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Prove the following trigonometric equation
\[\cos {24^ \circ } + \cos {55^ \circ } + \cos {125^ \circ } + \cos {204^ \circ } + \cos {300^ \circ } = \dfrac{1}{2}\]

Answer Verified Verified
Hint: - Break the angles as a sum of other angles with multiples of \[{90^ \circ }\].
Taking the L.H.S.
$ \Rightarrow \cos {24^ \circ } + \cos {55^ \circ } + \cos {125^ \circ } + \cos {204^ \circ } + \cos {300^ \circ }$ --- (1)
As we know that
$\left[ {\begin{array}{*{20}{c}}
  {\cos \left( {{{180}^ \circ } - {\theta ^ \circ }} \right) = - \cos {\theta ^ \circ }} \\
  {\cos \left( {{{180}^ \circ } + {\theta ^ \circ }} \right) = - \cos {\theta ^ \circ }} \\
  {\cos \left( {{{360}^ \circ } - {\theta ^ \circ }} \right) = \cos {\theta ^ \circ }}
\end{array}} \right]$
So we have:
\[
  \cos {125^ \circ } = \cos \left( {{{180}^ \circ } - {{55}^ \circ }} \right) = - \cos {55^ \circ } \\
  \cos {204^ \circ } = \cos \left( {{{180}^ \circ } + {{24}^ \circ }} \right) = - \cos {24^ \circ } \\
  \cos {300^ \circ } = \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) = \cos {60^ \circ } \\
\]
Putting these values in equation (1) we get,
\[
   \Rightarrow \cos {24^ \circ } + \cos {55^ \circ } - \cos {55^ \circ } - \cos {24^ \circ } + \cos {60^ \circ } \\
   \Rightarrow \cos {60^ \circ } \\
   \Rightarrow \dfrac{1}{2} = R.H.S. \\
\]
Hence the equation is proved.

Note - The following problem can also be solved by putting in the values of each of the terms, but it is easier to solve the problem by breaking the angles as a sum of other angles with multiple of \[{90^ \circ }\]. Also some of the common trigonometric identities must be remembered.

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