Complex notation question

Discussion in 'Math' started by boks, Nov 9, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
    218
    0
    According to my book, one can write exp(ikd sin\vartheta ) + exp(-ikd sin\vartheta ) = 2cos(kd sin \vartheta)? Why is this?
     
    Last edited: Nov 10, 2008
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    A*cos(x)+j*A*sin(x)=A*exp(j*x)

    A*cos(x)+j*A*sin(-x)=A*exp(-j*x)

    If you add the right hand sides of the above equations you get

    A*exp(j*x)+A*exp(-j*x)=A*cos(x)+j*A*sin(x)+A*cos(x)+j*A*sin(-x)

    sin(-x)=-sin(x) thus,

    A*exp(j*x)+A*exp(-j*x)=A*cos(x)+j*A*sin(x)+A*cos(x)-j*A*sin(x)

    thus

    A*exp(j*x)+A*exp(-j*x)=A*cos(x)+A*cos(x)

    thus

    2*A*cos(x)=A*exp(j*x)+A*exp(-j*x)
     
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