Question

Evaluate \[ \sum_{k=1}^{6}\sin\left(\frac{2\pi k}{7}\right) -i\cos\left(\frac{2\pi k}{7}\right) \]

• A. \(1\)
• B. \(-i\)
• C. \(i\)
• D. \(-1\)

Hint:

The sum of all \(n\)-th roots of unity is always zero: \[ 1+\omega+\omega^2+\cdots+\omega^{n-1}=0 \] where \[ \omega=e^{\frac{2\pi i}{n}} \] and \(\omega\neq 1\).

Answer 1

The Correct Option is C

Step 1: Rewrite the expression using Euler's form.
Recall that \[ \cos\theta+i\sin\theta=e^{i\theta} \] Also, \[ \sin\theta-i\cos\theta =-i(\cos\theta+i\sin\theta) \] Therefore, \[ \sin\theta-i\cos\theta =-ie^{i\theta} \] Hence, \[ \sum_{k=1}^{6} \left( \sin\frac{2\pi k}{7} -i\cos\frac{2\pi k}{7} \right) = -i\sum_{k=1}^{6} e^{\frac{2\pi i k}{7}} \]

Step 2: Use the roots of unity property.
The seventh roots of unity satisfy \[ 1+\sum_{k=1}^{6}e^{\frac{2\pi i k}{7}}=0 \] Thus, \[ \sum_{k=1}^{6}e^{\frac{2\pi i k}{7}}=-1 \] Substituting, \[ -i\sum_{k=1}^{6}e^{\frac{2\pi i k}{7}} =-i(-1) \] \[ =i \]

Step 3: Final conclusion.
Therefore, \[ \boxed{i} \]