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

- anonymous

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

\[\sum_{k=1}^{10} [(\sin 2k \pi/11) +i(\cos 2k \pi/11) ]\]

- anonymous

what do you want to find here? ^_^

- anonymous

the value of this summation=)

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## More answers

- anonymous

Well. First take 1/11 as a common factor and cut the summation to two parts.

- anonymous

sorry common factor??

- anonymous

You will see that the sum of (sin2k pi) is always zero for any value for k=1,2,3,..

- anonymous

Yeah.

- anonymous

sin (2k pi) is always zero for k=1,2,3,..*

- anonymous

yes.. so the sine part comes out to be zero. what about the cosine part?

- anonymous

on the contract cos (2k pi) is one for k=1,2,..

- anonymous

So the summation of cos (2k pi) is from k=1 to k=10 is equal to 10. right?

- anonymous

Are you following? :)

- anonymous

well the answer is supposed to be -i

- anonymous

\[{1 \over 11}\sum_{k=1}^{10}\sin(2k \pi)+{i \over 11}\sum_{k=1}^{10}\cos (2k \pi)=0+{i \over 11}(10)\]

- anonymous

Are you sure?

- anonymous

wait let me check again.

- anonymous

Is 11 part of the angle?

- anonymous

1/11 is not multiplied to sin but its multiplied to its angle

- anonymous

isn't cos(20 pi) = -1? which will give you -i/11?

- anonymous

LOL

- anonymous

what x_x

- anonymous

and 1/11 = not part of the angle

- anonymous

You should have put that in parenthesis >.<

- anonymous

why didn't you put it in the angle =_=...

- anonymous

Well, you got a solution for "a" problem :P

- anonymous

hey anwar, aren't you supposed to find the summation of both sin and cos from 1 - 10?

- anonymous

sorry about that guys

- anonymous

Yeah sstarica

- 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.

- anonymous

okay anwar

- anonymous

is it possible to solve using euler's representation of complex no.?

- 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

- anonymous

That's what I am thinking about right now.

- anonymous

lol

- anonymous

yeah all through 2,3,4 to 10

- anonymous

@sstarica: you just substitute for k=1,2,3,.. and add them up.

- anonymous

no need, there's a faster way to compute the sum without going through the 1 2 3 stages :)

- anonymous

take first term and add it with last term

- anonymous

and whats that?

- anonymous

something like that?

- anonymous

I don't know lol

- anonymous

Well yeah.. You just need to see a pattern.

- anonymous

lol..

- 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

- anonymous

z_z oh nvm me proceed.

- anonymous

\[i \sum_{k=1}^{10} [\cos(2k \pi/11) -i \sin(2k \pi/11)]\]
=\[i \sum_{k=1}^{10} (e ^{-2k \pi/11})\]

- anonymous

i dont know how to proceed any further

- anonymous

that's right, but there is an i in the power.

- anonymous

there is?

- anonymous

There should be :)

- anonymous

o.o

- anonymous

what?!

- anonymous

yea there should be.. otherwise its not a correct representation of the complex number

- anonymous

o..o

- anonymous

\[i \sum_{k=1}^{10} (e ^{-2k \pi i/11})\]
should look like this

- anonymous

=.=

- anonymous

hey star u workin on it?

- anonymous

I gtg now.. I'll BRB

- anonymous

bye ^_^

- anonymous

bye

- anonymous

I am back, but you're gone. :(

- anonymous

I think I am always getting the idea.

- anonymous

not always, almost :)

- anonymous

giv it a try..=)

- anonymous

sin(2 pi/11)+sin(20 pi/11)=?

- anonymous

Hello, anyone there?

- anonymous

sin x+sin y=2sin((x+y)/2)+....., right?

- anonymous

I have no idea lol

- anonymous

I'm trying to fix the DAMN PRINTER! >_<

- anonymous

Apply it to sin(2 pi/11)+sin(20 pi/11)

- anonymous

excuse my language ^^"

- anonymous

sin((2 * pi) / 11) + sin((20 * pi) / 11) = 0

- anonymous

It will give you 0. Same with sin(4 pi/11)+sin(18 pi/11), and so on.

- anonymous

yup

- anonymous

Do you see that the "sin" part of the summation will be 0?

- anonymous

agreed... wat abt the cosine part

- anonymous

trying it :).. just give me few minutes.

- anonymous

1,2,3,4,5,6,7,8,9........60

- anonymous

yup sure

- anonymous

what are you saying starica?

- anonymous

I'm counting ^_^

- anonymous

3 mins have passed

- anonymous

its still "few"

- anonymous

lol yeah~

- anonymous

=)
arent you trying?

- anonymous

Don't count -.-

- anonymous

cosx+cosy=2cos((x+y)/2)cos((x-y)/2)

- anonymous

I'm too tired to try and I love counting lol =P

- anonymous

it looks simple though u_u

- anonymous

well you should know..:P

- anonymous

If I try, I'll figure it out, but I'm tired :( ...

- anonymous

why dont you go take a nap for a change..

- anonymous

I'm in campus, and I can't sleep during the day

- anonymous

r u an undergrad student?

- anonymous

mhm, a first year ^_^

- anonymous

gr8.. ehat are you majoring in?

- anonymous

*what

- anonymous

IT , computer engineering hopefully.

- anonymous

I am tired :)

- anonymous

You said you want to go to chemical engineering, right?

- anonymous

isnt workin out anwar?:-(

- anonymous

Not really. My friend is distracting me :P .. He wants me to go with him get lunch :)

- anonymous

lol..
anyways i'd better leave now.. thanx for your effort anwar and sstarica..=)

- anonymous

No problem. If I get the idea, I will just post the solution here.

- anonymous

okay buddy.. bye

- anonymous

Bye!!

- anonymous

not really, I'm staying in IT

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