- #1

schapman22

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## 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.