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Math Help - Complex trig.

  1. #1
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    Complex trig.

    Recall that sin(x1+x2)=sinx1cosx2+cosx1sinx2 and cos(x1+x2)=cosx1cosx2-sinx1sinx2.

    Show that e^i(x1+x2)=e^(ix1)e^(ix2) using the formulas above.

    So I got e^i(x1+x2)=cos(x1+x2)+isin(x1+x2)=

    (cosx1cosx2-sinx1sinx2)+i(sinx1 cosx2+cosx1sinx2)=

    cos(x2)*(e^(ix1))+sin(x2)*(icosx1-sinx1), but this is where I am stumped.

    Any ideas how I can proceed.

    Thanks
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  2. #2
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    Hello, twittytwitter!

    Recall that: . \begin{array}{ccc}\ \sin(a+b)&=& \sin a\cos b+\cos a\sin b \\ \cos(a+b) &=& \cos a\cos b -\sin a\sin b \end{array}

    Show that: . e^{i(a+b)} \:=\:e^{ia}\cdot e^{ib}

    It's easier to start with the right side:

    \begin{array}{cccc}e^{ia}\cdot e^{ia} &=&\bigg[\cos a + i\sin a\bigg]\,\bigg[\cos b + i \sin b\bigg] \\ \\<br /> <br />
 & = & \cos a\cos b + i\sin a\cos b +   i \cos a \sin b + i^2\sin a\sin b \\ \\<br /> <br />
& =& (\cos a\cos b - \sin a\sin b) + i(\sin a\cos b + \cos a\sin b) \\ \\<br /> <br />
&=& \cos(a+b) + i\sin(a+b) \\ \\<br /> <br />
&=& e^{i(a+b)} \end{array}

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