Proving trig identities with euler's

[schapman22]

Homework Statement

Use Euler's identity to prove that cos(u)cos(v)=(1/2)[cos(u-v)+cos(u+v)]
and sin(u)cos(v)=(1/2)[sin(u+v)+sin(u-v)]

Homework Equations

e^ui=cos(u) + isin(u)
e^-ui=cos(u)-isin(u)

The Attempt at a Solution

I was able to this with other trig identities with no problem but this one I have hit a wall.
we are supposed to start with e^(u+v)i+e^(u-v)i=e^u(e^vi+e^-vi) which becomes.
cos(u+v)+isin(u+v)+cos(u-v)+isin(u-v)=e^u(cos(v)+isin(v)+cos(v)-isin(v)) then
equating the real parts
cos(u+v)+cos(u-v)=e^u(2cos(v)) then divide by 2
(1/2)[cos(u+v)+cos(u-v)]=e^u(cos(v))

I cannot figure out why I have an e^u and not a cos(u). Does anyone see where I have gone wrong or what I am missing? Thank you in advance.

 

[tiny-tim]

hi schapman22!

we are supposed to start with e^(u+v)i+e^(u-v)i=e^u(e^vi+e^-vi)

you're missing an "i" …

e^(u+v)i+e^(u-v)i=e^ui(e^vi+e^-vi) ​

 

[schapman22]

I'm looking at my worksheet and it says to use e^u(e^vi+e^-vi)
Are you certain of that. It could be a typo because my teacher hand writes all of our assignments.

 

[tiny-tim]

Are you certain of that.

yup!

look at it! ​

 

[schapman22]

Thanks, I really wish my teacher would use the book. This is like the 5th time I've spent hours on a problem only to find out there's a typo in it haha. I appreciate it.