Question
Question: If the given trigonometric equation is \(\cos \dfrac{{2\pi }}{3} - \cos \pi + \cos \dfrac{{4\pi }}{3...
If the given trigonometric equation is cos32π−cosπ+cos34π−cos35π+cos2π−cos37π+....+cos340π−cos341π=k, then the value of |6k| is
Solution
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 cosnπ, 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
cos32π−cosπ+cos34π−cos35π+cos2π−cos37π+....+cos340π−cos341π=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
cos32π−cos33π+cos34π−cos35π+cos36π−cos37π+....+cos340π−cos341π=k
(cos32π+cos34π+cos36π+....+cos340π)−(cos33π+cos35π+cos37π+....+cos341π)=k
So there are 20 terms in each of the above two series.
Now the above series is written as,
(cos2π+cos4π+cos6π+cos8π+cos10π+cos12π)+(cos32π+cos34π+....+cos340π) \-(cosπ+cos3π+cos5π+cos7π+cos9π+cos11π+cos13π)−(cos35π+cos37π+....+cos341π)=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, cos2π=cos4π=cos6π=cos8π=cos10π=cos12π=1
cosπ=cos3π=cos5π=cos7π=cos9π=cos11π=cos13π=−1
cos32π=cos34π=....=cos340π=2−1, (As cosine is negative in second and third quadrant)
And
cos35π=cos37π=....=cos341π=21, (As cosine is positive in first and fourth quadrant)
So the above equation is written as,
⇒6(cos2π)+14(cos32π)−7(cosπ)−13(cos35π)=k
Now substitute the values we have,
⇒6(1)+14(2−1)−7(−1)−13(21)=k
Now simplify we have,
⇒6+7−214+13=k
⇒13−227=k
⇒226−27=k
⇒k=2−1
Now we have to find out the value of |6k|
⇒∣6k∣=6×2−1=∣−3∣=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.