# Prove the following trigonometric equation

\[\cos {24^ \circ } + \cos {55^ \circ } + \cos {125^ \circ } + \cos {204^ \circ } + \cos {300^ \circ } = \dfrac{1}{2}\]

Answer

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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.

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.

Last updated date: 19th Sep 2023

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