How do you get cos{(a+b)}=cos{a}cos{b} - sin{a}sin{b}?

[Julia Coggins]

How do you get cos{a+b}=cos{a}cos{b} - sin{a}sin{b} from cos{a}cos{b} - sin{a}sin{b} + i(sin{a}cos{b} + cos{a}sin{b}?
From trying to use Euler's formula.
cos{a+b} + isin{a+b} = e^i(a+b)= e^ia + e^ib
(cos{a} + isin{a})(cos{b} + isin{b})

 

[BvU]

Hi Julia,
Simply carry out (write out) the multiplication: the real part is the $\cos(\alpha + \beta)$ expression and the imaginary part the $\sin$ one !

 

[Julia Coggins]

I'm just not sure how to put this in an equation that has sin(a + b) on one side or cos(a+b)

 

[Julia Coggins]

I know the real and imaginary parts, I just don't know how to show they're equal to cos{a + b} and sin{a+b} respectively.

 

[mathman]

$cos(a+b)=(e^{i(a+b)}+e^{-i(a+b)})/2$
$cos(a)cos(b)-sin(a)sin(b)=[(e^{ia}+e^{-ia})(e^{ib}+e^{-ib})+(e^{ia}-e^{-ia})(e^{ib}-e^{-ib})]/4 =(e^{i(a+b)}+e^{-i(a+b)})]/2$

 

[axmls]

e^i(a+b)= e^ia + e^ib

No, that's not how exponents work. It should be $$e^{i(a+b)} = e^{ia} e^{ib}$$

EDIT. I see you've written it correctly below that line. If you carry out the multiplication, you'll find that the real part is what you're looking for.

 

[Svein]

1. $e^{i(a+b)}= e^{ia}e^{ib}$
2. $e^{i(a+b)}= \cos(a+b) + i\sin(a+b)$
3. $e^{ia}e^{ib}=(\cos(a)+i\sin(a))\cdot (\cos(b)+i\sin(b))=(\cos(a) \cos(b)-sin(a)sin(b))+i(\sin(a)\cos(b)+\cos(a)sin(b))$

Now the real and imaginary parts must separately be equal to each other.

 

[BvU]

Hello Julia,

cos{a+b} + isin{a+b} = e^i(a+b)= e^ia + e^ib
(cos{a} + isin{a})(cos{b} + isin{b})

Sorry I missed the plus sign error. It should have been a multiplication. But on the next line you do write a multiplication, so I expect that + was just a typo.

Somewhat contrary to PF culture your helpers have given the full solution. Just to make sure: do you understand it all now ?

 

[Julia Coggins]

Yes, thanks all