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

  1. Apr 6, 2016 #1
    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})
  2. jcsd
  3. Apr 6, 2016 #2


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    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 !
  4. Apr 6, 2016 #3
    I'm just not sure how to put this in an equation that has sin(a + b) on one side or cos(a+b)
  5. Apr 6, 2016 #4
    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.
  6. Apr 6, 2016 #5


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  7. Apr 6, 2016 #6
    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.
  8. Apr 7, 2016 #7


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    1. [itex]e^{i(a+b)}= e^{ia}e^{ib} [/itex]
    2. [itex]e^{i(a+b)}= \cos(a+b) + i\sin(a+b) [/itex]
    3. [itex] 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))[/itex]
    Now the real and imaginary parts must separately be equal to each other.
  9. Apr 7, 2016 #8


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    Hello Julia,
    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 ?
  10. Apr 12, 2016 #9
    Yes, thanks all
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