Question

Evaluate $ \cos {{24}^{\circ }}+\cos {{55}^{\circ }}+\cos {{125}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }} $ .

Answer 1

Hint: In this question, we are given a trigonometric function in terms of cosine values and we need to find its value. For this we will use the formulas such as $ \cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta ,\cos \left( {{180}^{\circ }}+\theta \right)=-\cos \theta ,\cos \left( {{270}^{\circ }}+\theta \right)=\sin \theta $ to evaluate our answer. We will also use the value of $ \sin {{30}^{\circ }} $ from the trigonometric ratio table which is equal to $ \dfrac{1}{2} $ .

Complete step by step answer:
Here we are given the expression as $ \cos {{24}^{\circ }}+\cos {{55}^{\circ }}+\cos {{125}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }} $ .
As we can see, $ {{204}^{\circ }}-{{24}^{\circ }}={{180}^{\circ }} $ so let us write them together and also $ {{55}^{\circ }}+{{125}^{\circ }}={{180}^{\circ }} $ so write them together, we get,
 $ \cos {{24}^{\circ }}+\cos {{204}^{\circ }}+\cos {{55}^{\circ }}+\cos {{125}^{\circ }}+\cos {{300}^{\circ }} $ .
Now we know $ {{204}^{\circ }} $ can be written as $ {{180}^{\circ }}+{{24}^{\circ }} $ so we get,
 $ \cos {{24}^{\circ }}+\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)+\cos {{55}^{\circ }}+\cos {{125}^{\circ }}+\cos {{300}^{\circ }} $ .
We know that, $ {{125}^{\circ }} $ can be written as $ {{180}^{\circ }}-{{55}^{\circ }} $ so we get,
 $ \cos {{24}^{\circ }}+\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)+\cos {{55}^{\circ }}+\cos \left( {{180}^{\circ }}-{{55}^{\circ }} \right)+\cos {{300}^{\circ }} $ .
Also we can write $ {{300}^{\circ }} $ as $ {{270}^{\circ }}+{{30}^{\circ }} $ so we get,
 $ \cos {{24}^{\circ }}+\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)+\cos {{55}^{\circ }}+\cos \left( {{180}^{\circ }}-{{55}^{\circ }} \right)+\cos \left( {{270}^{\circ }}+{{30}^{\circ }} \right) $ .
Now we know the cosine function is negative in second quadrant and third quadrant. Also $ \left( {{180}^{\circ }}+{{24}^{\circ }} \right) $ lies in third quadrant, \[\left( {{180}^{\circ }}-{{55}^{\circ }} \right)\] lies in second quadrant. So $ \cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \text{ and }\cos \left( {{180}^{\circ }}+\theta \right)=-\cos \theta $ . But $ \cos \left( {{270}^{\circ }}+\theta \right) $ lies in fourth quadrant where cosine is positive and use of $ {{270}^{\circ }} $ changes trigonometric function. So, $ \cos \left( {{270}^{\circ }}+\theta \right)=\sin \theta $ . Using all these we get,
 $ \cos {{24}^{\circ }}-\cos {{24}^{\circ }}+\cos {{55}^{\circ }}-\cos {{55}^{\circ }}+\sin {{30}^{\circ }} $ .
Cancelling the cosine terms we are left with, $ \sin {{30}^{\circ }} $ .
From trigonometric ratio table we know that $ \sin {{30}^{\circ }}=\dfrac{1}{2} $ so our expression gets reduced to $ \dfrac{1}{2} $ which is our final answer.
Hence, $ \cos {{24}^{\circ }}+\cos {{55}^{\circ }}+\cos {{125}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}=\dfrac{1}{2} $ .

Note:
Students should take care of the signs while solving the sum. Keep in mind the signs of trigonometric functions in the four-quadrant. All trigonometric functions are positive in the first quadrant, only sine and cosecant are positive in the second quadrant, only tangent and cotangent are positive in the third quadrant and only cosine and secant are positive in the fourth quadrant.