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

Answer:

$cos(a+b)=\frac{e^{i(a-b)}+e^{i(-a+b)}}{2}$

Step-by-step explanation:

$cos(x)=\frac{e^{ix}+e^{-ix}}{2}$

$cos(a+b)$

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:

$sin(x)=\frac{e^{ix}-e^{-ix}}{2}$ (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

$cos(a+b)=(\frac{e^{ia}+e^{-ia}}{2})(\frac{e^{ib}+e^{-ib}}{2})-(\frac{e^{ia}-e^{-ia}}{2})(\frac{e^{ib}-e^{-ib}}{2})$

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

$cos(a+b)=\frac{e^{i(a+b)} + e^{i(a-b)}+e^{i(-a+b)} + e^{i(-a-b)}}{4}-\frac{e^{i(a+b)} - e^{i(a-b)}-e^{i(-a+b)}+e^{i(-a-b)}}{4}$

Now we can subtract these two terms.

$cos(a+b)=\frac{2e^{i(a-b)}+2e^{i(-a+b)}}{4}$

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

$cos(a+b)=\frac{e^{i(a-b)}+e^{i(-a+b)}}{2}$