Question

Find the value of $\cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}$.

Answer 1

Hint: We solve this question by first converting the value $\cos {{175}^{\circ }}$ into smaller angle using the formula $\cos \left( {{180}^{\circ }}-A \right)=-\cos A$. Then we substitute it in the given expression. Then we consider the value of $\cos {{65}^{\circ }}+\cos {{175}^{\circ }}$ and substitute the above obtained value of $\cos {{175}^{\circ }}$. Then we use the formula $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ and simplify it. Then we substitute the value $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$ in it and then we use the formula $\sin \left( {{90}^{\circ }}-A \right)=\cos A$ and find the value in terms of cosine. Then we substitute this value in the given expression and solve it to find the required value.

Complete step-by-step solution
We are asked to find the value of $\cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}$.
First let us consider the value of $\cos {{175}^{\circ }}$.
Now let us consider the formula,
$\cos \left( {{180}^{\circ }}-A \right)=-\cos A$
Using this formula, we can write $\cos {{175}^{\circ }}$ as,
$\begin{align}
  & \Rightarrow \cos {{175}^{\circ }}=\cos \left( {{180}^{\circ }}-{{5}^{\circ }} \right) \\
 & \Rightarrow \cos {{175}^{\circ }}=-\cos {{5}^{\circ }} \\
\end{align}$
So, substituting this in the given expression we can write it as,
$\Rightarrow \cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}=\cos {{55}^{\circ }}+\cos {{65}^{\circ }}-\cos {{5}^{\circ }}.............\left( 1 \right)$
Now let us consider the trigonometric expression,
$\Rightarrow \cos {{65}^{\circ }}+\cos {{175}^{\circ }}=\cos {{65}^{\circ }}-\cos {{5}^{\circ }}$
Now let us consider the formula,
$\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
So, using this formula we can write the above expression as,
$\begin{align}
  & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-2\sin \left( \dfrac{{{65}^{\circ }}+{{5}^{\circ }}}{2} \right)\sin \left( \dfrac{{{65}^{\circ }}-{{5}^{\circ }}}{2} \right) \\
 & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-2\sin \left( \dfrac{{{70}^{\circ }}}{2} \right)\sin \left( \dfrac{{{60}^{\circ }}}{2} \right) \\
 & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-2\sin \left( {{35}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right) \\
\end{align}$
Now let us substitute the value of $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$ in the above equation. Then we get,
$\begin{align}
  & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-2\sin \left( {{35}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right) \\
 & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-2\sin \left( {{35}^{\circ }} \right)\times \dfrac{1}{2} \\
 & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-\sin {{35}^{\circ }} \\
 & \Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-\sin \left( {{90}^{\circ }}-{{55}^{\circ }} \right) \\
\end{align}$
Now let us consider the formula,
$\sin \left( {{90}^{\circ }}-A \right)=\cos A$
Using it we can write the above equation as,
$\Rightarrow \cos {{65}^{\circ }}-\cos {{5}^{\circ }}=-\cos {{55}^{\circ }}............\left( 2 \right)$
Substituting the value obtained in equation (2) in the equation (1) we get,
$\begin{align}
  & \Rightarrow \cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}=\cos {{55}^{\circ }}+\cos {{65}^{\circ }}-\cos {{5}^{\circ }} \\
 & \Rightarrow \cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}=\cos {{55}^{\circ }}-\cos {{55}^{\circ }} \\
 & \Rightarrow \cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}=0 \\
\end{align}$
So, we get the value of $\cos {{55}^{\circ }}+\cos {{65}^{\circ }}+\cos {{175}^{\circ }}$ as 0. Hence the answer is 0.

Note: The common mistakes one makes while solving this question is one might take the formulas wrongly as, $\cos \left( {{180}^{\circ }}-A \right)=\cos A$ but it should have the negative sign and might write the formula as $\cos A-\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ by confusing with the formula $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$.