Question

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

Answer 1

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.