Find the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$?
(a) 1
(b) 2
(c) 3
(d) 4
Answer
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Hint: We start solving the problem by assigning the variable for the required value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$. We then make the arrangements in the numerator for the angles present inside the sine terms. We then recall the relation between the sine and cosine functions in the first quadrant as $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $. We use this relation and make the necessary calculations to get the required value.
Complete step by step answer:
According to the problem, we need to find the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$.
Let us assume the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$ be ‘d’.
So, we have $d=\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$.
\[\Rightarrow d=\dfrac{16\sin \left( {{90}^{\circ }}-{{10}^{\circ }} \right)\sin \left( {{90}^{\circ }}-{{25}^{\circ }} \right)\sin \left( {{90}^{\circ }}-{{55}^{\circ }} \right)}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}\] ---(1).
We know that the relation between sine and cosine functions present in the first quadrant is defined as $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $. We use this relation in equation (1).
So, we get \[d=\dfrac{16\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}\].
$\Rightarrow d=\dfrac{16}{4}$.
$\Rightarrow d=4$.
So, we have found the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$ as 4.
So, the correct answer is “Option d”.
Note: We can also use the relation $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ by making the arrangements for the angles present in the denominator. We can also solve this problem as follows.
$\Rightarrow d=\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$.
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\left( 2\sin {{65}^{\circ }}\sin {{35}^{\circ }} \right)}{2\cos {{10}^{\circ }}\left( 2\cos {{55}^{\circ }}\cos {{25}^{\circ }} \right)}$ ---(2).
From the product to sum trigonometric identities, we know that $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ and $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$. We use these relations in equation (2).
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\left( \cos \left( {{65}^{\circ }}-{{35}^{\circ }} \right)-\cos \left( {{65}^{\circ }}+{{35}^{\circ }} \right) \right)}{2\cos {{10}^{\circ }}\left( \cos \left( {{55}^{\circ }}+{{25}^{\circ }} \right)+\cos \left( {{55}^{\circ }}-{{25}^{\circ }} \right) \right)}$.
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\left( \cos {{30}^{\circ }}-\cos {{100}^{\circ }} \right)}{2\cos {{10}^{\circ }}\left( \cos {{80}^{\circ }}+\cos {{30}^{\circ }} \right)}$.
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\cos {{30}^{\circ }}-8\sin {{80}^{\circ }}\cos {{100}^{\circ }}}{2\cos {{10}^{\circ }}\cos {{80}^{\circ }}+2\cos {{10}^{\circ }}\cos {{30}^{\circ }}}$.
$\Rightarrow d=\dfrac{4\left( 2\sin {{80}^{\circ }}\cos {{30}^{\circ }} \right)-4\left( 2\cos {{100}^{\circ }}\sin {{80}^{\circ }} \right)}{2\cos {{80}^{\circ }}\cos {{10}^{\circ }}+2\cos {{30}^{\circ }}\cos {{10}^{\circ }}}$ ---(3).
From the product to sum trigonometric identities, we know that $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ and $2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)$. We use these relations in equation (3).
$\Rightarrow d=\dfrac{4\left( \sin \left( {{80}^{\circ }}+{{30}^{\circ }} \right)+\sin \left( {{80}^{\circ }}-{{30}^{\circ }} \right) \right)-4\left( \sin \left( {{100}^{\circ }}+{{80}^{\circ }} \right)-\sin \left( {{100}^{\circ }}-{{80}^{\circ }} \right) \right)}{\left( \cos \left( {{80}^{\circ }}+{{10}^{\circ }} \right)+\cos \left( {{80}^{\circ }}-{{10}^{\circ }} \right) \right)+\left( \cos \left( {{30}^{\circ }}+{{10}^{\circ }} \right)+\cos \left( {{30}^{\circ }}-{{10}^{\circ }} \right) \right)}$.
\[\Rightarrow d=\dfrac{4\left( \sin {{110}^{\circ }}+\sin {{50}^{\circ }} \right)-4\left( \sin {{180}^{\circ }}-\sin {{30}^{\circ }} \right)}{\left( \cos {{90}^{\circ }}+\cos {{70}^{\circ }} \right)+\left( \cos {{40}^{\circ }}+\cos {{20}^{\circ }} \right)}\].
\[\Rightarrow d=\dfrac{4\left( \sin {{110}^{\circ }}+\sin {{50}^{\circ }}+\sin {{30}^{\circ }} \right)}{\cos {{70}^{\circ }}+\cos {{40}^{\circ }}+\cos {{20}^{\circ }}}\].
\[\Rightarrow d=\dfrac{4\left( \sin \left( {{90}^{\circ }}+{{20}^{\circ }} \right)+\sin \left( {{90}^{\circ }}-{{20}^{\circ }} \right)+\sin \left( {{90}^{\circ }}-{{40}^{\circ }} \right) \right)}{\cos {{70}^{\circ }}+\cos {{40}^{\circ }}+\cos {{20}^{\circ }}}\].
We know that $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ and $\sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta $.
\[\Rightarrow d=\dfrac{4\left( \cos {{70}^{\circ }}+\cos {{20}^{\circ }}+\cos {{40}^{\circ }} \right)}{\cos {{70}^{\circ }}+\cos {{40}^{\circ }}+\cos {{20}^{\circ }}}\].
\[\Rightarrow d=4\].
