Q47 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics
$\sum_{k=1}^6 \sin \left(\frac{2 \pi k}{7}\right)-i \cos \left(\frac{2 \pi k}{7}\right)=$
- A$1$
- B$-i$
- Correct answerC$i$
- D$-1$
Solution
$\sum_{k=1}^6-i\left[\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}\right]$
$\cos \theta+i \sin \theta=e^{i \theta}$
$(-i) \sum_{k=1}^6 e^{i \frac{2 \pi k}{7}}$
$=(-i)\left[e^{i \frac{2 \pi}{7}}+e^{i \frac{4 \pi}{7}}+e^{i \frac{6 \pi}{7}}+\ldots+e^{i \frac{12 \pi}{7}}\right]$
$=-i\left\{e^{i \frac{2 \pi}{7}} \frac{\left(1-\left(e^{i \frac{2 \pi}{7}}\right)^6\right)}{1-e^{i \frac{2 \pi}{7}}}\right\}$ (sum of GP)
$=-i\left[\frac{e^{i \frac{2 \pi}{7}}-e^{i \frac{14 \pi}{7}}}{1-e^{i \frac{2 \pi}{7}}}\right]$
$=-i\left[\frac{e^{i \frac{2 \pi}{7}}-e^{i 2 \pi}}{1-e^{i \frac{2 \pi}{7}}}\right]\left\{\because e^{i 2 \pi}=\cos 2 \pi+i \sin 2 \pi=1\right.$
$=-i\left[\frac{e^{i \frac{2 \pi}{7}}-1}{1-e^{i \frac{2 \pi}{7}}}\right]$
$=-i(-1)=i$
Hence, the correct answer is option: 'C'.