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Mimi_x3
 one year ago
im not able to figure this out
Mimi_x3
 one year ago
im not able to figure this out

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Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0how does sin (...) sin(..) = cos(...)cos(...)

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0apparently its accordance to this https://gyazo.com/0df138c31be5b67523ca6f3ea60ca281

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0YES IT IS ALMOST THERE

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0oh it was my browser tha tbroke, sorry hah

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0well ya IM LIKE 5 WEEKS BEHIND LOL

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's just product to sum formulas.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0i need to know this https://gyazo.com/883588ab3191a7dbb78de630c9b1549c

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0miwi review http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0actually the top of the link at http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0hell yeah, pauls notebook is awesome

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0I WANNA DROP THIS COURSE

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Arthur Matuk has awesome video series for DE on MIT OCW, got me an A using that series ...

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0thanks for info dan  gonna take ODEs next semester :o

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0it still doesnt tell me why sin(..)sin(...) = cos(..)  cos(..)

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex1 doesn't help?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3@Mimi_x3 as Concentrationalizing mentioned, it's the trig product to sum formula:\[\large\rm \sin(a)\sin(b)=\frac{1}{2}\left[\cos(ab)\cos(a+b)\right]\]

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0why does my prof say this https://gyazo.com/55c13f04ae511455aa39b45ddc333030

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0(havent done maths for 3 years)

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0that's a trig fromula from precalc

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3the angle addition formula? Hmm not sure why he included that... thinking

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0the minus on top of the plus means the signs are opposite

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.0zepdrix do you think he was using it to derive the formulas you linked too?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hmm :\ I don't think so.\[\large\rm \cos(AB)=\cos(A) \cos(B)+\sin(A) \sin(B)\]Subtracting cos(A)cos(B) from each side,\[\large\rm \cos(AB)\cos(A) \cos(B)=\sin(A) \sin(B)\]Which doesn't quite get us there :d Hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Oh wait wait wait.. maybe this.\[\large\rm (1)\qquad \cos(AB)=\cos(A) \cos(B)+\sin(A) \sin(B)\]\[\large\rm (2)\qquad \cos(A+B)=\cos(A) \cos(B)\sin(A) \sin(B)\]I'm going to combine these equations like this: \(\large\rm (1)(2)\), giving us:\[\large\rm \cos(AB)\cos(A+B)=\begin{align}&~~~~\cos(A) \cos(B)+\sin(A) \sin(B)\\&\cos(A) \cos(B)+\sin(A) \sin(B)\end{align}\]Ah yes, I guess it does lead to the identity I mentioned :)\[\large\rm \cos(AB)\cos(A+B)=2\sin(A) \sin(B)\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0what do u wanna know aobut inner product

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.0\[e^{ix} = \cos x + i \sin x\]\[e^{ix}=\cos x  i \sin x\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[\begin{align} \int\limits_{\pi}^\pi\sin(mx)\sin(nx)\text dx &=\frac12\int\limits_{\pi}^\pi\cos((mn)x)\cos((m+n)x)\text dx\\ &=\int\limits_{0}^\pi\cos((mn)x)\cos((m+n)x)\text dx\qquad (\text{even integrand})\\\\ &=\frac{\sin((mn)x)}{mn}\frac{\sin((m+n)x)}{m+n}\Big_0^\pi\\ &=\frac{\sin(\pi(mn))}{mn}\frac{\sin(\pi(m+n))}{m+n}\\ &\hspace{16em} p=mn\\ &\hspace{16em} q=m+n\\ \text{I for }m\neq n &\qquad=\frac{\sin(\pi p)}{p}\frac{\sin(\pi q)}{q}\\ &\qquad=\frac 0p\frac0q\\ &\qquad=0\\ \\ \text{II for }m=n=0 &\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}\frac{\sin(2\pi m)}{2m}\\ &\qquad\stackrel{\text{l'H}}=\pi\frac{\cos(\pi p)}{1}2\pi\frac{\cos(2\pi m)}{2}\\ &\qquad=\pi\frac{2\pi}2\\ &\qquad=0\\ \\ \text{III for }m=n\neq0&\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}\frac{\sin(\pi q)}{q}\\ &\qquad\stackrel{\text{l'H}}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos(\pi p)}{1}\frac 0q\\ &\qquad=\pi0\\ &\qquad=\pi\\ \\ &=\pi\delta_{m,n}(1\delta_{m,0}) \end{align}\]
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