Thread: how to solve the following using Euler identity .

show that

exp (i (x + y)) = exp (ix) exp (iy)

Euler's identity is:

hence:



Now, use the angle-sum identities for sine and cosine and see if you can arrange the terms to get the desired result.

Originally Posted by MarkFL2

Euler's identity is:

hence:



Now, use the angle-sum identities for sine and cosine and see if you can arrange the terms to get the desired result.

thanks for your reply. i ended up with this :
exp(i(x+y))= cos(x+y)+isin(x+y)
= cos(x)+cos(y)+isin(x)+isin(y)
=[cos(x)+isin(x)]+[cos(y)+isin(y)]
where cos(x)+isin(x)= exp(i(x))
after substitution we get
=exp(i(x))+exp(i(y))

the confusing part is the "+" sign that i'm getting it should be " * "

any suggestions..

You need to use the identities:





These are the angle-sum identities for sine and cosine.

= cos(x)+cos(y)+isin(x)+isin(y)
=[cos(x)+isin(x)]+[cos(y)+isin(y)]
where cos(x)+isin(x)= exp(i(x))
after substitution we get
=exp(i(x))+exp(i(y))
\Quote

yes i'm using these identities.

It looks to me like you are using:





and these are not correct, and will give you the wrong result, as you have found.

please take a look at this
The Way of Analysis - Robert S. Strichartz - Google Books

I'm using the same method but my answer is ... exp(i(x))+exp(i(y)) which is not correct

Access is restricted to that page.

Use the correct identities I gave above, and you will get the correct result. If you get stuck, I will be glad to help.

thank you for your help i got it

cos(x+y)=cosx.cosy-sinx.siny

sin(x+y)=sinx.cosy+cosx.siny

Now, cos(x+y)+isin(x+y)

=(cosx.siny-sinx.cosy)+i(sinx.cosy+cosx.sinx)

= cosx.siny-sinx.cosy+isinx.cosy+icosx.sinx

=(cosx+isinx)(cosy+isiny)[i^2=-1]

Yes, good work!