Mimi_x3
  • Mimi_x3
im not able to figure this out
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mimi_x3
  • Mimi_x3
https://gyazo.com/883588ab3191a7dbb78de630c9b1549c
Mimi_x3
  • Mimi_x3
how does sin (...) sin(..) = cos(...)-cos(...)
DanJS
  • DanJS
link is broken...

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Mimi_x3
  • Mimi_x3
apparently its accordance to this https://gyazo.com/0df138c31be5b67523ca6f3ea60ca281
Mimi_x3
  • Mimi_x3
itrs not borken
inkyvoyd
  • inkyvoyd
FOURIER SERIES
Mimi_x3
  • Mimi_x3
NOT YET
inkyvoyd
  • inkyvoyd
YES IT IS ALMOST THERE
DanJS
  • DanJS
oh it was my browser tha tbroke, sorry hah
Mimi_x3
  • Mimi_x3
well ya IM LIKE 5 WEEKS BEHIND LOL
anonymous
  • anonymous
It's just product to sum formulas.
inkyvoyd
  • inkyvoyd
http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3
Mimi_x3
  • Mimi_x3
i need to know this https://gyazo.com/883588ab3191a7dbb78de630c9b1549c
inkyvoyd
  • inkyvoyd
miwi review http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3
inkyvoyd
  • inkyvoyd
actually the top of the link at http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx
DanJS
  • DanJS
hell yeah, pauls notebook is awesome
Mimi_x3
  • Mimi_x3
oh FML
Mimi_x3
  • Mimi_x3
I WANNA DROP THIS COURSE
DanJS
  • DanJS
Arthur Matuk has awesome video series for DE on MIT OCW, got me an A using that series ...
inkyvoyd
  • inkyvoyd
thanks for info dan - gonna take ODEs next semester :o
Mimi_x3
  • Mimi_x3
it still doesnt tell me why sin(..)sin(...) = cos(..) - cos(..)
inkyvoyd
  • inkyvoyd
http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex1 doesn't help?
zepdrix
  • zepdrix
@Mimi_x3 as Concentrationalizing mentioned, it's the trig product to sum formula:\[\large\rm \sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right]\]
Mimi_x3
  • Mimi_x3
why does my prof say this https://gyazo.com/55c13f04ae511455aa39b45ddc333030
Mimi_x3
  • Mimi_x3
(havent done maths for 3 years)
inkyvoyd
  • inkyvoyd
that's a trig fromula from precalc
zepdrix
  • zepdrix
the angle addition formula? Hmm not sure why he included that... thinking
inkyvoyd
  • inkyvoyd
the minus on top of the plus means the signs are opposite
inkyvoyd
  • inkyvoyd
zepdrix do you think he was using it to derive the formulas you linked too?
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
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 :)\[\large\rm \cos(A-B)-\cos(A+B)=2\sin(A) \sin(B)\]
dan815
  • dan815
what do u wanna know aobut inner product
dan815
  • dan815
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
dan815
  • dan815
|dw:1443255318087:dw|
Empty
  • Empty
\[e^{ix} = \cos x + i \sin x\]\[e^{-ix}=\cos x - i \sin x\]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align} \int\limits_{-\pi}^\pi\sin(mx)\sin(nx)\text dx &=\frac12\int\limits_{-\pi}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\\ &=\int\limits_{0}^\pi\cos((m-n)x)-\cos((m+n)x)\text dx\qquad (\text{even integrand})\\\\ &=\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big|_0^\pi\\ &=\frac{\sin(\pi(m-n))}{m-n}-\frac{\sin(\pi(m+n))}{m+n}\\ &\hspace{16em} p=m-n\\ &\hspace{16em} q=m+n\\ \text{I for }m\neq n &\qquad=\frac{\sin(\pi p)}{p}-\frac{\sin(\pi q)}{q}\\ &\qquad=\frac 0p-\frac0q\\ &\qquad=0\\ \\ \text{II for }m=n=0 &\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}-\frac{\sin(2\pi m)}{2m}\\ &\qquad\stackrel{\text{l'H}}=\pi\frac{\cos(\pi p)}{1}-2\pi\frac{\cos(2\pi m)}{2}\\ &\qquad=\pi-\frac{2\pi}2\\ &\qquad=0\\ \\ \text{III for }m=n\neq0&\qquad=\lim\limits_{p\rightarrow 0}\frac{\sin(\pi p)}{p}-\frac{\sin(\pi q)}{q}\\ &\qquad\stackrel{\text{l'H}}=\pi\lim\limits_{p\rightarrow 0}\frac{\cos(\pi p)}{1}-\frac 0q\\ &\qquad=\pi-0\\ &\qquad=\pi\\ \\ &=\pi\delta_{m,n}(1-\delta_{m,0}) \end{align}\]

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