Question

$\mathrm{The}\mathrm{value}\mathrm{of}\cos \left(\frac{\pi }{11}\right)+\cos \left(\frac{3\pi }{11}\right)+\cos \left(\frac{5\pi }{11}\right)+\cos \left(\frac{7\pi }{11}\right)+\cos \left(\frac{9\pi }{11}\right)\mathrm{is}$

• A.
  0
• B.
  0.5
• C.
  1
• D.
  2

Answer 1

The correct option is B

0.5
To solve this problem, we will use the concept of sum of cosine series.
Here, let us assume:
$S=\cos \left(\frac{\pi }{11}\right)+\cos \left(\frac{3\pi }{11}\right)+\cos \left(\frac{5\pi }{11}\right)+\cos \left(\frac{7\pi }{11}\right)+\cos \left(\frac{9\pi }{11}\right).$Now, we know the sum of cosines
$S=\cos \left(\alpha \right)+\cos \left(\alpha +\beta \right)+\cos \left(\alpha +2\beta \right)+....+\cos \left(\alpha +(n-1)\beta \right)\Rightarrow S=\frac{\sin \frac{\mathrm{n\beta }}{2}}{\sin \frac{\beta }{2}}\cos \left(\alpha +\left(n-1\right)\frac{\beta }{2}\right].$
For our cosine series, first term first term $\alpha =\frac{\pi }{11}$ and common difference
$\beta =\frac{2\pi }{11}.$
So, substituting we get
$S=\frac{\sin \left(5\frac{2\pi }{2\times 11}\right)}{\sin \left(\frac{2\pi }{2\times 11}\right)}\cos \left(\frac{\pi }{11}+(5-1)\frac{2\pi }{2\times 11}\right)\Rightarrow S=\frac{\sin \left(\frac{5\pi }{11}\right)}{\sin \left(\frac{\pi }{11}\right)}\cos \left(\frac{5\pi }{11}\right).$
Now, multiplying by 2 in both numerator and denominator we get,
$S=\frac{2\sin \left(\frac{5\pi }{11}\right)\cos \left(\frac{5\pi }{11}\right)}{2\sin \left(\frac{\pi }{11}\right)}\Rightarrow S=\frac{\sin \left(\frac{10\pi }{11}\right)}{2\sin \left(\frac{\pi }{11}\right)}\Rightarrow S=\frac{\sin \left(\pi -\frac{\pi }{11}\right)}{2\sin \left(\frac{\pi }{11}\right)}\Rightarrow S=\frac{\sin \left(\frac{\pi }{11}\right)}{2\sin \left(\frac{\pi }{11}\right)}=\frac{1}{2}$
Thus S = 0.5.
Hence, Option b. is correct.