Complete step by step answer:
According to the problem, we need to find the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$.
Let us assume the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$ be ‘d’.
So, we have $d=\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$.
\[\Rightarrow d=\dfrac{16\sin \left( {{90}^{\circ }}-{{10}^{\circ }} \right)\sin \left( {{90}^{\circ }}-{{25}^{\circ }} \right)\sin \left( {{90}^{\circ }}-{{55}^{\circ }} \right)}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}\] ---(1).
We know that the relation between sine and cosine functions present in the first quadrant is defined as $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $. We use this relation in equation (1).
So, we get \[d=\dfrac{16\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}\].
$\Rightarrow d=\dfrac{16}{4}$.
$\Rightarrow d=4$.
So, we have found the value of $\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$ as 4.
So, the correct answer is “Option d”.
Note: We can also use the relation $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ by making the arrangements for the angles present in the denominator. We can also solve this problem as follows.
$\Rightarrow d=\dfrac{16\sin {{80}^{\circ }}\sin {{65}^{\circ }}\sin {{35}^{\circ }}}{4\cos {{10}^{\circ }}\cos {{25}^{\circ }}\cos {{55}^{\circ }}}$.
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\left( 2\sin {{65}^{\circ }}\sin {{35}^{\circ }} \right)}{2\cos {{10}^{\circ }}\left( 2\cos {{55}^{\circ }}\cos {{25}^{\circ }} \right)}$ ---(2).
From the product to sum trigonometric identities, we know that $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ and $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$. We use these relations in equation (2).
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\left( \cos \left( {{65}^{\circ }}-{{35}^{\circ }} \right)-\cos \left( {{65}^{\circ }}+{{35}^{\circ }} \right) \right)}{2\cos {{10}^{\circ }}\left( \cos \left( {{55}^{\circ }}+{{25}^{\circ }} \right)+\cos \left( {{55}^{\circ }}-{{25}^{\circ }} \right) \right)}$.
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\left( \cos {{30}^{\circ }}-\cos {{100}^{\circ }} \right)}{2\cos {{10}^{\circ }}\left( \cos {{80}^{\circ }}+\cos {{30}^{\circ }} \right)}$.
$\Rightarrow d=\dfrac{8\sin {{80}^{\circ }}\cos {{30}^{\circ }}-8\sin {{80}^{\circ }}\cos {{100}^{\circ }}}{2\cos {{10}^{\circ }}\cos {{80}^{\circ }}+2\cos {{10}^{\circ }}\cos {{30}^{\circ }}}$.
$\Rightarrow d=\dfrac{4\left( 2\sin {{80}^{\circ }}\cos {{30}^{\circ }} \right)-4\left( 2\cos {{100}^{\circ }}\sin {{80}^{\circ }} \right)}{2\cos {{80}^{\circ }}\cos {{10}^{\circ }}+2\cos {{30}^{\circ }}\cos {{10}^{\circ }}}$ ---(3).
From the product to sum trigonometric identities, we know that $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ and $2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)$. We use these relations in equation (3).
$\Rightarrow d=\dfrac{4\left( \sin \left( {{80}^{\circ }}+{{30}^{\circ }} \right)+\sin \left( {{80}^{\circ }}-{{30}^{\circ }} \right) \right)-4\left( \sin \left( {{100}^{\circ }}+{{80}^{\circ }} \right)-\sin \left( {{100}^{\circ }}-{{80}^{\circ }} \right) \right)}{\left( \cos \left( {{80}^{\circ }}+{{10}^{\circ }} \right)+\cos \left( {{80}^{\circ }}-{{10}^{\circ }} \right) \right)+\left( \cos \left( {{30}^{\circ }}+{{10}^{\circ }} \right)+\cos \left( {{30}^{\circ }}-{{10}^{\circ }} \right) \right)}$.
\[\Rightarrow d=\dfrac{4\left( \sin {{110}^{\circ }}+\sin {{50}^{\circ }} \right)-4\left( \sin {{180}^{\circ }}-\sin {{30}^{\circ }} \right)}{\left( \cos {{90}^{\circ }}+\cos {{70}^{\circ }} \right)+\left( \cos {{40}^{\circ }}+\cos {{20}^{\circ }} \right)}\].
\[\Rightarrow d=\dfrac{4\left( \sin {{110}^{\circ }}+\sin {{50}^{\circ }}+\sin {{30}^{\circ }} \right)}{\cos {{70}^{\circ }}+\cos {{40}^{\circ }}+\cos {{20}^{\circ }}}\].
\[\Rightarrow d=\dfrac{4\left( \sin \left( {{90}^{\circ }}+{{20}^{\circ }} \right)+\sin \left( {{90}^{\circ }}-{{20}^{\circ }} \right)+\sin \left( {{90}^{\circ }}-{{40}^{\circ }} \right) \right)}{\cos {{70}^{\circ }}+\cos {{40}^{\circ }}+\cos {{20}^{\circ }}}\].
We know that $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ and $\sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta $.
\[\Rightarrow d=\dfrac{4\left( \cos {{70}^{\circ }}+\cos {{20}^{\circ }}+\cos {{40}^{\circ }} \right)}{\cos {{70}^{\circ }}+\cos {{40}^{\circ }}+\cos {{20}^{\circ }}}\].
\[\Rightarrow d=4\].
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