Question

Use Euler’s formula for exp(ix) and exp(-ix) to write cos(x) as a combination of exp(ix) and exp(-ix)

Answer = (cos(x) = (exp(ix)+exp(-ix))/2)

For real a and b, use the previous answer to find write both cos(a+b) and cos(a)cos(b) in terms of exp. Throughout the rest you will probably use exp(x+y)=exp(x)exp(y).

Answer 1

`cos(a+b)=(e^(i(a-b))+e^(i(-a+b)))/(2)`

Explanation:

`cos(x)=(e^(ix)+e^(-ix))/(2)`

`cos(a+b)`

We need to expand cos(a+b) using the cos addition formula.

`cos(a+b)=cos(a)cos(b)-sin(a)sin(b)`

We know that we also need to use Euler's formula for sin, which is:

`sin(x)=(e^(ix)-e^(-ix))/(2)` (you can get this from a similar way of getting the first result, of simply just expanding
`e^(ix)=cosx+isinx` and seeing the necessary result)

We can now substitute our cos's and sin's for e's

`cos(a+b)=((e^(ia)+e^(-ia))/(2))((e^(ib)+e^(-ib))/(2))-((e^(ia)-e^(-ia))/(2))((e^(ib)-e^(-ib))/(2))`

Now lets multiply out both of our terms, I'm using the exponent multiplication identity here (
`e^(x+y)=e^xe^y`)

`cos(a+b)=(e^(i(a+b)) + e^(i(a-b))+e^(i(-a+b)) + e^(i(-a-b)))/(4)-(e^(i(a+b)) - e^(i(a-b))-e^(i(-a+b))+e^(i(-a-b)))/(4)`

Now we can subtract these two terms.

`cos(a+b)=(2e^(i(a-b))+2e^(i(-a+b)))/(4)`

This is starting to look a lot tidier, let's cancel the 2

`cos(a+b)=(e^(i(a-b))+e^(i(-a+b)))/(2)`