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Question: If \( \sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ } = \dfrac{1}{{2a}} \) . Find \...

If sin⁡6∘sin⁡42∘sin⁡66∘sin⁡78∘=12a\sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ } = \dfrac{1}{{2a}} . Find aa

Explanation

Solution

Hint : To find aa, 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 aa. To simplify the left-hand side, we will be using a cofunction identity: sin⁡x=cos⁡(90−x)\sin x = \cos (90 - x) and a double angle formula: sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x.

Complete step by step solution:
The given question is sin⁡6∘sin⁡42∘sin⁡66∘sin⁡78∘=12a\sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ } = \dfrac{1}{{2a}} . We have to find the value of aa .
First, let us only consider Left-hand side,
That is, sin⁡6∘sin⁡42∘sin⁡66∘sin⁡78∘\sin {6^ \circ }\sin {42^\circ }\sin {66^\circ }\sin {78^\circ }
To this let us multiply and divide by 2cos⁡6∘2\cos {6^\circ }
⇒2cos⁡6∘sin⁡6∘sin⁡42∘sin⁡66∘sin⁡78∘2cos⁡6∘\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,
⇒(2sin⁡6∘cos⁡6∘)sin⁡42∘sin⁡66∘sin⁡78∘2cos⁡6∘\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=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
So, the above equation becomes,
⇒(sin⁡2(6∘))sin⁡42∘sin⁡66∘sin⁡78∘2cos⁡6∘\Rightarrow \dfrac{{(\sin 2({6^ \circ }))\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}}
⇒sin⁡12∘sin⁡42∘sin⁡66∘sin⁡78∘2cos⁡6∘\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)\sin x = \cos (90 - x)
So, sin⁡78∘\sin {78^\circ } can be written as:

sin⁡78∘=cos⁡(90−78)∘ ⇒sin⁡78∘=cos⁡12∘   \sin {78^\circ } = \cos {(90 - 78)^\circ } \\\ \Rightarrow \sin {78^\circ } = \cos {12^\circ } \;

Substituting this in sin⁡12∘sin⁡42∘sin⁡66∘sin⁡78∘2cos⁡6∘\dfrac{{\sin {{12}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }\sin {{78}^\circ }}}{{2\cos {6^\circ }}} , we get:
⇒sin⁡12∘sin⁡42∘sin⁡66∘cos⁡12∘2cos⁡6∘\Rightarrow \dfrac{{\sin {{12}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }\cos {{12}^\circ }}}{{2\cos {6^\circ }}}
Again, rearranging the terms, we get
⇒sin⁡12∘cos⁡12∘sin⁡42∘sin⁡66∘2cos⁡6∘\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 22 ,
⇒2sin⁡12∘cos⁡12∘sin⁡42∘sin⁡66∘2×2cos⁡6∘\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=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
⇒sin⁡2×12∘sin⁡42∘sin⁡66∘4cos⁡6∘ ⇒sin⁡24∘sin⁡42∘sin⁡66∘4cos⁡6∘   \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∘\sin 66^\circ as
sin⁡66∘=cos⁡(90−66) ⇒sin⁡66∘=cos⁡24∘   \sin 66^\circ = \cos (90 - 66) \\\ \Rightarrow \sin 66^\circ = \cos {24^\circ } \;
Using this, sin⁡24∘sin⁡42∘sin⁡66∘4cos⁡6∘\dfrac{{\sin {{24}^ \circ }\sin {{42}^\circ }\sin {{66}^\circ }}}{{4\cos {6^\circ }}} becomes:
⇒sin⁡24∘sin⁡42∘cos⁡24∘4cos⁡6∘\Rightarrow \dfrac{{\sin {{24}^ \circ }\sin {{42}^\circ }\cos {{24}^\circ }}}{{4\cos {6^\circ }}}
Again, multiplying and dividing by 22 and rearranging the terms, we write:
⇒2sin⁡24∘cos⁡24∘sin⁡42∘2×4cos⁡6∘\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=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x

⇒sin⁡2×24∘sin⁡42∘8cos⁡6∘ ⇒sin⁡48∘sin⁡42∘8cos⁡6∘   \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∘\sin {48^\circ } can again be written as:
sin⁡48∘=cos⁡(90−48)∘ ⇒sin⁡48∘=cos⁡42∘   \sin {48^\circ } = \cos {(90 - 48)^\circ } \\\ \Rightarrow \sin {48^\circ } = \cos {42^\circ } \;
On substituting this, we get
cos⁡42∘sin⁡42∘8cos⁡6∘\dfrac{{\cos {{42}^ \circ }\sin {{42}^\circ }}}{{8\cos {6^\circ }}}
Multiplying and dividing by 22 ,
2sin⁡42∘cos⁡42∘2×8cos⁡6∘\dfrac{{2\sin {{42}^\circ }\cos {{42}^ \circ }}}{{2 \times 8\cos {6^\circ }}}
using double angle formula sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x , we get:

sin⁡2×42∘16cos⁡6∘ ⇒sin⁡84∘16cos⁡6∘   \dfrac{{\sin 2 \times{{42}^\circ }}}{{16\cos {6^\circ }}} \\\ \Rightarrow \dfrac{{\sin {{84}^\circ }}}{{16\cos {6^\circ }}} \;

sin⁡84∘\sin {84^\circ } can be written as
sin⁡84∘=cos⁡(90−84)∘ ⇒sin⁡84∘=cos⁡6∘   \sin {84^\circ } = \cos {(90 - 84)^\circ } \\\ \Rightarrow \sin {84^\circ } = \cos {6^\circ } \;
Substitute this in the previous step,
⇒cos⁡6∘16cos⁡6∘\Rightarrow \dfrac{{\cos {6^\circ }}}{{16\cos {6^\circ }}}
Since, cos⁡6∘\cos {6^\circ } is a common term in both numerator and denominator, they get cancelled off.
Thus, in the left-hand side we have, 116\dfrac{1}{{16}}
Equating left-hand side and right-hand side, we get:
116=12a\dfrac{1}{{16}} = \dfrac{1}{{2a}}
Taking reciprocal of both the sides and rearranging them, we get
2a=162a = 16
Dividing both the sides by 22 ,
⇒2a2=162 ⇒a=8   \Rightarrow \dfrac{{2a}}{2} = \dfrac{{16}}{2} \\\ \Rightarrow a = 8 \;
Thus, the value of aa is 88 .
So, the correct answer is “a=8”.

Note : Solving the above given problem using the transformation formula sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\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.