A) \[1+i\]
B) \[1-i\]
C) \[1+\frac{i}{\sqrt{2}}\]
D) \[\frac{1-i}{\sqrt{2}}\]
Correct Answer: D
Solution :
\[\sum\limits_{k=33}^{65}{\left( \sin \frac{2k\pi }{8}-i\cos \frac{2k\pi }{8} \right)}\] \[=\left[ \sin \frac{33\pi }{4}\sin \frac{34\pi }{4}+...+\sin \frac{65\pi }{4} \right]\] \[-i\left[ \cos \frac{33\pi }{4}+\cos \frac{34\pi }{4}+...+\cos \frac{65\pi }{4} \right]\] \[=\sin \frac{\pi }{4}-i\cos \frac{\pi }{4}\] \[\sin \alpha +\sin (\alpha +\beta )+\sin (\alpha +2\beta )+...+\sin [\alpha +(n-1)\beta ]\] \[=\frac{\sin \left\{ a+(n-1)\frac{\beta }{2} \right\}.\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\] and \[\cos (\alpha )+cos(\alpha +\beta )+...+cos(\alpha +(n-1)\beta )\] \[=\frac{\cos \left\{ a+(n-1)\frac{\beta }{2} \right\}\sin \left( \frac{n\beta }{2} \right)}{\sin \frac{\beta }{2}}\] \[=-\left( \frac{1+i}{\sqrt{2}} \right)=\frac{1-i}{\sqrt{2}}\]
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