Question

If \( \cos A + \cos(A + B) + \cos(A + 2B) + \cdots \) upto \( n \) terms \( = \cos \left( \frac{2A + (n-1)B}{2} \right) \frac{\sin \frac{nB}{2}}{\sin \frac{B}{2}} \), then \( \cos \frac{3\pi}{19} + \cos \frac{5\pi}{19} + \cos \frac{7\pi}{19} + \cdots + \cos \frac{17\pi}{19} = \)

• A. \( 1 \)
• B. \( -\frac{1}{2} \)
• C. \( \frac{1}{2} \)
• D. \( 0 \)

Hint:

Recognize the sum as a series of cosines with angles in arithmetic progression. Identify the first term's angle \( A \), the common difference \( B \), and the number of terms \( n \). Then, directly apply the given formula for the sum of such a series.

Answer 1

The Correct Option is C

The given sum is a sum of cosines in an arithmetic progression of angles.
The formula for the sum of \( n \) terms of \( \cos(A) + \cos(A + B) + \cdots + \cos(A + (n-1)B) \) is given as \( \cos \left( A + \frac{(n-1)B}{2} \right) \frac{\sin \frac{nB}{2}}{\sin \frac{B}{2}} \).
The series we need to evaluate is \( \cos \frac{3\pi}{19} + \cos \frac{5\pi}{19} + \cos \frac{7\pi}{19} + \cdots + \cos \frac{17\pi}{19} \).
Here, the first term has angle \( A = \frac{3\pi}{19} \), and the common difference is \( B = \frac{5\pi}{19} - \frac{3\pi}{19} = \frac{2\pi}{19} \).
To find the number of terms \( n \), let the \( n^{th} \) term be \( \cos \frac{17\pi}{19} \).
The angle of the \( n^{th} \) term is \( A + (n-1)B = \frac{3\pi}{19} + (n-1) \frac{2\pi}{19} = \frac{3\pi + 2n\pi - 2\pi}{19} = \frac{(2n + 1)\pi}{19} \).
We have \( \frac{(2n + 1)\pi}{19} = \frac{17\pi}{19} \), so \( 2n + 1 = 17 \implies 2n = 16 \implies n = 8 \).
There are 8 terms in the series.
Now, using the formula for the sum with \( A = \frac{3\pi}{19} \), \( B = \frac{2\pi}{19} \), and \( n = 8 \): Sum \( = \cos \left( \frac{3\pi}{19} + \frac{(8-1) \frac{2\pi}{19}}{2} \right) \frac{\sin \left( \frac{8 \cdot \frac{2\pi}{19}}{2} \right)}{\sin \left( \frac{\frac{2\pi}{19}}{2} \right)} \) Sum \( = \cos \left( \frac{3\pi}{19} + \frac{7 \cdot \frac{2\pi}{19}}{2} \right) \frac{\sin \left( \frac{8\pi}{19} \right)}{\sin \left( \frac{\pi}{19} \right)} \) Sum \( = \cos \left( \frac{3\pi}{19} + \frac{7\pi}{19} \right) \frac{\sin \frac{8\pi}{19}}{\sin \frac{\pi}{19}} = \cos \left( \frac{10\pi}{19} \right) \frac{\sin \frac{8\pi}{19}}{\sin \frac{\pi}{19}} \) We know \( \sin(\pi - x) = \sin x \), so \( \sin \frac{8\pi}{19} = \sin(\pi - \frac{11\pi}{19}) = \sin \frac{11\pi}{19} \).
Also, \( \cos(\pi - x) = -\cos x \), so \( \cos \frac{10\pi}{19} = -\cos(\pi - \frac{9\pi}{19}) = -\cos \frac{9\pi}{19} \).
Sum \( = -\cos \frac{9\pi}{19} \frac{\sin \frac{8\pi}{19}}{\sin \frac{\pi}{19}} \) Multiply and divide by \( 2 \sin \frac{\pi}{19} \): Sum \( = \frac{2 \sin \frac{\pi}{19} \cos \frac{10\pi}{19} \sin \frac{8\pi}{19}}{2 \sin^2 \frac{\pi}{19}} \)