Question

Find the value of the equation $ \cos \theta + \cos (\pi + \theta ) + \cos (2\pi + \theta ) + ............... $ up to 2n terms.
A. 0
B. 2n $ \cos \theta $
C. $ \cos \theta $
D. 1

Answer 1

Hint: In order to solve this problem we need to use the sine of cosine function in the 1st and 3rd quadrant. Putting those values in the given equation will provide you the right answer. We know that in the 1st quadrant all the trigonometric ratios are positive and in the 3rd quadrant only tangent and cotangent are positive. We will use this property.

Complete step-by-step answer:
So as we know the value of cosine function in the first quadrant is positive, in the second quadrant its negative, in the third quadrant its negative and in the fourth quadrant its positive.
The given equation is,
 $ \cos \theta \, + \,\cos (\,\pi \, + \,\theta \,)\, + \,\cos \,(\,2\pi \, + \,\theta \,)\, + ............... $ where we can say $ \theta \,{\text{ < }}\,{\text{90}} $ if it would had been greater than 90 degrees then it would lie in another quadrant.
So, we know $ \cos \theta $ is in first quadrant therefore $ {\text{cos}}\theta $ = $ {\text{cos}}\theta $
 $ \cos (\,\pi \, + \,\theta \,) $ is in third quadrant so $ \cos (\,\pi \, + \,\theta \,) $ =- $ {\text{cos}}\theta $
And $ \cos (2\,\pi \, + \,\theta \,) $ is in first quadrant so $ \cos (2\,\pi \, + \,\theta \,) $
Then
 $
  \cos (3\pi + \theta ) = - \cos \theta \\
  \cos (4\pi + \theta ) = \cos \theta \\
 $
Following the same procedure sequentially,
The second last term will be,
 $ \cos ((2n - 1)\pi + \theta ) = - \cos \theta $
And the last term will be
 $ \cos (2n\pi + \theta ) = \cos \theta $
On putting those values in the given equation we get the equation as,
 $ \Rightarrow \,\cos \theta $ - $ \cos \theta $ + $ \cos \theta $ - $ \cos \theta $ + $ \cos \theta $ - $ \cos \theta $ +…………+ $ \cos \theta $ - $ \cos \theta $ + $ \cos \theta $
 $ \Rightarrow \,\cos \theta $
We can see the trend that the first term and second term are the same in magnitude but different in sign similarly to the third and fourth term and so on. So, this will cancel out all the terms but there is no term left to cancel out the 2nth term whose value is $ \cos \theta $ . So, the answer to this question is $ \cos \theta $ since it is the last term left after cancelling out all the terms.
Hence, the answer of this question is $ \Rightarrow \,\cos \theta $ .
So, the correct option is C.

Note: Whenever you face such types of problems, you have to use the values and signs of trigonometric functions. Here we have used the signs of cosine functions in different quadrants to solve this problem. We should know that in the 1st quadrant all the functions are positive, in the 2nd quadrant only sine and cosec is positive, in the 3rd quadrant only tangent and cotangent functions are positive, in the fourth quadrant cosecant and secant functions are positive. Knowing this will take you to reach the right answer.