Question
Answers

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 Verified Verified
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,
$\cos {{92}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{87}^{{}^\circ }} \right)=-\cos {{87}^{{}^\circ }}$
,, ,, ,,
,, ,, ,,
,, ,, ,,
,, ,, ,,
,, ,, ,,
$\begin{align}
  & \cos {{179}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{1}^{{}^\circ }} \right)=-\cos {{1}^{{}^\circ }} \\
 & \cos {{180}^{{}^\circ }}=\cos \left( {{180}^{{}^\circ }}-{{0}^{{}^\circ }} \right)=-\cos {{0}^{{}^\circ }} \\
\end{align}$
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,
$\begin{align}
  & S=\cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+...+\cos {{179}^{{}^\circ
 }}+\cos {{180}^{{}^\circ }} \\
 & S=\left( \cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+.....\cos
 {{90}^{{}^\circ }} \right)+\left( \cos {{91}^{{}^\circ }}+\cos {{92}^{{}^\circ }}+...+\cos
 {{179}^{{}^\circ }}+\cos {{180}^{{}^\circ }} \right) \\
 & S=\left( \cos {{1}^{{}^\circ }}+\cos {{2}^{{}^\circ }}+\cos {{3}^{{}^\circ }}+.....\cos
{{90}^{{}^\circ }} \right)+S' \\
\end{align}$
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{align}
  & 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{align}$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
\[\begin{align}
  & S=\cos A+\cos (A+D)+\cos (A+2D)+......\cos (A+(n-1)D) \\
 & S=\dfrac{\cos \left( A+\dfrac{(n-1)}{2}D \right)\sin \dfrac{nD}{2}}{\sin \dfrac{D}{2}} \\
\end{align}\]
From the given series we have n=180, A=1, D=1
\[S=\dfrac{\cos \left( 1+\dfrac{179}{2}\left( 1 \right) \right)\sin \dfrac{180}{2}\left( 1
 \right)}{\sin \left( \dfrac{1}{2} \right)}\]
\[\begin{align}
  & S=\dfrac{\cos \left( \dfrac{181}{2} \right)\sin 90}{\sin \left( \dfrac{1}{2} \right)}=\dfrac{\cos \left( 90+\dfrac{1}{2} \right)}{\sin \left( \dfrac{1}{2} \right)}-\dfrac{-\sin
 \left( \dfrac{1}{2} \right)}{\sin \left( \dfrac{1}{2} \right)} \\
 & S=-1 \\
\end{align}\]
One can go wrong if he/she converts
$\cos \theta $ to sin form like;
$\begin{align}
  & \cos {{91}^{{}^\circ }}=\cos \left( 90+1 \right)=-\sin 1 \\
 & \cos {{92}^{{}^\circ }}=\cos \left( 90+2 \right)=-\sin 2 \\
\end{align}$
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.

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