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OpenStudy (anonymous):
wait let me check again.
OpenStudy (anonymous):
Is 11 part of the angle?
OpenStudy (anonymous):
1/11 is not multiplied to sin but its multiplied to its angle
OpenStudy (anonymous):
isn't cos(20 pi) = -1? which will give you -i/11?
OpenStudy (anonymous):
LOL
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OpenStudy (anonymous):
what x_x
OpenStudy (anonymous):
and 1/11 = not part of the angle
OpenStudy (anonymous):
You should have put that in parenthesis >.<
OpenStudy (anonymous):
why didn't you put it in the angle =_=...
OpenStudy (anonymous):
Well, you got a solution for "a" problem :P
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OpenStudy (anonymous):
hey anwar, aren't you supposed to find the summation of both sin and cos from 1 - 10?
OpenStudy (anonymous):
sorry about that guys
OpenStudy (anonymous):
Yeah sstarica
OpenStudy (anonymous):
It's OK.. But, I am not sure if I have enough time to do the problem. Just let me think about it quickly.
OpenStudy (anonymous):
okay anwar
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OpenStudy (anonymous):
is it possible to solve using euler's representation of complex no.?
OpenStudy (anonymous):
like this :
\[[\sum_{k=1}^{10}\sin(\frac{2 \pi}{11}) + i \cos(\frac{2\pi}{11}) ] + [\sum_{k=1}^{10}\sin(\frac{2(10)\pi}{11}) + i \cos(\frac{2(10) \pi}{11})]\]
? .-. right
OpenStudy (anonymous):
That's what I am thinking about right now.
OpenStudy (anonymous):
lol
OpenStudy (anonymous):
yeah all through 2,3,4 to 10
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OpenStudy (anonymous):
@sstarica: you just substitute for k=1,2,3,.. and add them up.
OpenStudy (anonymous):
no need, there's a faster way to compute the sum without going through the 1 2 3 stages :)
OpenStudy (anonymous):
take first term and add it with last term
OpenStudy (anonymous):
and whats that?
OpenStudy (anonymous):
something like that?
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OpenStudy (anonymous):
I don't know lol
OpenStudy (anonymous):
Well yeah.. You just need to see a pattern.
OpenStudy (anonymous):
lol..
OpenStudy (anonymous):
my prof did that last time, he took the first term and added it up with the last term, then got the answer, or it's prolly one of my hallucinations? I just remembered he did something like that
OpenStudy (anonymous):
z_z oh nvm me proceed.
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