Question

im not able to figure this out

Answer 1

https://gyazo.com/883588ab3191a7dbb78de630c9b1549c

Answer 2

how does sin (...) sin(..)
=
cos(...)-cos(...)

Answer 3

link is broken...

Answer 4

apparently its accordance to this
https://gyazo.com/0df138c31be5b67523ca6f3ea60ca281

Answer 5

itrs not borken

Answer 6

FOURIER SERIES

Answer 7

NOT YET

Answer 8

YES IT IS ALMOST THERE

Answer 9

oh it was my browser tha tbroke, sorry hah

Answer 10

well ya
IM LIKE 5 WEEKS BEHIND LOL

Answer 11

It's just product to sum formulas.

Answer 12

http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3

Answer 13

i need to know this https://gyazo.com/883588ab3191a7dbb78de630c9b1549c

Answer 14

miwi review http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex3

Answer 15

actually the top of the link at http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx

Answer 16

hell yeah, pauls notebook is awesome

Answer 17

oh FML

Answer 18

I WANNA DROP THIS COURSE

Answer 19

Arthur Matuk has awesome video series for DE on MIT OCW, got me an A using that series ...

Answer 20

thanks for info dan - gonna take ODEs next semester :o

Answer 21

it still doesnt tell me why
sin(..)sin(...)
= cos(..) - cos(..)

Answer 22

http://tutorial.math.lamar.edu/Classes/DE/PeriodicOrthogonal.aspx#BVPFourier_Orthog_Ex1
doesn't help?

Answer 23

@Mimi_x3 as Concentrationalizing mentioned, it's the trig product to sum formula:\[\large\rm \sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right]\]

Answer 24

why does my prof say this
https://gyazo.com/55c13f04ae511455aa39b45ddc333030

Answer 25

(havent done maths for 3 years)

Answer 26

that's a trig fromula from precalc

Answer 27

the angle addition formula? Hmm not sure why he included that... thinking

Answer 28

the minus on top of the plus means the signs are opposite

Answer 29

zepdrix do you think he was using it to derive the formulas you linked too?

Answer 30

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

Answer 31

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 :)\[\large\rm \cos(A-B)-\cos(A+B)=2\sin(A) \sin(B)\]

Answer 32

what do u wanna know aobut inner product

Answer 33

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