The value of $\displaystyle\sum_{k=1}^{6}\!\left(\sin\frac{2\pi k}{7} - i\cos\frac{2\pi k}{7}\right)$ is
Step-by-Step Solution
Key Concept: The identity $\sin\theta - i\cos\theta = -ie^{i\theta}$ converts the sum into $-i\sum_{k=1}^{6}\zeta^k$ where $\zeta=e^{2\pi i/7}$, and the sum of all 7th roots of unity is zero.
**Step 1: Rewrite each term**
$\sin\theta - i\cos\theta = -i(\cos\theta + i\sin\theta) = -ie^{i\theta}$. So each term equals $-ie^{2\pi ik/7}$.
**Step 2: Factor out −i**
$\displaystyle\sum_{k=1}^{6}(-i)e^{2\pi ik/7} = -i\sum_{k=1}^{6}e^{2\pi ik/7}$.
**Step 3: Use sum of 7th roots of unity**
$\displaystyle\sum_{k=0}^{6}e^{2\pi ik/7} = 0 \Rightarrow \sum_{k=1}^{6}e^{2\pi ik/7} = -1$. So the answer $= -i\cdot(-1) = i$.
Correct Answer: 4