Question

Find the exact value of expression \[\dfrac{{\sin {{40}^ \circ }}}{{\sin {{80}^ \circ }}} + \dfrac{{\sin {{80}^ \circ }}}{{\sin {{20}^ \circ }}} - \dfrac{{\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}\]

Answer 1

Hint: In this we can see trigonometric functions for different values of angles. We can use the following formulas to proceed
$
  \sin 2x = 2\sin x\cos x \\
  2\cos a\cos b = \cos \left( {a + b} \right) + \cos \left( {a - b} \right) \\
  \sin ({90^ \circ } - x) = \cos x \\
 $

Complete answer:
Given,
\[\dfrac{{\sin {{40}^ \circ }}}{{\sin {{80}^ \circ }}} + \dfrac{{\sin {{80}^ \circ }}}{{\sin {{20}^ \circ }}} - \dfrac{{\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}\]
We can Rewrite the equation in simple form
$
  \sin 2x = 2\sin x\cos x \\
   = \dfrac{{\sin {{40}^ \circ }}}{{\sin {{80}^ \circ }}} + \dfrac{{\sin {{80}^ \circ }}}{{\sin {{20}^ \circ }}} - \dfrac{{\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}} \\
   = \dfrac{{\sin {{40}^ \circ }}}{{2\sin {{40}^ \circ }\cos {{40}^ \circ }}} + \dfrac{{2\sin {{20}^ \circ }\cos {{20}^ \circ }\cos {{40}^ \circ }}}{{\sin {{20}^ \circ }}} - \dfrac{{\sin {{20}^ \circ }}}{{2\sin {{20}^ \circ }\cos {{20}^ \circ }}} \\
 $
By further solving after cancelling the like terms in numerator and denominator we will get,
$
   = \dfrac{1}{{2\cos {{40}^ \circ }}} + \dfrac{{4\cos {{20}^ \circ }\cos {{40}^ \circ }}}{1} - \dfrac{1}{{2\cos {{20}^ \circ }}} \\
   = \dfrac{1}{2}\left( {\dfrac{{\cos {{20}^ \circ } - \cos {{40}^ \circ }}}{{\cos {{20}^ \circ }\cos {{40}^ \circ }}}} \right) + 2(\cos {60^ \circ } + \cos {20^ \circ }) \\
 $
As we know that the identity,
$ \cos a - \cos b = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)$
We can solve the equation further,
\[ = \dfrac{1}{2}\left( {\dfrac{{2\sin {{30}^ \circ }\sin {{10}^ \circ }}}{{\cos {{20}^ \circ }\cos {{40}^ \circ }}}} \right) + 2(\cos {20^ \circ }) + 1\]
By multiplying numerator and denominator by $ \sin {20^ \circ }$
\[
   = = \dfrac{1}{2}\left( {\dfrac{{\sin {{20}^ \circ }\sin {{10}^ \circ }}}{{\sin {{20}^ \circ }\cos {{20}^ \circ }\cos {{40}^ \circ }}}} \right) + 2(\cos {20^ \circ }) + 1 \\
  use\,identity, \\
  \,\sin 2x = 2\sin x\cos x\,\, \\
   = \left( {\dfrac{{2\sin {{10}^ \circ }\sin {{20}^ \circ }}}{{\sin {{80}^ \circ }}}} \right) + 2\cos {20^ \circ } + 1 \\
 \]
By using identities of trigonometric function
\[
   = \left( {\dfrac{{2\cos {{80}^ \circ }\sin {{20}^ \circ } + 2\cos {{20}^ \circ }\sin {{10}^ \circ }}}{{\sin {{80}^ \circ }}}} \right) + 1 \\
  use\,identity,\, \\
  \sin ({90^ \circ } - x) = \cos x \\
   = \left( {\dfrac{{2\cos {{10}^ \circ }}}{{\sin {{80}^ \circ }}}} \right) + 1 = 3 \\
 \]
Hence solution of given problem \[\dfrac{{\sin {{40}^ \circ }}}{{\sin {{80}^ \circ }}} + \dfrac{{\sin {{80}^ \circ }}}{{\sin {{20}^ \circ }}} - \dfrac{{\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}\]= 3

Note:
Students must be well equipped with the trigonometric identities they are going to use. Every time by seeing the equation one can understand which equation to reduce in order to get the simpler form.
The reduction of the equation is important to find a step wise solution.