Complex Numbers And Quadratic Equations question 627
Question: $ \sum\limits_{k=33}^{65}{( \sin \frac{2k\pi }{8}-i\cos \frac{2k\pi }{8} )} $
Options:
A) $ 1+i $
B) $ 1-i $
C) $ 1+\frac{i}{\sqrt{2}} $
D) $ \frac{1-i}{\sqrt{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \sum\limits_{k=33}^{65}{( \sin \frac{2k\pi }{8}-i\cos \frac{2k\pi }{8} )} $ $ =[ \sin \frac{33\pi }{4}\sin \frac{34\pi }{4}+…+\sin \frac{65\pi }{4} ] $ $ -i[ \cos \frac{33\pi }{4}+\cos \frac{34\pi }{4}+…+\cos \frac{65\pi }{4} ] $ $ =\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 { a+(n-1)\frac{\beta }{2} }.\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}} $ and $ \cos (\alpha )+cos(\alpha +\beta )+…+cos(\alpha +(n-1)\beta ) $ $ =\frac{\cos { a+(n-1)\frac{\beta }{2} }\sin ( \frac{n\beta }{2} )}{\sin \frac{\beta }{2}} $ $ =-( \frac{1+i}{\sqrt{2}} )=\frac{1-i}{\sqrt{2}} $
BETA
