Question & Answer
QUESTION

The value of $\cos {{12}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{72}^{{}^\circ }}\times cos{{96}^{{}^\circ }}$ equals
   (A) $-\dfrac{1}{{{2}^{6}}}$
   (B) $\dfrac{1}{{{2}^{8}}}$
   (C) $\dfrac{1}{{{2}^{7}}}$
   (D) $-\dfrac{1}{{{2}^{7}}}$

ANSWER Verified Verified
Hint: We will use the formula $2\times \sin \theta \times \cos \theta =\sin 2\theta $ to evaluate the whole expression. Also, we will use directly the values of following cosine angles, $\cos {{36}^{{}^\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$ and $\cos {{72}^{{}^\circ }}=\dfrac{\left( \sqrt{5}+1 \right)}{4}$.

Complete step by step answer:
In the given question, we have to find the values of the expression $\cos {{12}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{72}^{{}^\circ }}\times cos{{96}^{{}^\circ }}$. We can write the equation as follows by just rearranging the cosine terms so as to get better view and the path of next procedure \[\cos {{12}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}\].
Now we multiply both numerator and denominator by $2\times \sin {{12}^{{}^\circ }}$ to get the modified equation as $\dfrac{2\times \sin {{12}^{{}^\circ }}\times \cos {{12}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{2\times \sin {{12}^{{}^\circ }}}$. We know that $2\times \sin \theta \times \cos \theta =\sin 2\theta $, therefore applying this formula we get $2\times \sin {{12}^{{}^\circ }}\times \cos {{12}^{{}^\circ }}=\sin {{24}^{{}^\circ }}$, putting in the above modified equation, we get $\dfrac{\sin {{24}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{2\sin {{12}^{{}^\circ }}}$. Now, multiplying the numerator and denominator by 2, we get $\dfrac{2\times \sin {{24}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{2\times 2\times \sin {{12}^{{}^\circ }}}$.
Again using the formulae $2\times \sin \theta \times \cos \theta =\sin 2\theta $, we get $2\times \sin {{24}^{{}^\circ }}\times \cos {{24}^{{}^\circ }}=\sin {{48}^{{}^\circ }}$, Putting it in main equation we get, $\dfrac{\sin {{48}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{4\times \sin {{12}^{{}^\circ }}}$. Similarly, we multiply both numerator and denominator by a factor 2, we get equation as $\dfrac{2\times \sin {{48}^{{}^\circ }}\times cos{{48}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{2\times 4\times \sin {{12}^{{}^\circ }}}$. Again we use the formulae $2\times \sin \theta \times \cos \theta =\sin 2\theta $, we get $2\times \sin {{48}^{{}^\circ }}\times \cos {{48}^{{}^\circ }}=\sin {{96}^{{}^\circ }}$, Putting it in main equation we get, $\dfrac{\sin {{96}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{8\times \sin {{12}^{{}^\circ }}}$.
Now, again we multiply both numerator and denominator by a factor 2, we get equation as $\dfrac{2\times \sin {{96}^{{}^\circ }}\times cos{{96}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{2\times 8\times \sin {{12}^{{}^\circ }}}$. Again we use the formulae $2\times \sin \theta \times \cos \theta =\sin 2\theta $, we get $2\times \sin {{96}^{{}^\circ }}\times \cos {{96}^{{}^\circ }}=\sin {{192}^{{}^\circ }}$, Putting it in main equation we get, $\dfrac{\sin {{192}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{16\times \sin {{12}^{{}^\circ }}}$. Now we can write $\sin {{192}^{{}^\circ }}$ as $\sin \left( {{180}^{{}^\circ }}+{{12}^{{}^\circ }} \right)=-\sin \left( {{12}^{{}^\circ }} \right)$. Putting this value in the equation $\dfrac{\sin {{192}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{16\times \sin {{12}^{{}^\circ }}}$ we get, $\dfrac{-\sin {{12}^{{}^\circ }}\times cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{16\times \sin {{12}^{{}^\circ }}}$, cancelling the sine term $\sin {{12}^{{}^\circ }}$ from numerator and denominator to get the remaining equation as $\dfrac{-cos{{36}^{{}^\circ }}\times cos{{72}^{{}^\circ }}}{16}$.
 Putting the values of $\cos {{36}^{{}^\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$ and $\cos {{72}^{{}^\circ }}=\dfrac{\left( \sqrt{5}+1 \right)}{4}$ in the equation we get $\dfrac{-1}{16}\times \left( \dfrac{\left( \sqrt{5}-1 \right)}{4} \right)\times \left( \dfrac{\left( \sqrt{5}+1 \right)}{4} \right)$ which is equal to $\dfrac{-1}{16}\times \dfrac{1}{16}\left( {{\left( \sqrt{5} \right)}^{2}}-{{\left( 1 \right)}^{2}} \right)$ = $-\dfrac{1}{256}\times (5-1)=-\dfrac{4}{256}=-\dfrac{1}{64}=-\dfrac{1}{{{2}^{6}}}$. Therefore, the correct answer is option (a).


Note: Student may not separate the values, $\cos {{36}^{{}^\circ }}=\dfrac{\left( \sqrt{5}-1 \right)}{4}$ and $\cos {{72}^{{}^\circ }}=\dfrac{\left( \sqrt{5}+1 \right)}{4}$, if they do not know the values of these angles. As a result it will increase their efforts to solve the same question. It is required for the students to memorize the standard values of some of the angles like involved here.