Question

The value of the expression\[(\sin {36^o})(\sin {72^o})(\sin {108^o})(\sin {144^o})\] is
A.$\dfrac{1}{4}$
B.$\dfrac{1}{{16}}$
C.$\dfrac{3}{4}$
D.$\dfrac{5}{{16}}$

Answer 1

Hint: First we will convert all angles to the first quadrant using the properties;
\[\sin ({180^o} - x) = \sin x\]
We will further simplify the expression to convert in the form of cos36° and sin18° as they are defined as having the values
$
  \cos {36^o} = \dfrac{{\sqrt 5 + 1}}{4} \\
  \sin {18^o} = \dfrac{{\sqrt 5 - 1}}{4} \\
  $

Complete step-by-step answer:
Given \[(\sin {36^o})(\sin {72^o})(\sin {108^o})(\sin {144^o})\]
We first use \[\sin ({180^o} - x) = \sin x\], we get,
\[ \Rightarrow (\sin {36^o})(\sin {72^o})[\sin ({180^o} - {72^o})][\sin ({180^o} - {36^o})]\]
\[
   \Rightarrow (\sin {36^o})(\sin {72^o})(\sin {72^o})(\sin {36^o}) \\
   \Rightarrow {\left[ {(\sin {{36}^o})(\sin {{72}^o})} \right]^2} \\
  \]
On Multiplying and dividing by 4, we get,
\[ \Rightarrow \dfrac{1}{4}{\left[ {2(\sin {{36}^o})(\sin {{72}^o})} \right]^2}\]
Now on using \[2\sin A\sin B = [\cos (A - B) - \cos (A + B)]\], we get,
\[
   \Rightarrow \dfrac{1}{4}{\left[ {\cos {{36}^o} - \cos {{108}^o}} \right]^2} \\
   \Rightarrow \dfrac{1}{4}{\left[ {\cos {{36}^o} - \cos ({{90}^o} + {{18}^o})} \right]^2} \\
  \]
On using \[\cos \left( {{{90}^o} + x} \right){\text{ }} = {\text{ }} - \sin x\], we get,

On substituting the value of \[\cos {36^o} = \dfrac{{\sqrt 5 + 1}}{4}\]and \[\sin {18^o} = \dfrac{{\sqrt 5 - 1}}{4}\]we get,
\[
   \Rightarrow \dfrac{1}{4}{\text{ }}{\left[ {\dfrac{{\sqrt 5 {\text{ }} + {\text{ }}1}}{4}{\text{ }} + {\text{ }}\dfrac{{\sqrt 5 {\text{ }} - {\text{ }}1}}{4}} \right]^2} \\
   \Rightarrow \dfrac{1}{4}{\text{ }}{\left[ {\dfrac{{\sqrt 5 {\text{ }} + {\text{ }}1 + \sqrt 5 {\text{ }} - {\text{ }}1}}{4}{\text{ }}} \right]^2} \\
   \Rightarrow \dfrac{1}{4}{\left[ {\dfrac{{2\sqrt 5 }}{4}} \right]^2} \\
  On{\text{ }}squaring{\text{ }}we{\text{ }}get, \\
   \Rightarrow \dfrac{1}{4}\left[ {\dfrac{{4(5)}}{{16}}} \right] \\
   \Rightarrow \dfrac{1}{4}\left( {\dfrac{5}{4}} \right) \\
   \Rightarrow \dfrac{5}{{16}} \\
  \]
Hence, option (D) is correct.

Note: Whenever solving trigonometric expressions if there is any angle not lying in the first quadrant then try to make it in the first quadrant using the formulas and then try to simplify further, it will make the problem easier.
An alternative method to solve is,
\[
   = (\sin {36^o})(\sin {72^o})(\sin {108^o})(\sin {144^o}) \\
   = (\sin {36^o})(\sin {72^o})[\sin ({180^o} - {72^o})][\sin ({180^o} - {36^o})] \\
  u\sin g,{\text{ }}\sin ({180^o} - x) = \sin x \\
   = (\sin {36^o})(\sin {72^o})(\sin {72^o})(\sin {36^o}) \\
   = {\left[ {(\sin {{36}^o})(\sin {{72}^o})} \right]^2} \\
   = {\left[ {(\sin {{36}^o})(\sin ({{90}^o} - {{18}^o}))} \right]^2} \\
  using,{\text{ }}\sin x = \sqrt {1 - {{\cos }^2}x} {\text{ }}and,{\text{ }}\sin ({90^o} - x) = \cos x \\
   = {\left[ {(\sqrt {1 - {{\cos }^2}{{36}^o}} )(\cos {{18}^o})} \right]^2} \\
  using,{\text{ }}\cos x = \sqrt {1 - {{\sin }^2}x} \\
   = {\left[ {(\sqrt {1 - {{\cos }^2}{{36}^o}} )(\sqrt {1 - {{\sin }^2}{{18}^o}} )} \right]^2} \\
  putting{\text{ }}the{\text{ }}value{\text{ }}of{\text{ }}\cos {36^o} = \dfrac{{\sqrt 5 + 1}}{4}{\text{ }}and,{\text{ }}\sin {18^o} = \dfrac{{\sqrt 5 - 1}}{4} \\
   = {\left[ {\sqrt {1 - {{\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)}^2}} \sqrt {1 - {{\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)}^2}} } \right]^2} \\
   = {\left[ {\sqrt {\dfrac{{16 - 5 - 1 - 2\sqrt 5 }}{{16}}} \sqrt {\dfrac{{16 - 5 - 1 + 2\sqrt 5 }}{{16}}} } \right]^2} \\
   = {\left[ {\dfrac{1}{{16}}\sqrt {10 - 2\sqrt 5 } \sqrt {10 + 2\sqrt 5 } } \right]^2} \\
   = \dfrac{{{{10}^2} - {{\left( {2\sqrt 5 } \right)}^2}}}{{256}} \\
   = \dfrac{{100 - 20}}{{256}} \\
   = \dfrac{{80}}{{256}} \\
   = \dfrac{5}{{16}} \\
  \]