Question

If A = $\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left( {\sin {{96}^0} + \sin {{24}^0}} \right)$ and B = $\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left( {\cos {{48}^0} - \cos {{72}^0}} \right)$, then what is the value of $\dfrac{{\text{A}}}{{\text{B}}}$?
$
  {\text{A}}{\text{. }} - 1 \\
  {\text{B}}{\text{. 0}} \\
  {\text{C}}{\text{. 1}} \\
  {\text{D}}{\text{. 2}} \\
 $

Answer 1

Hint- Here, we will proceed by using the formulas $\sin \left( {{{90}^0} + \theta } \right) = \cos \theta $, $\cos \left( {{\text{A}} + {\text{B}}} \right) = \cos {\text{A}}\cos {\text{B}} - \sin {\text{A}}\sin {\text{B}}$ and $\cos \left( {{\text{A}} - {\text{B}}} \right) = \cos {\text{A}}\cos {\text{B}} + \sin {\text{A}}\sin {\text{B}}$ in order to simplify the expression $\dfrac{{\text{A}}}{{\text{B}}}$.

Complete step-by-step answer:
Given, A = $\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left( {\sin {{96}^0} + \sin {{24}^0}} \right)$ and B = $\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left( {\cos {{48}^0} - \cos {{72}^0}} \right)$
The expression whose value is needed can be obtained by dividing A by B as shown under $
  \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left( {\sin {{96}^0} + \sin {{24}^0}} \right)}}{{\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left( {\cos {{48}^0} - \cos {{72}^0}} \right)}} \\
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left[ {\sin \left( {{{90}^0} + {6^0}} \right) + \sin {{24}^0}} \right]}}{{\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left[ {\cos \left( {{{60}^0} - {{12}^0}} \right) - \cos \left( {{{60}^0} + {{12}^0}} \right)} \right]}}{\text{ }} \to {\text{(1)}} \\
 $
Using the formula $\sin \left( {{{90}^0} + \theta } \right) = \cos \theta $, we can write
$\sin \left( {{{90}^0} + {6^0}} \right) = \cos {6^0}{\text{ }} \to {\text{(2)}}$
As we know that $\cos \left( {{\text{A}} + {\text{B}}} \right) = \cos {\text{A}}\cos {\text{B}} - \sin {\text{A}}\sin {\text{B}}$ and $\cos \left( {{\text{A}} - {\text{B}}} \right) = \cos {\text{A}}\cos {\text{B}} + \sin {\text{A}}\sin {\text{B}}$
Using the above formulas, we can write
$\cos \left( {{{60}^0} + {{12}^0}} \right) = \cos {60^0}\cos {12^0} - \sin {60^0}\sin {12^0}{\text{ }} \to {\text{(3)}}$ and $\cos \left( {{{60}^0} - {{12}^0}} \right) = \cos {60^0}\cos {12^0} + \sin {60^0}\sin {12^0}{\text{ }} \to {\text{(4)}}$
By subtracting equation (3) from equation (4), we get
$
   \Rightarrow \cos \left( {{{60}^0} - {{12}^0}} \right) - \cos \left( {{{60}^0} + {{12}^0}} \right) = \cos {60^0}\cos {12^0} + \sin {60^0}\sin {12^0} - \left( {\cos {{60}^0}\cos {{12}^0} - \sin {{60}^0}\sin {{12}^0}} \right) \\
   \Rightarrow \cos \left( {{{60}^0} - {{12}^0}} \right) - \cos \left( {{{60}^0} + {{12}^0}} \right) = \cos {60^0}\cos {12^0} + \sin {60^0}\sin {12^0} - \cos {60^0}\cos {12^0} + \sin {60^0}\sin {12^0} \\
   \Rightarrow \cos \left( {{{60}^0} - {{12}^0}} \right) - \cos \left( {{{60}^0} + {{12}^0}} \right) = 2\sin {60^0}\sin {12^0}{\text{ }} \to {\text{(5)}} \\
 $
By substituting equations (2) and (5) in equation (1), we have
$
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left[ {\cos {6^0} + \sin {{24}^0}} \right]}}{{\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left[ {2\sin {{60}^0}\sin {{12}^0}} \right]}} \\
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left[ {\cos {6^0} + \sin \left( {{{90}^0} - {{66}^0}} \right)} \right]}}{{\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left[ {2\sin {{60}^0}\sin {{12}^0}} \right]}}{\text{ }} \to {\text{(6)}} \\
 $
