Question

What is the value $ \cos {24^ \circ } + \cos {5^ \circ } + \cos {175^ \circ } + \cos {204^ \circ } + \cos {300^ \circ } = $ ?
A. $ {1 {\left/
 {\vphantom {1 2}} \right.
 } 2} $
B. $ - {1 {\left/
 {\vphantom {1 2}} \right.
 } 2} $
C. $ {{\sqrt 3 } {\left/
 {\vphantom {{\sqrt 3 } 2}} \right.
 } 2} $
D. $ 1 $

Answer 1

Hint: This question can be solved by using the trigonometric functions and relations. As we know that the value of trigonometric functions changes from quadrant to quadrant. So, some relations are derived from these changes and we will use these relations to solve this problem. For example, trigonometrical relations for angles $ \left( {{{180}^ \circ } - \theta } \right) $ and $ \left( {{{360}^ \circ } - \theta } \right) $.

Formula Used:
 $ \cos \left( {{{180}^ \circ } - \theta } \right) = - \cos \theta $
 $ - \cos \theta $ because the value of $ {\rm{cosine}} $ is negative in the second quadrant.
 $ \cos \left( {{{180}^ \circ } + \theta } \right) = \cos \theta $
 $ \cos \left( {{{360}^ \circ } - \theta } \right) = \cos \theta $
 $ \cos \theta $ because the value of $ {\rm{cosine}} $ is positive in the third quadrant.

Complete step-by-step answer:
 $ \cos {24^ \circ } + \cos {5^ \circ } + \cos {175^ \circ } + \cos {204^ \circ } + \cos {300^ \circ } $
We can write the above expression as,
 $
\Rightarrow \cos \left( {{{175}^ \circ }} \right) = \cos \left( {{{180}^ \circ } - {5^ \circ }} \right)\\
\Rightarrow \cos \left( {{{204}^ \circ }} \right) = \cos \left( {{{180}^ \circ } + {{24}^ \circ }} \right)
 $
\[\Rightarrow \cos \left( {{{300}^ \circ }} \right) = \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right)\]
On putting all the values into the above expression we get,
 $ = \cos {24^ \circ } + \cos {5^ \circ } + \cos \left( {{{180}^ \circ } - {5^ \circ }} \right) + \cos \left( {{{180}^ \circ } + {{24}^ \circ }} \right) + \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) $
Now, we know from the trigonometric relations we know that, the term $ \cos \left( {{{180}^ \circ } - {5^ \circ }} \right) $ is in the second quadrant and the value of $ {\rm{cosine}} $ is negative in the second quadrant. So, we get,
 $\Rightarrow \cos \left( {{{180}^ \circ } - {5^ \circ }} \right) = - \cos {5^ \circ } $
Similarly, the term $ \cos \left( {{{180}^ \circ } + {{24}^ \circ }} \right) $ is in the third quadrant and the value of the $ {\rm{cosine}} $ is negative in the third quadrant. So, we get,
 $\Rightarrow \cos \left( {{{180}^ \circ } + {{24}^ \circ }} \right) = - \cos {24^ \circ } $
Also, the term $ \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) $ is in the fourth quadrant and the value of the $ {\rm{cosine}} $ is positive in the fourth quadrant. So, we get,
 $\Rightarrow \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) = \cos {60^ \circ } $
Now, on putting these values into the expression, we get,
 $
 = \cos {24^ \circ } + \cos {5^ \circ } + \left( { - \cos {5^ \circ }} \right) + \left( { - \cos {{24}^ \circ }} \right) + \cos {60^ \circ }\\
 = \cos {60^ \circ }\\
 = \dfrac{1}{2}
 $
Hence the correct answer is option A.

Note: We do not need to use trigonometric relations for all of the terms in this question because this was a trick question. The trick part in the question is that in order to solve it we have to use the relations for only three of the terms and leave the rest terms. This way the question becomes easier and gets solved quickly. If we solve it normally it would make the calculations even more complex.