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Roo2
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Deriving addition of sines from Euler's formula (edit: please ignore)
I'd like to derive the value of sin(2πkt + θ) so that I can convince myself that the Fourier sum in terms of sines and cosines works. I found the derivation on Wolfram by using the full Euler's formula and equating reals and imaginaries, but I'd like to do it from just sine. I think I did it but I'm confused about the implications. Could someone straighten me out?
e^{ix} = cos(x) + isin(x)
sin(x) = \frac{e^{ix} - e^{-ix}}{2i}
sin(2πkt + θ) = \frac{e^{i(2πkt + θ)} - e^{-i(2πkt + θ)}}{2i}
= \frac{e^{i2πkt}e^{iθ} - e^{-i2πkt}e^{-iθ}}{2i}
= \frac{ae^{i2πkt} - be^{-i2πkt}}{2i}
= \frac{acos(2πkt) + aisin(2πkt) - bcos(2πkt) + bisin(2πkt)}{2i}
= \frac{(a-b)cos(2πkt) + (a+b)isin(2πkt)}{2i}
= \frac{(a-b)}{2i}cos(2πkt) + \frac{(a+b)}{2}sin(2πkt)
= \frac{i(b-a)}{2}cos(2πkt) + \frac{(a+b)}{2}sin(2πkt)
= \frac{i(e^{-iθ}-e^{iθ})}{2}cos(2πkt) + \frac{(e^{iθ}+e^{-iθ})}{2}sin(2πkt)
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Edit: Wow, I'm an idiot. I didn't realize that the first coefficient evaluates to sin(θ) while the second evaluates to cos(θ), thereby proving the sum of angles rule. Please ignore my questions. I'll refrain from deleting this thread so that if someone Googles a similar question they'll find the above derivation.
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I have two questions now:
1: Is my math right? Did I arrive at the correct result? The lecture I'm looking at only says that sin(2πkt + θ) = acos(2πkt) + bsin(2πkt), but I wanted to derive the whole thing so that I could appreciate it.
2: It seems to me that the coefficients of the cos + sin representation should be complex except for in special cases. How, then, can the sum of such waves represent a real wave? Perhaps you could only take the real part of each component of the sum, but that leaves "extra" information in the complex planes. How is this avoided?
Thank you.
Homework Statement
I'd like to derive the value of sin(2πkt + θ) so that I can convince myself that the Fourier sum in terms of sines and cosines works. I found the derivation on Wolfram by using the full Euler's formula and equating reals and imaginaries, but I'd like to do it from just sine. I think I did it but I'm confused about the implications. Could someone straighten me out?
Homework Equations
e^{ix} = cos(x) + isin(x)
sin(x) = \frac{e^{ix} - e^{-ix}}{2i}
The Attempt at a Solution
sin(2πkt + θ) = \frac{e^{i(2πkt + θ)} - e^{-i(2πkt + θ)}}{2i}
= \frac{e^{i2πkt}e^{iθ} - e^{-i2πkt}e^{-iθ}}{2i}
= \frac{ae^{i2πkt} - be^{-i2πkt}}{2i}
= \frac{acos(2πkt) + aisin(2πkt) - bcos(2πkt) + bisin(2πkt)}{2i}
= \frac{(a-b)cos(2πkt) + (a+b)isin(2πkt)}{2i}
= \frac{(a-b)}{2i}cos(2πkt) + \frac{(a+b)}{2}sin(2πkt)
= \frac{i(b-a)}{2}cos(2πkt) + \frac{(a+b)}{2}sin(2πkt)
= \frac{i(e^{-iθ}-e^{iθ})}{2}cos(2πkt) + \frac{(e^{iθ}+e^{-iθ})}{2}sin(2πkt)
------------------------
Edit: Wow, I'm an idiot. I didn't realize that the first coefficient evaluates to sin(θ) while the second evaluates to cos(θ), thereby proving the sum of angles rule. Please ignore my questions. I'll refrain from deleting this thread so that if someone Googles a similar question they'll find the above derivation.
------------------------
I have two questions now:
1: Is my math right? Did I arrive at the correct result? The lecture I'm looking at only says that sin(2πkt + θ) = acos(2πkt) + bsin(2πkt), but I wanted to derive the whole thing so that I could appreciate it.
2: It seems to me that the coefficients of the cos + sin representation should be complex except for in special cases. How, then, can the sum of such waves represent a real wave? Perhaps you could only take the real part of each component of the sum, but that leaves "extra" information in the complex planes. How is this avoided?
Thank you.
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