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# Prove the following trigonometric equation$\cos {24^ \circ } + \cos {55^ \circ } + \cos {125^ \circ } + \cos {204^ \circ } + \cos {300^ \circ } = \dfrac{1}{2}$  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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CBSE Class 11 Maths Chapter 3 - Trigonometric Functions Formulas  Trigonometric Functions  Trigonometric Ratios  Trigonometric Identities  Trigonometric Equations  Trigonometry Values  Inverse Trigonometric Functions  Trigonometric Identities - Class 10  Integration of Trigonometric Functions  Trigonometric Ratios of Complementary Angles  