times 8\cos {6^\circ }}}\]
using double angle formula $ \sin 2x = 2\sin x\cos x $ , we get:
\[
  \dfrac{{\sin 2 \times{{42}^\circ }}}{{16\cos {6^\circ }}} \\
   \Rightarrow \dfrac{{\sin {{84}^\circ }}}{{16\cos {6^\circ }}} \;
 \]
 $ \sin {84^\circ } $ can be written as
 $
  \sin {84^\circ } = \cos {(90 - 84)^\circ } \\
   \Rightarrow \sin {84^\circ } = \cos {6^\circ } \;
 $
Substitute this in the previous step,
\[ \Rightarrow \dfrac{{\cos {6^\circ }}}{{16\cos {6^\circ }}}\]
Since, \[\cos {6^\circ }\] is a common term in both numerator and denominator, they get cancelled off.
Thus, in the left-hand side we have, $ \dfrac{1}{{16}} $
Equating left-hand side and right-hand side, we get:
 $ \dfrac{1}{{16}} = \dfrac{1}{{2a}} $
Taking reciprocal of both the sides and rearranging them, we get
 $ 2a = 16 $
Dividing both the sides by $ 2 $ ,
 $
   \Rightarrow \dfrac{{2a}}{2} = \dfrac{{16}}{2} \\
   \Rightarrow a = 8 \;
 $
Thus, the value of $ a $ is $ 8 $ .
So, the correct answer is “a=8”.

Note: Solving the above given problem using the transformation formula $ \sin A\sin B = \dfrac{1}{2}\left[ {\cos (A - B) - \cos (A + B)} \right] $ will make the solution more complicated. Hence, it is recommended to follow the above method to solve these types of problems.