Using the formula $\sin \left( {{{90}^0} - \theta } \right) = \cos \theta $, we can write
$\sin \left( {{{90}^0} - {{66}^0}} \right) = \cos {66^0}{\text{ }} \to {\text{(7)}}$
By substituting equation (7) in equation (6), we get
$
 \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {\cos {{12}^0} - \cos {{36}^0}} \right)\left[ {\cos {6^0} + \cos {{66}^0}} \right]}}{{\left( {\sin {{60}^0} - \sin {{12}^0}} \right)\left[ {2\sin {{60}^0}\sin {{12}^0}} \right]}}{\text{ }} \to {\text{(8)}}$
Using the formulas $\cos {\text{A}} + \cos {\text{B}} = 2\cos \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$, $\cos {\text{A}} - \cos {\text{B}} = - 2\sin \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\sin \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ and $\sin {\text{A}} - \sin {\text{B}} = 2\cos \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\sin \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ , we can write
$
   \Rightarrow \cos {6^0} + \cos {66^0} = 2\cos \left( {\dfrac{{{6^0} + {{66}^0}}}{2}} \right)\cos \left( {\dfrac{{{6^0} - {{66}^0}}}{2}} \right) \\
   \Rightarrow \cos {6^0} + \cos {66^0} = 2\cos \left( {\dfrac{{{{72}^0}}}{2}} \right)\cos \left( {\dfrac{{ - {{60}^0}}}{2}} \right) \\
   \Rightarrow \cos {6^0} + \cos {66^0} = 2\cos {36^0}\cos \left( { - {{30}^0}} \right) \\
 $
Since, $\cos \left( { - \theta } \right) = \cos \theta $
$ \Rightarrow \cos {6^0} + \cos {66^0} = 2\cos {36^0}\cos {30^0}{\text{ }} \to {\text{(9)}}$
Similarly,
$
  \cos {12^0} - \cos {36^0} = - 2\sin \left( {\dfrac{{{{12}^0} + {{36}^0}}}{2}} \right)\sin \left( {\dfrac{{{{12}^0} - {{36}^0}}}{2}} \right) \\
   \Rightarrow \cos {12^0} - \cos {36^0} = - 2\sin \left( {\dfrac{{{{48}^0}}}{2}} \right)\sin \left( {\dfrac{{ - {{24}^0}}}{2}} \right) \\
   \Rightarrow \cos {12^0} - \cos {36^0} = - 2\sin {24^0}\sin \left( { - {{12}^0}} \right) \\
 $
Using the formula $\sin \left( { - \theta } \right) = - \sin \theta $ in the above equation, we get
$ \Rightarrow \cos {12^0} - \cos {36^0} = 2\sin {24^0}\sin {12^0}{\text{ }} \to {\text{(10)}}$
Similarly,
$
  \sin {60^0} - \sin {12^0} = 2\cos \left( {\dfrac{{{{60}^0} + {{12}^0}}}{2}} \right)\sin \left( {\dfrac{{{{60}^0} - {{12}^0}}}{2}} \right) \\
   \Rightarrow \sin {60^0} - \sin {12^0} = 2\cos \left( {\dfrac{{{{72}^0}}}{2}} \right)\sin \left( {\dfrac{{{{48}^0}}}{2}} \right) \\
   \Rightarrow \sin {60^0} - \sin {12^0} = 2\cos {36^0}\sin {24^0}{\text{ }} \to {\text{(11)}} \\
 $
By substituting equations (9), (10) and (11) in equation (8), we get
 \[
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {2\sin {{24}^0}\sin {{12}^0}} \right)\left[ {2\cos {{36}^0}\cos {{30}^0}} \right]}}{{\left( {2\cos {{36}^0}\sin {{24}^0}} \right)\left[ {2\sin {{60}^0}\sin {{12}^0}} \right]}} \\
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\cos {{30}^0}}}{{\sin {{60}^0}}} \\
 \]
Using $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ and $\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$ in the above equation, we get
$
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}} \\
   \Rightarrow \dfrac{{\text{A}}}{{\text{B}}} = 1 \\
 $
Therefore, the value of $\dfrac{{\text{A}}}{{\text{B}}}$ is equal to 1.
Hence, option C is correct.

Note- In this particular problem, all the unknown angles in the expression $\dfrac{{\text{A}}}{{\text{B}}}$ (which are not available in the general trigonometric table) needs to be removed so that the final expression for $\dfrac{{\text{A}}}{{\text{B}}}$ is in terms of the angles given in the general trigonometric table where the trigonometric functions (i.e., sine, cosine, etc) of these angles are known.