Question

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

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
$\left[\begin{array}{c}\cos \left({180}^{\circ }-{\theta }^{\circ }\right)=-\cos {\theta }^{\circ }\\ \cos \left({180}^{\circ }+{\theta }^{\circ }\right)=-\cos {\theta }^{\circ }\\ \cos \left({360}^{\circ }-{\theta }^{\circ }\right)=\cos {\theta }^{\circ }\end{array}\right]$
So we have:
$$\begin{gathered} \cos {125}^{\circ }=\cos \left({180}^{\circ }-{55}^{\circ }\right)=-\cos {55}^{\circ } \\ \cos {204}^{\circ }=\cos \left({180}^{\circ }+{24}^{\circ }\right)=-\cos {24}^{\circ } \\ \cos {300}^{\circ }=\cos \left({360}^{\circ }-{60}^{\circ }\right)=\cos {60}^{\circ } \end{gathered}$$
Putting these values in equation (1) we get,
$$\begin{gathered} \Rightarrow \cos {24}^{\circ }+\cos {55}^{\circ }-\cos {55}^{\circ }-\cos {24}^{\circ }+\cos {60}^{\circ } \\ \Rightarrow \cos {60}^{\circ } \\ \Rightarrow {\displaystyle \frac{1}{2}}=R.H.S. \end{gathered}$$
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.