Complex notation question

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boks

Joined Oct 10, 2008
218
According to my book, one can write \[exp(ikd sin\vartheta ) + exp(-ikd sin\vartheta ) = 2cos(kd sin \vartheta)\]? Why is this?
 
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mik3

Joined Feb 4, 2008
4,846
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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