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anonymous
 5 years ago
sum of (sin 2kx/11 + i cos2kx/11) from k=1 to k=10
where i=square root of negative one
anonymous
 5 years ago
sum of (sin 2kx/11 + i cos2kx/11) from k=1 to k=10 where i=square root of negative one

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{10} [(\sin 2k \pi/11) +i(\cos 2k \pi/11) ]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you want to find here? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the value of this summation=)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well. First take 1/11 as a common factor and cut the summation to two parts.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry common factor??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You will see that the sum of (sin2k pi) is always zero for any value for k=1,2,3,..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin (2k pi) is always zero for k=1,2,3,..*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes.. so the sine part comes out to be zero. what about the cosine part?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on the contract cos (2k pi) is one for k=1,2,..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the summation of cos (2k pi) is from k=1 to k=10 is equal to 10. right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you following? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the answer is supposed to be i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{1 \over 11}\sum_{k=1}^{10}\sin(2k \pi)+{i \over 11}\sum_{k=1}^{10}\cos (2k \pi)=0+{i \over 11}(10)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait let me check again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is 11 part of the angle?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/11 is not multiplied to sin but its multiplied to its angle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't cos(20 pi) = 1? which will give you i/11?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 1/11 = not part of the angle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should have put that in parenthesis >.<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why didn't you put it in the angle =_=...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you got a solution for "a" problem :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey anwar, aren't you supposed to find the summation of both sin and cos from 1  10?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry about that guys

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's OK.. But, I am not sure if I have enough time to do the problem. Just let me think about it quickly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it possible to solve using euler's representation of complex no.?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's what I am thinking about right now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah all through 2,3,4 to 10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@sstarica: you just substitute for k=1,2,3,.. and add them up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no need, there's a faster way to compute the sum without going through the 1 2 3 stages :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take first term and add it with last term

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well yeah.. You just need to see a pattern.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0z_z oh nvm me proceed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[i \sum_{k=1}^{10} [\cos(2k \pi/11) i \sin(2k \pi/11)]\] =\[i \sum_{k=1}^{10} (e ^{2k \pi/11})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know how to proceed any further

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's right, but there is an i in the power.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea there should be.. otherwise its not a correct representation of the complex number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[i \sum_{k=1}^{10} (e ^{2k \pi i/11})\] should look like this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey star u workin on it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am back, but you're gone. :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I am always getting the idea.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not always, almost :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin(2 pi/11)+sin(20 pi/11)=?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin x+sin y=2sin((x+y)/2)+....., right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying to fix the DAMN PRINTER! >_<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Apply it to sin(2 pi/11)+sin(20 pi/11)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0excuse my language ^^"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin((2 * pi) / 11) + sin((20 * pi) / 11) = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It will give you 0. Same with sin(4 pi/11)+sin(18 pi/11), and so on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you see that the "sin" part of the summation will be 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0agreed... wat abt the cosine part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0trying it :).. just give me few minutes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01,2,3,4,5,6,7,8,9........60

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what are you saying starica?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cosx+cosy=2cos((x+y)/2)cos((xy)/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm too tired to try and I love counting lol =P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it looks simple though u_u

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you should know..:P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I try, I'll figure it out, but I'm tired :( ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why dont you go take a nap for a change..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm in campus, and I can't sleep during the day

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r u an undergrad student?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mhm, a first year ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gr8.. ehat are you majoring in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0IT , computer engineering hopefully.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You said you want to go to chemical engineering, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt workin out anwar?:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not really. My friend is distracting me :P .. He wants me to go with him get lunch :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol.. anyways i'd better leave now.. thanx for your effort anwar and sstarica..=)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No problem. If I get the idea, I will just post the solution here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not really, I'm staying in IT
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