Question

If $A = \cos {20^0}\cos {40^0}\cos {60^0}\cos {80^0}$
$B = \cos {6^0}\cos {42^0}\cos {66^0}\cos {78^0}$
$C = \cos {36^0}\cos {72^0}\cos {108^0}\cos {144^0}$
then which of the following conditions is true?
A. A > B > C
B. B > C > A
C. C > A > B
D. A = B = C

Answer 1

Hint: The main approach we use in this question is that we will take the expressions one by one and simplify them. For simplifying each expression we will just multiply and divide by whatever factor we might be needing for the expression to get converted into property of $2\sin A\cos A = \sin 2A$ , $2\cos A\cos B = \cos (A + B) + \cos (A - B)$ , or any other property related to trigonometry and then we will just compare the simplified numerical value to arrive at the answer.

Complete step-by-step answer:
It is given that,
$A = \cos {20^0}\cos {40^0}\cos {60^0}\cos {80^0}$
On Multiplying and divide by $2\sin {20^0}$ in numerator and denominator we get,
$A = \dfrac{{(2\sin {{20}^0}\cos {{20}^0})\cos {{40}^0}\cos {{60}^0}\cos {{80}^0}}}{{2\sin {{20}^0}}}$
Using the property, $2\sin A\cos A = \sin 2A$
$A = \dfrac{{\sin {{40}^0}\cos {{40}^0}\cos {{60}^0}\cos {{80}^0}}}{{2\sin {{20}^0}}}$
Now, on multiplying and divide by 2 we get,
$A = \dfrac{{(2\sin {{40}^0}\cos {{40}^0})\cos {{60}^0}\cos {{80}^0}}}{{4\sin {{20}^0}}}$
Again, using the property, $2\sin A\cos A = \sin 2A$
$A = \dfrac{{\sin {{80}^0}\cos {{60}^0}\cos {{80}^0}}}{{4\sin {{20}^0}}} = \dfrac{{\sin {{80}^0}\cos {{80}^0}\cos {{60}^0}}}{{4\sin {{20}^0}}}$
Now, on multiplying and divide by 2 we get,
$A = \dfrac{{2\sin {{80}^0}\cos {{80}^0}\cos {{60}^0}}}{{8\sin {{20}^0}}}$
Again, using the property, $2\sin A\cos A = \sin 2A$
$A = \dfrac{{2\sin {{80}^0}\cos {{80}^0}\cos {{60}^0}}}{{8\sin {{20}^0}}} = \dfrac{{\sin {{160}^0}\cos {{60}^0}}}{{8\sin {{20}^0}}}$
On substituting the value of $\cos {60^0} = \dfrac{1}{2}$
$A = \dfrac{{\sin {{160}^0}}}{{16\sin {{20}^0}}} = \dfrac{{\sin ({{180}^0} - {{20}^0})}}{{16\sin {{20}^0}}}$
We know that, $\sin ({180^0} - A) = \sin A$
$A = \dfrac{{\sin {{20}^0}}}{{16\sin {{20}^0}}} = \dfrac{1}{{16}}$
$\therefore A = \dfrac{1}{{16}}$

Now, $B = \cos {6^0}\cos {42^0}\cos {66^0}\cos {78^0}$
On Multiplying and divide by $4$ in numerator and denominator and also split the 4 in numerator as $2 \times 2$ we get,
$B = \dfrac{{(2\cos {6^0}\cos {{66}^0})(2\cos {{42}^0}\cos {{78}^0})}}{4}$
Using the property, $2\cos A\cos B = \cos (A + B) + \cos (A - B)$
$B = \dfrac{{(\cos ({6^0} + {{66}^0}) + \cos ({6^0} - {{66}^0}))(\cos ({{42}^0} + {{78}^0}) + \cos ({{42}^0} - {{78}^0}))}}{4}$
$B = \dfrac{{(\cos {{72}^0} + \cos ( - {{60}^0}))(\cos {{120}^0} + \cos ( - {{36}^0}))}}{4}$
We know that, $\cos ( - A) = \cos (A)$
$B = \dfrac{{(\cos {{72}^0} + \cos {{60}^0})(\cos {{120}^0} + \cos {{36}^0})}}{4}$
On substituting the value of $\cos {60^0} = \dfrac{1}{2},\cos {72^0} = \dfrac{{\left( {\sqrt 5 - 1} \right)}}{4},\cos {36^0} = \dfrac{{\left( {\sqrt 5 + 1} \right)}}{4},\cos {120^0} = - \left( {\dfrac{1}{2}} \right)$
$B = \dfrac{{\left( {\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right) + \dfrac{1}{2}} \right)\left( {\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right) - \dfrac{1}{2}} \right)}}{4} = \dfrac{{\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 - 1} \right)}}{{64}}$
Using $(a + b)(a - b) = {a^2} - {b^2}$
$B = \dfrac{{{{\left( {\sqrt 5 } \right)}^2} - {1^2}}}{{64}} = \dfrac{{5 - 1}}{{64}} = \dfrac{4}{{64}} = \dfrac{1}{{16}}$
$\therefore B = \dfrac{1}{{16}}$

Now, $C = \cos {36^0}\cos {72^0}\cos {108^0}\cos {144^0}$
Using the property, $\cos ({180^0} - A) = \cos A$
$C = \cos {36^0}\cos {72^0}\cos ({180^0} - {72^0})\cos ({180^0} - {36^0})$
$C = \cos {36^0}\cos {72^0}\cos {72^0}\cos {36^0}$
$C = {\left( {\cos {{36}^0}\cos {{72}^0}} \right)^2}$
On substituting, $\cos {36^0} = \dfrac{{\sqrt 5 + 1}}{4},\cos {72^0} = \dfrac{{\sqrt 5 - 1}}{4}$
$C = {\left( {\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right) \times \left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)} \right)^2}$
Using the property, $(a + b)(a - b) = {a^2} - {b^2}$
$C = {\left( {\dfrac{{{{\left( {\sqrt 5 } \right)}^2} - {1^2}}}{{16}}} \right)^2} = {\left( {\dfrac{4}{{16}}} \right)^2} = {\left( {\dfrac{1}{4}} \right)^2} = \dfrac{1}{{16}}$
$\therefore C = \dfrac{1}{{16}}$
So, now we get to know that $A = B = C = \dfrac{1}{{16}}$
Hence, we can say that $A = B = C$
$\therefore$ Option D. $A = B = C$ is our correct answer.

Note: For such types of questions, we just have to simplify the expressions and compare them. For this we have to use the following properties,
$2\sin A\cos A = \sin 2A$
$\sin ({180^0} - A) = \sin A$
$\cos ({180^0} - A) = \cos A$
$2\cos A\cos B = \cos (A + B) + \cos (A - B)$
$\cos {60^0} = \dfrac{1}{2},\cos {72^0} = \dfrac{{\left( {\sqrt 5 - 1} \right)}}{4},\cos {36^0} = \dfrac{{\left( {\sqrt 5 + 1} \right)}}{4},\cos {120^0} = - \left( {\dfrac{1}{2}} \right)$
$\cos ( - A) = \cos (A)$
The following properties can simplify the expression very well.