Question

If the given trigonometric equation is $\cos \dfrac{{2\pi }}{3} - \cos \pi + \cos \dfrac{{4\pi }}{3} - \cos \dfrac{{5\pi }}{3} + \cos 2\pi - \cos \dfrac{{7\pi }}{3} + .... + \cos \dfrac{{40\pi }}{3} - \cos \dfrac{{41\pi }}{3} = k$, then the value of |6k| is

Answer 1

Hint: In this particular question first find out the number of terms in the series than separate negative and positive terms and calculate the number of terms in both of the negative and positive series then again separate the integral multiples of $\cos n\pi $, where n is an integer number, so use these concepts to reach the solution of the given question.

Complete step by step answer:
Given trigonometric equation
$\cos \dfrac{{2\pi }}{3} - \cos \pi + \cos \dfrac{{4\pi }}{3} - \cos \dfrac{{5\pi }}{3} + \cos 2\pi - \cos \dfrac{{7\pi }}{3} + .... + \cos \dfrac{{40\pi }}{3} - \cos \dfrac{{41\pi }}{3} = k$
Now in the above equation there are 40 terms present (starting from 2 and end on 41 so there are total 40 terms)
We can also write the above equation like this
 $\cos \dfrac{{2\pi }}{3} - \cos \dfrac{{3\pi }}{3} + \cos \dfrac{{4\pi }}{3} - \cos \dfrac{{5\pi }}{3} + \cos \dfrac{{6\pi }}{3} - \cos \dfrac{{7\pi }}{3} + .... + \cos \dfrac{{40\pi }}{3} - \cos \dfrac{{41\pi }}{3} = k$
$\left( {\cos \dfrac{{2\pi }}{3} + \cos \dfrac{{4\pi }}{3} + \cos \dfrac{{6\pi }}{3} + .... + \cos \dfrac{{40\pi }}{3}} \right) - \left( {\cos \dfrac{{3\pi }}{3} + \cos \dfrac{{5\pi }}{3} + \cos \dfrac{{7\pi }}{3} + .... + \cos \dfrac{{41\pi }}{3}} \right) = k$
So there are 20 terms in each of the above two series.
Now the above series is written as,
$
  \left( {\cos 2\pi + \cos 4\pi + \cos 6\pi + \cos 8\pi + \cos 10\pi + \cos 12\pi } \right) + \left( {\cos \dfrac{{2\pi }}{3} + \cos \dfrac{{4\pi }}{3} + .... + \cos \dfrac{{40\pi }}{3}} \right) \\
   - \left( {\cos \pi + \cos 3\pi + \cos 5\pi + \cos 7\pi + \cos 9\pi + \cos 11\pi + \cos 13\pi } \right) - \left( {\cos \dfrac{{5\pi }}{3} + \cos \dfrac{{7\pi }}{3} + .... + \cos \dfrac{{41\pi }}{3}} \right) = k \\
$
Now in the first series there are 6 terms, in the second series there are 14 terms, in the third series there are 7 terms and in the fourth series there are 13 terms.
Now as we know that, $\cos 2\pi = \cos 4\pi = \cos 6\pi = \cos 8\pi = \cos 10\pi = \cos 12\pi = 1$
$\cos \pi = \cos 3\pi = \cos 5\pi = \cos 7\pi = \cos 9\pi = \cos 11\pi = \cos 13\pi = - 1$
$\cos \dfrac{{2\pi }}{3} = \cos \dfrac{{4\pi }}{3} = .... = \cos \dfrac{{40\pi }}{3} = \dfrac{{ - 1}}{2}$, (As cosine is negative in second and third quadrant)
And
$\cos \dfrac{{5\pi }}{3} = \cos \dfrac{{7\pi }}{3} = .... = \cos \dfrac{{41\pi }}{3} = \dfrac{1}{2}$, (As cosine is positive in first and fourth quadrant)

So the above equation is written as,
$ \Rightarrow 6\left( {\cos 2\pi } \right) + 14\left( {\cos \dfrac{{2\pi }}{3}} \right) - 7\left( {\cos \pi } \right) - 13\left( {\cos \dfrac{{5\pi }}{3}} \right) = k$
Now substitute the values we have,
$ \Rightarrow 6\left( 1 \right) + 14\left( {\dfrac{{ - 1}}{2}} \right) - 7\left( { - 1} \right) - 13\left( {\dfrac{1}{2}} \right) = k$
Now simplify we have,
$ \Rightarrow 6 + 7 - \dfrac{{14 + 13}}{2} = k$
$ \Rightarrow 13 - \dfrac{{27}}{2} = k$
$ \Rightarrow \dfrac{{26 - 27}}{2} = k$
$ \Rightarrow k = \dfrac{{ - 1}}{2}$
Now we have to find out the value of |6k|
$ \Rightarrow \left| {6k} \right| = \left| {6 \times \dfrac{{ - 1}}{2}} \right| = \left| { - 3} \right| = 3$, (as the modulus of a negative number is always positive or we can say we have to take the absolute value).

So 3 is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that cosine is negative in second and third quadrant and cosine is positive in first and fourth quadrant and also remember that the modulus of a negative number is always positive or we can say we have to take the absolute value.