Question

Simplify equation:

See attachment

Answer 1

i know that \[\cos(n \pi) = (-1)^{n}\]

but how did they get (-1)^(n-1) is the cos is: cos((1+n)t)

Answer 2

assess it from 0 to pi right?

Answer 3

good morning

Answer 4

the boundaries are from 0 to pi indeed

Answer 5

when t=0 and when t=pi is what I see

Answer 6

helloo :)

Answer 7

howdy

Answer 8

i think it is just shifting to make the plus and minus correct

Answer 9

cos(n*0) = 1

cos(n*pi) = .... flips back and forth from 1 to -1 right?

Answer 10

as a matter of fact i am sure it is. check what happens when you replace t by pi

Answer 11

if cos(x) is pi, then it will always be -1

but if cos(0) then this will always be 1 so why is it (-1)^(n-1)

Answer 12

you should write it out doubled; there is a step missing

Answer 13

but

\[(-1)^{n-1}=(-1)^{n+1}\]

right?

Answer 14

yes

Answer 15

why did they choose for n-1 and not for n+1?

Answer 16

cant tell yet, but I think its in the missing step ....

Answer 17

exactly is miss that step, so thats why i dont understand it :(

Answer 18

\[\frac{1}{pi}\left(\frac{-1}{1+n}cos((1+n)t)+\frac{-1}{1-n}cos((1-n)t)\right)_{0}^{pi}\]
\[\frac{1}{pi}\left(\frac{-1}{1+n}cos((1+n)pi)+\frac{-1}{1-n}cos((1-n)pi)-(\frac{-1}{1+n}cos(0)+\frac{-1}{1-n}cos(0))\right)\]
\[\frac{1}{pi}\left(\frac{-1}{1+n}cos((1+n)pi)+\frac{-1}{1-n}cos((1-n)pi)-(\frac{-1}{1+n}+\frac{-1}{1-n})\right)\]
\[\frac{1}{pi}\left(\frac{-1}{1+n}cos((1+n)pi)+\frac{-1}{1-n}cos((1-n)pi)+\frac{1}{1+n}+\frac{1}{1-n}\right)\]

so far

Answer 19

might as well see if we can combine those fractions on the right and see if we can match their work

Answer 20

or hold off on it :)

Answer 21

i see it, i think

Answer 22

they combined the denominators; so the cos(n*pi) is connected to both of them

Answer 23

\[\frac{1}{pi}\left(\frac{-1}{1+n}cos((1+n)pi)+\frac{-1}{1-n}cos((1-n)pi)+\frac{1}{1+n}+\frac{1}{1-n}\right)\]
\[\frac{1}{pi}\left(\frac{1}{1+n}+\frac{-1}{1+n}cos((1+n)pi)+\frac{1}{1-n}+\frac{-1}{1-n}cos((1-n)pi)\right)\]
\[\frac{1}{pi}\left(\frac{1}{1+n}-\frac{cos((1+n)pi)}{1+n}+\frac{1}{1-n}-\frac{cos((1-n)pi)}{1-n}\right)\]
\[\frac{1}{pi}\left(\frac{1-cos((1+n)pi)}{1+n}+\frac{1-cos((1-n)pi)}{1-n}\right)\]
and the cos(pi) stuff alternates

Answer 24

uhhh... and now they combined the denominators..

so multiplying with (1+n) and with (1-n)

Answer 25

right, the finalized it after words into one scheme

Answer 26

ahh oke.. let me try that :)
thank you for your help

Answer 27

youre welcome

Answer 28

amistre64,

I still dont know why it is (-1)^(n-1) in stead of (-1)^(n+1)

Answer 29

the result is the same, but they choose for (-1)^(n-1)

Answer 30

the reason is becasue they had to choose one them and thats the one they choose.

Answer 31

now, if n starts a specific number, then we can choose the best one that helps us out

Answer 32

but in this case it just looks like htey had to pick one and choose n-1

Answer 33

ahhh oke :) thank you for your explanation :)