Question

Evaluate the expression $\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{180}^{{}^\circ }}$ is equal to
(A) 1
(B) 0
(C) 2
(D) -1

Answer 1

Hint: Convert all the angles to acute angles i.e. between 0 to ${{90}^{{}^\circ }}$.

Here, we need to find summation of series between 0 to ${{90}^{{}^\circ }}$.
$\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{180}^{{}^\circ }}$
Let us suppose the series is denoted by S.
$S=\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{180}^{{}^\circ }}............\left( 1 \right)$
We need to know about trigonometric conversions (i.e. one function to another) by changing
the angles.
We have,
$S=\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{180}^{{}^\circ }}$
Let us write another series which has only cosine with obtuse angles.
$S'=\cos {{91}^{{}^\circ }}+\cos {{92}^{{}^\circ }}+\cos {{93}^{{}^\circ }}+...+\cos {{179}^{{}^\circ }}+\cos {{180}^{{}^\circ }}$
Let us convert one by one to acute angle cosines;
We can write $\cos {{91}^{{}^\circ }}$as $\cos \left( {{180}^{{}^\circ }}-{{89}^{{}^\circ }} \right)\text{ or }\cos \left( \pi -{{89}^{{}^\circ }} \right)$. As angle is in the form of multiple of
$\pi$ , which means no conversion of function only signs can be changed. $\left( \pi -{{89}^{{}^\circ }} \right)\text{ or }{{91}^{{}^\circ }}$lies in second quadrant where cos is
negative, so we can write
$\cos {{91}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{89}^{{}^\circ }} \right)=-\cos {{89}^{{}^\circ }}$
Similarly, $\cos {{92}^{{}^\circ }}$can be substitute as $\cos \left( {{180}^{{}^\circ }}-{{88}^{{}^\circ }} \right)=-\cos {{88}^{{}^\circ }}$
Similarly,
$\cos {{92}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{87}^{{}^\circ }} \right)=-\cos {{87}^{{}^\circ }}$
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$\begin{aligned} & \cos {{179}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{1}^{{}^\circ }} \right)=-\cos {{1}^{{}^\circ }} \\\ & \cos {{180}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{0}^{{}^\circ }} \right)=-\cos {{0}^{{}^\circ }} \\\ \end{aligned}$
And, hence series S’ can be written as;
$S'=-\cos {{0}^{{}^\circ }}-\cos {{1}^{{}^\circ }}-\cos {{2}^{{}^\circ }}-...-\cos {{89}^{{}^\circ }}........\left( 2 \right)$
Now, let us put the value of series S’ in series S.
We have,
$\begin{aligned} & S=\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{179}^{{}^\circ }}+\cos {{180}^{{}^\circ }} \\\ & S=\left( \cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+.....\cos {{90}^{{}^\circ }} \right)+\left( \cos {{91}^{{}^\circ }}+\cos {{92}^{{}^\circ }}+...+\cos {{179}^{{}^\circ }}+\cos {{180}^{{}^\circ }} \right) \\\ & S=\left( \cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+.....\cos {{90}^{{}^\circ }} \right)+S' \\\ \end{aligned}$
Putting value of S’ from equation (2) we get;
$S=\left( \cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+.....\cos {{90}^{{}^\circ }} \right)-\left( \cos {{0}^{{}^\circ }}+\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+.....+\cos {{89}^{{}^\circ }} \right)$
Taking similar terms in one bracket, we get;
$\begin{aligned}
& S=\left( \cos {{1}^{{}^\circ }}-\cos {{1}^{{}^\circ }} \right)+\left( \cos {{2}^{{}^\circ }}-\cos
{{2}^{{}^\circ }} \right)+\left( \cos {{3}^{{}^\circ }}-\cos {{3}^{{}^\circ }} \right)+.....\left( \cos
{{89}^{{}^\circ }}-\cos {{89}^{{}^\circ }} \right)+\left( \cos {{90}^{{}^\circ }}-\cos {{0}^{{}^\circ }}

\right) \\
& S=0+0+0......0+0-1 \\
& S=-1 \\
\end{aligned}$Hence, summation of series S is -1. Option (D) is the correct answer. Note: Another approach for the given question would be like;$S=\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{179}^{{}^\circ

}}+\cos {{180}^{{}^\circ }}$
We have direct formula of cosine series

& S=\cos A+\cos (A+D)+\cos (A+2D)+......\cos (A+(n-1)D) \\\ & S=\dfrac{\cos \left( A+\dfrac{(n-1)}{2}D \right)\sin \dfrac{nD}{2}}{\sin \dfrac{D}{2}} \\\ \end{aligned}$$ From the given series we have n=180, A=1, D=1 $$S=\dfrac{\cos \left( 1+\dfrac{179}{2}\left( 1 \right) \right)\sin \dfrac{180}{2}\left( 1 \right)}{\sin \left( \dfrac{1}{2} \right)}$$ $$\begin{aligned} & S=\dfrac{\cos \left( \dfrac{181}{2} \right)\sin 90}{\sin \left( \dfrac{1}{2} \right)}=\dfrac{\cos \left( 90+\dfrac{1}{2} \right)}{\sin \left( \dfrac{1}{2} \right)}-\dfrac{-\sin \left( \dfrac{1}{2} \right)}{\sin \left( \dfrac{1}{2} \right)} \\\ & S=-1 \\\ \end{aligned}$$ One can go wrong if he/she converts $\cos \theta $ to sin form like; $\begin{aligned} & \cos {{91}^{{}^\circ }}=\cos \left( 90+1 \right)=-\sin 1 \\\ & \cos {{92}^{{}^\circ }}=\cos \left( 90+2 \right)=-\sin 2 \\\ \end{aligned}$ Here, one more step needs to do that convert all cos acute angles to sin by subtracting ${{90}^{{}^\circ }}$ to that. Hence, the answer will be the same but takes longer than the given solution. Converting one trigonometric function to another is a key point of the question and needs to be visualized very well.