Question

Evaluate:

$\sum _{k=1}^6[\sin \frac{2k\pi }{7}-i\cos \frac{2k\pi }{7}]$

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

Answer: (C)

$-i$

Since, $cos\theta +isin\theta =e^{i\theta }$

$\sum _{k=1}^6[\sin \frac{2k\pi }{7}-i\cos \frac{2k\pi }{7}]=-i\sum _{k=1}^6\left[cos\right(\frac{2k\pi }{7})+isin(\frac{2k\pi }{7}\left)\right]=-i\sum _{k=1}^6{e}^{i\frac{2k\pi }{7}}$

Solving the index part only which is

$i\frac{2k\pi }{7}=i\frac{2\pi }{7}(1+2+3+\cdots +6)$......[putting the values of $k$]

$=i6\pi$

So,

$-i{\sum }_{k=1}^6{e}^{i\frac{2k\pi }{7}}=-i{e}^{i6\pi }=-i(cos6\pi +sin6\pi )=-i$