Question:

**Step 1:** Rewrite each term using Euler's formula. The expression $\sin\theta + i\cos\theta$ can be rewritten by factoring out $i$: $$ \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta) $$ Using Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$, we know that $\cos\theta - i\sin\theta = e^{-i\theta}$. Thus, each term in the sum can be expressed as: $$ \sin\theta + i\cos\theta = i e^{-i\theta} $$ **Step 2:** Apply this transformation to the given sum. The sum becomes: $$ \sum_{k=1}^{6} \left(\sin\frac{2\pi k}{7} + i\cos\frac{2\pi k}{7}\right) = \sum_{k=1}^{6} i e^{-i\frac{2\pi k}{7}} $$ Factor out the constant $i$: $$ i \sum_{k=1}^{6} e^{-i\frac{2\pi k}{7}} $$ **Step 3:** Introduce the roots of unity. Let $\omega = e^{i\frac{2\pi}{7}}$. This is a primitive 7th root of unity. Then, $e^{-i\frac{2\pi k}{7}}$ can be written as $(e^{-i\frac{2\pi}{7}})^k = (\omega^{-1})^k$. The sum can now be expressed in terms of powers of $\omega^{-1}$: $$ i \sum_{k=1}^{6} (\omega^{-1})^k $$ **Step 4:** Use the fundamental property of roots of unity. For any primitive $n$-th root of unity $\zeta$, the sum of its powers from $k=0$ to $n-1$ is zero: $$ \sum_{k=0}^{n-1} \zeta^k = 0 $$ In this case, $n=7$, and $\omega^{-1}$ is also a primitive 7th root of unity. Therefore: $$ \sum_{k=0}^{6} (\omega^{-1})^k = 0 $$ Expanding this sum: $$ (\omega^{-1})^0 + \sum_{k=1}^{6} (\omega^{-1})^k = 0 $$ Since $(\omega^{-1})^0 = 1$: $$ 1 + \sum_{k=1}^{6} (\omega^{-1})^k = 0 $$ Solving for the sum from $k=1$ to $6$: $$ \sum_{k=1}^{6} (\omega^{-1})^k = -1 $$ **Step 5:** Substitute this result back into the expression from Step 3. $$ i \sum_{k=1}^{6} (\omega^{-1})^k = i \times (-1) = -i $$ Thus, the value of the given sum is $-i$.