im not able to figure this out

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im not able to figure this out

Mathematics
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https://gyazo.com/883588ab3191a7dbb78de630c9b1549c
how does sin (...) sin(..) = cos(...)-cos(...)
link is broken...

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apparently its accordance to this https://gyazo.com/0df138c31be5b67523ca6f3ea60ca281
itrs not borken
FOURIER SERIES
NOT YET
YES IT IS ALMOST THERE
oh it was my browser tha tbroke, sorry hah
well ya IM LIKE 5 WEEKS BEHIND LOL
It's just product to sum formulas.
http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3
i need to know this https://gyazo.com/883588ab3191a7dbb78de630c9b1549c
miwi review http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3
actually the top of the link at http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx
hell yeah, pauls notebook is awesome
oh FML
I WANNA DROP THIS COURSE
Arthur Matuk has awesome video series for DE on MIT OCW, got me an A using that series ...
thanks for info dan - gonna take ODEs next semester :o
it still doesnt tell me why sin(..)sin(...) = cos(..) - cos(..)
http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex1 doesn't help?
@Mimi_x3 as Concentrationalizing mentioned, it's the trig product to sum formula:\[\large\rm \sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right]\]
why does my prof say this https://gyazo.com/55c13f04ae511455aa39b45ddc333030
(havent done maths for 3 years)
that's a trig fromula from precalc
the angle addition formula? Hmm not sure why he included that... thinking
the minus on top of the plus means the signs are opposite
zepdrix do you think he was using it to derive the formulas you linked too?
Hmm :\ I don't think so.\[\large\rm \cos(A-B)=\cos(A) \cos(B)+\sin(A) \sin(B)\]Subtracting cos(A)cos(B) from each side,\[\large\rm \cos(A-B)-\cos(A) \cos(B)=\sin(A) \sin(B)\]Which doesn't quite get us there :d Hmm
Oh wait wait wait.. maybe this.\[\large\rm (1)\qquad \cos(A-B)=\cos(A) \cos(B)+\sin(A) \sin(B)\]\[\large\rm (2)\qquad \cos(A+B)=\cos(A) \cos(B)-\sin(A) \sin(B)\]I'm going to combine these equations like this: \(\large\rm (1)-(2)\), giving us:\[\large\rm \cos(A-B)-\cos(A+B)=\begin{align}&~~~~\cos(A) \cos(B)+\sin(A) \sin(B)\\&-\cos(A) \cos(B)+\sin(A) \sin(B)\end{align}\]Ah yes, I guess it does lead to the identity I mentioned :)\[\large\rm \cos(A-B)-\cos(A+B)=2\sin(A) \sin(B)\]
what do u wanna know aobut inner product
you know how to take inner / dot produt of 2 vecotrs now think of it like taking infinite dimensional dot product where sin(x) is telling you like the value at eacch dimension in your vector
|dw:1443255318087:dw|
\[e^{ix} = \cos x + i \sin x\]\[e^{-ix}=\cos x - i \sin x\]
\[\begin{align} \int\limits_{-\pi}^\pi\sin(mx)\sin(nx)\text dx &=\frac12\int\limits_{-\pi}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\\ &=\int\limits_{0}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\qquad (\text{even integrand})\\\\ &=\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big|_0^\pi\\ &=\frac{\sin(\pi(m-n))}{m-n}-\frac{\sin(\pi(m+n))}{m+n}\\ &\hspace{16em} p=m-n\\ &\hspace{16em} q=m+n\\ \text{I for }m\neq n &\qquad=\frac{\sin(\pi p)}{p}-\frac{\sin(\pi q)}{q}\\ &\qquad=\frac 0p-\frac0q\\ &\qquad=0\\ \\ \text{II for }m=n=0 &\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}-\frac{\sin(2\pi m)}{2m}\\ &\qquad\stackrel{\text{l'H}}=\pi\frac{\cos(\pi p)}{1}-2\pi\frac{\cos(2\pi m)}{2}\\ &\qquad=\pi-\frac{2\pi}2\\ &\qquad=0\\ \\ \text{III for }m=n\neq0&\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}-\frac{\sin(\pi q)}{q}\\ &\qquad\stackrel{\text{l'H}}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos(\pi p)}{1}-\frac 0q\\ &\qquad=\pi-0\\ &\qquad=\pi\\ \\ &=\pi\delta_{m,n}(1-\delta_{m,0}) \end{align}\]

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