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Mathematics 84 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

what do you want to find here? ^_^

OpenStudy (anonymous):

the value of this summation=)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

sorry common factor??

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Yeah.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Are you following? :)

OpenStudy (anonymous):

well the answer is supposed to be -i

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Are you sure?

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

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

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

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

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?

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i dont know how to proceed any further

OpenStudy (anonymous):

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

OpenStudy (anonymous):

there is?

OpenStudy (anonymous):

There should be :)

OpenStudy (anonymous):

o.o

OpenStudy (anonymous):

what?!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

o..o

OpenStudy (anonymous):

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

OpenStudy (anonymous):

=.=

OpenStudy (anonymous):

hey star u workin on it?

OpenStudy (anonymous):

I gtg now.. I'll BRB

OpenStudy (anonymous):

bye ^_^

OpenStudy (anonymous):

bye

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I think I am always getting the idea.

OpenStudy (anonymous):

not always, almost :)

OpenStudy (anonymous):

giv it a try..=)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Hello, anyone there?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I have no idea lol

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

excuse my language ^^"

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

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

OpenStudy (anonymous):

agreed... wat abt the cosine part

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yup sure

OpenStudy (anonymous):

what are you saying starica?

OpenStudy (anonymous):

I'm counting ^_^

OpenStudy (anonymous):

3 mins have passed

OpenStudy (anonymous):

its still "few"

OpenStudy (anonymous):

lol yeah~

OpenStudy (anonymous):

=) arent you trying?

OpenStudy (anonymous):

Don't count -.-

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it looks simple though u_u

OpenStudy (anonymous):

well you should know..:P

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

r u an undergrad student?

OpenStudy (anonymous):

mhm, a first year ^_^

OpenStudy (anonymous):

gr8.. ehat are you majoring in?

OpenStudy (anonymous):

*what

OpenStudy (anonymous):

IT , computer engineering hopefully.

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