Question

Prove the following expression:
$\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$

Answer 1

Hint: We can write the terms together such that they make the pair of $\sin {{10}^{{}^\circ }}\sin {{70}^{{}^\circ }}$ and $\sin {{30}^{{}^\circ }}\sin {{50}^{{}^\circ }}$ now, multiply numerator and denominator of both the pairs by 2. And now, we can simplify them using $2\sin a\times \sin b=\cos \left( a-b \right)-\cos \left( a+b \right)$, $2\cos a\times \cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$ and we can use $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\times \cos \left( \dfrac{a-b}{2} \right)$.

Complete step-by-step answer:
In this question, we have to prove that, $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$. To prove this, first we will consider left hand side, that is, $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}$, which we can write as $\left( \sin {{10}^{\circ }}\sin {{70}^{\circ }} \right)\left( \sin {{30}^{\circ }}\sin {{50}^{\circ }} \right)$
We know that $2\sin a\times \sin b=\cos \left( a-b \right)-\cos \left( a+b \right)$
So to apply this formula we need 2 in numerator with both the terms and to get that we will multiply and divide both terms by 2, so we will get,
$\left( \dfrac{2}{2}\sin {{10}^{\circ }}\sin {{70}^{\circ }} \right)\left( \dfrac{2}{2}\sin {{30}^{\circ }}\sin {{50}^{\circ }} \right)$
Now, applying the formula, $2\sin a\times \sin b=\cos \left( a-b \right)-\cos \left( a+b \right)$, so, we will get L.H.S. as,
$\dfrac{1}{4}\left( \cos \left( {{70}^{\circ }}-{{10}^{\circ }} \right)-\cos \left( {{70}^{\circ }}+{{10}^{\circ }} \right) \right)\left( \cos \left( {{50}^{\circ }}-{{30}^{\circ }} \right)-\cos \left( {{50}^{\circ }}+{{30}^{\circ }} \right) \right)$
$\Rightarrow \dfrac{1}{4}\left( \cos \left( {{60}^{\circ }} \right)-\cos \left( {{80}^{\circ }} \right) \right)\left( \cos \left( {{20}^{\circ }} \right)-\cos \left( {{80}^{\circ }} \right) \right)$
Now, we will open brackets to get more simplified form
$\Rightarrow \dfrac{1}{4}\left( \cos \left( {{60}^{\circ }} \right)\cos \left( {{20}^{\circ }} \right)-\cos \left( {{60}^{\circ }} \right)\cos \left( {{80}^{\circ }} \right)-\cos \left( {{20}^{\circ }} \right)\cos \left( {{80}^{\circ }} \right)+{{\cos }^{2}}\left( {{80}^{\circ }} \right) \right)$
Now, we know $2\cos a\times \cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$, so to apply this formula, we will multiply and divide each term of above expression by 2, so we will get L.H.S. as,
$\dfrac{1}{4}\left( \dfrac{2}{2}\cos \left( {{60}^{\circ }} \right)\cos \left( {{20}^{\circ }} \right)-\dfrac{2}{2}\cos \left( {{60}^{\circ }} \right)\cos \left( {{80}^{\circ }} \right)-\dfrac{2}{2}\cos \left( {{20}^{\circ }} \right)\cos \left( {{80}^{\circ }} \right)+\dfrac{2}{2}{{\cos }^{2}}\left( {{80}^{\circ }} \right) \right)$
Now, we are applying the formula, $2\cos a\times \cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$, so we get L.H.S. as,
$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{\circ }}+\cos {{40}^{\circ }}-\cos {{140}^{\circ }}-\cos {{20}^{\circ }}-\cos {{100}^{\circ }}-\cos {{60}^{\circ }}+\cos {{0}^{\circ }}+\cos {{160}^{\circ }} \right)$
Now, we know that, $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta$, and we can write L.H.S. as $\dfrac{1}{8}\left( \cos {{80}^{\circ }}+\cos {{40}^{\circ }}-\cos \left( {{180}^{\circ }}-{{40}^{\circ }} \right)-\cos {{20}^{\circ }}-\cos \left( {{180}^{\circ }}-{{80}^{\circ }} \right)-\cos {{60}^{\circ }}+\cos {{0}^{\circ }}+\cos \left( {{180}^{\circ }}-{{20}^{\circ }} \right) \right)$
Therefore, L.H.S.= $\dfrac{1}{8}\left( \cos {{80}^{\circ }}+\cos {{40}^{\circ }}+\cos {{40}^{\circ }}-\cos {{20}^{\circ }}+\cos {{80}^{\circ }}-\cos {{60}^{\circ }}+\cos {{0}^{\circ }}-\cos {{20}^{\circ }} \right)$
We know that, $\cos {{60}^{\circ }}=\dfrac{1}{2}$ and $\cos {{0}^{\circ }}=1$, therefore, we can write L.H.S. as,
$\dfrac{1}{8}\left( \cos {{80}^{\circ }}+\cos {{40}^{\circ }}+\cos {{40}^{\circ }}-\cos {{20}^{\circ }}+\cos {{80}^{\circ }}-\dfrac{1}{2}+1-\cos {{20}^{\circ }} \right)$
$\Rightarrow \dfrac{1}{8}\left( 2\cos {{80}^{\circ }}+2\cos {{40}^{\circ }}-2\cos {{20}^{\circ }}+\dfrac{1}{2} \right)$
Here, we will take 2 common from every term inside the bracket, so we will get,
$\Rightarrow \dfrac{2}{8}\left( \cos {{80}^{\circ }}+\cos {{40}^{\circ }}-\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
Now, we will apply $\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\times \cos \left( \dfrac{a-b}{2} \right)$ formula among $\cos {{80}^{\circ }}$and $\cos {{40}^{\circ }}$, so we will get L.H.S. as,
$\dfrac{1}{4}\left( 2\cos \left( \dfrac{{{80}^{\circ }}+{{40}^{\circ }}}{2} \right)\times \cos \left( \dfrac{{{80}^{\circ }}-{{40}^{\circ }}}{2} \right)-\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
On further simplifying this, we will get,
$\Rightarrow \dfrac{1}{4}\left( 2\cos \left( {{60}^{{}^\circ }} \right)\times \cos \left( {{20}^{{}^\circ }} \right)-\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
$\Rightarrow \dfrac{1}{4}\left( \left( 2\cos \left( {{60}^{{}^\circ }} \right)-1 \right)\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
$\Rightarrow \dfrac{1}{4}\left( \left( 2\times \dfrac{1}{2}-1 \right)\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
$\Rightarrow \dfrac{1}{4}\left( \left( 1-1 \right)\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
$\Rightarrow \dfrac{1}{4}\left( \left( 0 \right)\cos {{20}^{\circ }}+\dfrac{1}{4} \right)$
$\Rightarrow \dfrac{1}{4}\left( 0+\dfrac{1}{4} \right)$
$\Rightarrow \dfrac{1}{16}$
= R.H.S.
Hence, we have proved that $\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$

Note: The possible mistakes one can make in the question is that mistake in writing formulas which will mislead our solution. So, we will not be able to prove the desired result. Also, considering the wrong terms while applying formulas will make the question more complicated and lengthier.