Question

Simplify from euler to cos

Answer 1

From step one to step 2

Answer 2

could someone please help me with this question?

Answer 3

i am going to have to do this with pen and paper, but my guess is to write it in terms of sines and cosines and you will see the double angle formula for cosine
it will take a while to figure out which ones go together but i will try it

Answer 4

hi satellite73 :)
i dont see the combination of those exponent.. because they all contain a e^-j term.. and to turn it in a cos or sin.. you need to have e^j +/- e^-j

I really hope you can help me with this :)

Answer 5

ok first question
is i an index or is this
\[j^2=-1\] i.e. j is what we usually call "i"?

Answer 6

must be i otherwise it doesn't make any sense

Answer 7

owhh oke.. sorry.. i always use 'j' instead of 'i'.. but it is 'i' ;)

Answer 8

i is complex

Answer 9

ok give me a few minutes, i am sure this is doable, but i cannot do it on the fly

Answer 10

oke.. no problem.. im already glad you take the time to help me :)

Answer 11

ok it is not so bad
\[2e^{\frac{\pi k}{3}i}=2(\cos(\frac{\pi k}{3})+i\sin(\frac{\pi k}{3}))\]
\[2e^{\frac{ 5 \pi k}{3}i}=2(\cos(\frac{5 \pi k}{3})+i\sin(\frac{5 \pi k}{3})\]

Answer 12

now we notice that
\[\cos(\frac{\pi k}{3})+\cos(\frac{5 \pi k}{3})=2\cos(\frac{\pi k}{3})\] draw the points on the unit circle and it will probably be cleare.

Answer 13

are you saying cos(pi*k/3) is the same as cos(5*pi/3) ?

Answer 14

and also that
\[\sin(\frac{\pi k}{3})+\sin(\frac{ 5 \pi k}{3})=0\]

Answer 15

\[

Answer 16

i still dont get it :S

Answer 17

that is the picture for when k = 1 or k = 3 etc. you see that both the cosines are the same but the sines add up to zero

Answer 18

cos(πk/3)+cos(5πk/3)=2cos(πk/3)

This one i dont understand..

Answer 19

lets try it with specific values of k
k = 1
\[\cos(\frac{\pi}{3})=\cos(\frac{5\pi}{3})=\frac{1}{2}\]

Answer 20

i understand the k=1 :)

Answer 21

so you just fill in a random integer?

Answer 22

for k = 2 you get
\[\cos(\frac{2\pi}{3})=\cos(\frac{10\pi}{3})=-\frac{1}{2}\] think of where you are on the unit circle

Answer 23

if k is odd you are at these two points on the unit circle

Answer 24

and from the picture you see that both cosines are the same, but the sines add up to zero. that is why they are gone in the formula

Answer 25

hmm... oke :O

Answer 26

if k is even you are at one of these two points, and again the cosines are the same and the sines add up to zero

Answer 27

in fact we can be more specific. if k is odd then
\[\cos(\frac{k\pi}{3})=\cos(\frac{5k\pi}{3})=\frac{1}{2}\]

Answer 28

and similarly if k is even they are both -1/2

Answer 29

hmm... i need to take the time.. to read this back.. its a lot of information ;)

Answer 30

likewise for
\[e^{\frac{2k\pi}{3}i}\] and
\[e^{\frac{4k\pi}{3}i}\] write in terms of sine and cosine. see that the cosines are the same, but the sines add up to zero (cancel)
don't be afraid to try it with specific values of k, for example k = 1, 2, 3 and see exactly what you get and where you are on the unit circle. that will make it clear

Answer 31

okeee.. i will give it another try with your information..

Thank you somuch for taking the time to explain it to me :)

Answer 32

i hope this step was clear
\[2e^{\frac{\pi k}{3}i}=2(\cos(\frac{\pi k}{3})+i\sin(\frac{\pi k}{3}))\] if so y ou should be in good shape
post again if it is still not clear

Answer 33

okee.. i will thanks :)