Undergrad How can I go from sine to cosine using exponential numbers?

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The discussion focuses on converting sine and cosine functions using exponential forms. The key formulas provided are for cosine and sine in terms of exponentials: cos(ω) = (e^(jω) + e^(-jω))/2 and sin(ω) = (e^(jω) - e^(-jω))/(2j). A challenge is presented in manipulating the expression for cos(ω - π/2) into the form of sin(ω) and vice versa. The solution involves recognizing that e^(j(ω - π/2)) can be expressed as e^(jω) * e^(-jπ/2), which simplifies the conversion. The discussion concludes with a note on LaTeX formatting for trigonometric functions, emphasizing the simplicity of the notation.
amama
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##cos(\omega)## is

$$\frac{e^{j \omega } + e^{-j \omega }}{2}$$

##sin(\omega)## is

$$\frac{e^{j \omega } - e^{-j \omega }}{2 j}$$

I also know that ##cos(\omega - \pi / 2) = sin(\omega)##.

I've been trying to show this using exponentials, but I can't seem to manipulate one form into the other. Specifically, I can't figure out how to manipulate

$$\frac{e^{j (\omega - \pi / 2)} + e^{-j( \omega - \pi / 2)}}{2}$$

into

$$\frac{e^{j \omega } - e^{-j \omega }}{2 j}$$

or the other way around.

I seem to remember from back in school that there was some trick to this, but I can't remember it. I keep getting stuck on the fact that one has a plus between the exponential terms, and one has a minus.
 
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Write the exponentials as ##e^{j\omega} \cdot e^{...}## and then evaluate the second term.
 
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mfb said:
Write the exponentials as ##e^{j\omega} \cdot e^{...}## and then evaluate the second term.

So simple... thanks! I kept trying to convert everything to exponentials, but after evaluating ##e^{\pi /2}## and ##e^{-\pi /2}## to ##j## and ##-j## it all worked out.
 
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Just for future reference. The circular trigonometric functions have the neat LaTex codes \sin, \cos, \tan, \cot. So compare ## sin x## or ##sin(x)## to ##\sin x##. The brackets to denote the variable are superfluous.
 
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