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    So
     e^{j \omega t} = cos \omega t + jsin \omega t

    My brain is just not workin atm, how do u derive the following from the above :

     cos \omega t = \frac{e^{j \omega t} + e^{-j \omega t}    } {2}

    And the sine identity as well plz?


    Thanks!
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    (Original post by wizz_kid)
    So
     e^{j \omega t} = cos \omega t + jsin \omega t

    My brain is just not workin atm, how do u derive the following from the above :

     cos \omega t = \frac{e^{j \omega t} + e^{-j \omega t}    } {2}

    And the sine identity as well plz?


    Thanks!
    You know what e^{j\omega t} is. What do you get for e^{-j\omega t}? (What do you know about even and odd functions - is cos even or odd, same question for sin)

    What do you get if you add e^{j\omega t} and e^{-j\omega t}? (And likewise, if you subtract them)?
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    (Original post by wizz_kid)
    So
     e^{j \omega t} = cos \omega t + jsin \omega t

    My brain is just not workin atm, how do u derive the following from the above :

     cos \omega t = \frac{e^{j \omega t} + e^{-j \omega t}    } {2}

    And the sine identity as well plz?


    Thanks!
    e^{-j \omega t} = cos (-\omega t) + jsin(-\omega t) = cos \omega t - jsin \omega t

    Adding this to e^{j \omega t} gives the required result as the positive sine and the negative sine cancel. Subtracting the two will yield the result for sines.
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    (Original post by Phil_Waite)
    e^{-j \omega t} = cos (-\omega t) + jsin(-\omega t) = cos \omega t - jsin \omega t

    Adding this to e^{j \omega t} gives the required result as the positive sine and the negative sine cancel. Subtracting the two will yield the result for sines.
    Yes! Its all about thinking in terms of odd and even functions! Thanks mate!!
 
 
 
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