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Astrophysics

  • one year ago

How do you derive this using vector algebra

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  1. Astrophysics
    • one year ago
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    \[\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\] and \[\sin( \alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\]

  2. ganeshie8
    • one year ago
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    try \[\cos(\alpha-\beta)=\dfrac{<\cos\alpha, \sin\alpha> \cdot <\cos\beta, \sin \beta>}{\|<\cos\alpha, \sin\alpha>\|\| <\cos\beta, \sin \beta>\|}\]

  3. Astrophysics
    • one year ago
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    I know for second one we can do \[\sin^2(\alpha - \beta) = 1-\cos^2(\alpha - \beta)\] I think this may work

  4. Astrophysics
    • one year ago
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    Ok let me try xD

  5. ganeshie8
    • one year ago
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    |dw:1436340353271:dw|

  6. ganeshie8
    • one year ago
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    let them be unit vectors, then their position vectors are given by \(<\cos\alpha, \sin\alpha>\) and \(<\cos\beta, \sin \beta>\) next we simply use the dot product to find the angle between them : \(\alpha-\beta\)

  7. ganeshie8
    • one year ago
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    * looks i had the angles wrong \(\beta-\alpha\)

  8. Astrophysics
    • one year ago
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    That was much simpler than I thought it would be

  9. Astrophysics
    • one year ago
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    Thanks haha, and a way I gooooo.....|dw:1436340890359:dw|

  10. Astrophysics
    • one year ago
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    Looks we can also use law of cosines

  11. Astrophysics
    • one year ago
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    Oh forgot we derive that from law of cosines xD

  12. ganeshie8
    • one year ago
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    there are different ways to go about these, my fav is always to use complex numbers

  13. Astrophysics
    • one year ago
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    Yeah, I'm not very good with complex numbers, never really learnt them, so I have to use khan academy haha

  14. ganeshie8
    • one year ago
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    btw you can derive sin(a-b) similarly by replacing a by pi/2-a and b by -b

  15. Astrophysics
    • one year ago
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    Ooh nice

  16. ganeshie8
    • one year ago
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    as you can see, vectors proof is much nicer and simpler than using trig

  17. ganeshie8
    • one year ago
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    similarly using complex numbers is much simpler than vectors

  18. Astrophysics
    • one year ago
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    Yeah, I would've thought it was harder, but these are easier and funner

  19. ganeshie8
    • one year ago
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    you just need to see it done ocne

  20. ganeshie8
    • one year ago
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    Here is a derivation for the 4 angle sum/addition identities : \[\begin{align}\cos(\alpha\pm\beta) + i\sin(\alpha\pm\beta) &= e^{i(\alpha\pm\beta)} \\~\\ &= e^{i\alpha}e^{\pm i\beta} \\~\\ &=[\cos(\alpha)+i\sin(\alpha)][\cos(\pm\beta)+i\sin(\pm \beta)] \end{align}\] compare real, imaginary parts both sides and we're done!

  21. Astrophysics
    • one year ago
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    Does that have anything to do with Euler's formula by any chance

  22. Astrophysics
    • one year ago
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    Oh yes it does, I see!

  23. ganeshie8
    • one year ago
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    it is using euler formula all the way yeah

  24. Astrophysics
    • one year ago
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    Yeah I just realized it haha, that's ncie

  25. Astrophysics
    • one year ago
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    Wow that's pretty awesome with this we can get both plus minus just by considering them odd/ even functions

  26. ganeshie8
    • one year ago
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    Hmm we're just comparing real, imaginary parts both sides real part of left hand side is \(\cos(\alpha\pm \beta)\), real part of right hand side need to be worked

  27. Astrophysics
    • one year ago
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    Ah ok I get it, I was just thinking of \[\cos(\alpha - \beta)\]

  28. ganeshie8
    • one year ago
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    \[\begin{align}&\color{blue}{\cos(\alpha\pm\beta) }+ i\color{purple}{\sin(\alpha\pm\beta)} \\~\\ &= e^{i(\alpha\pm\beta)} \\~\\ &= e^{i\alpha}e^{\pm i\beta} \\~\\ &=[\cos(\alpha)+i\sin(\alpha)][\cos(\pm\beta)+i\sin(\pm \beta)] \\~\\ &=\color{blue}{cos(\alpha)\cos(\pm \beta)- \sin(\alpha)\sin(\pm\beta)} + i[\color{purple}{\sin(\alpha)\cos(\pm\beta)+ \cos( \alpha)\sin(\pm\beta)}] \end{align}\]

  29. Astrophysics
    • one year ago
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    Haha, great now I know I can derive it with complex, vector algebra, and trig!

  30. ganeshie8
    • one year ago
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    each in increasing order of pain

  31. Astrophysics
    • one year ago
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    XDDDDD

  32. Astrophysics
    • one year ago
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    Thanks haha, that was fun

  33. ganeshie8
    • one year ago
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    Here is another fun identity, very useful in differential eqns : \[a\cos t+b\sin t=C\cos(t-\phi)\] where \(C = \sqrt{a^2+b^2}\) and \(\phi=\arctan(b/a)\)

  34. ganeshie8
    • one year ago
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    you may use trig/vectors/complex to prove it

  35. Astrophysics
    • one year ago
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    Looks like fun, I will give it a try tomorrow

  36. ganeshie8
    • one year ago
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    ok remember, trig is for highschoolers vectors is for undergrads complex is for real men ;p

  37. Astrophysics
    • one year ago
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    Ok, sounds good, I will try all though!

  38. Empty
    • one year ago
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    Hahahaha I am loving this whole question ty ganeshie

  39. Empty
    • one year ago
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    Another great thing to mention is deriving rotation matrices. A pure rotation (without stretching) is represented by simply multiplying by a complex number of length 1. So if we start with the complex number (2D vector) \(z=x+iy\) we can rotate it by this length 1 complex number to rotate it by an angle \(e^{i \theta} \). (Oh yeah how do we find length of complex numbers, let's just check real quick for both of these: \[\sqrt{z^*z} = \sqrt{(x-iy)(x+iy)} = \sqrt{x^2+y^2}\] \[\sqrt{(e^{i \theta})( e^{i \theta})^* } =\sqrt{e^{i \theta}e^{-i \theta} }= \sqrt{e^{i \theta - i \theta}} = \sqrt{e^0} = 1 \] Cool. Ok so let's actually rotate (and then you can check the new number you get will also not be changed length. \[(a+bi)e^{i \theta} = (a+bi)(\cos \theta + i \sin \theta) = (a \cos \theta - b \sin \theta) + i (a \sin \theta + b \cos \theta ) \] So what we've really done is a linear transformation where we have a system of equations by separating it out into real and imaginary parts: \[a \rightarrow a \cos \theta - b \sin \theta \] \[ b \rightarrow a \sin \theta + b \cos \theta\] Which in matrix form is just: \[\left[ \begin{array}c \cos \theta & - \sin \theta\\\sin \theta & \cos \theta\\\end{array} \right]\left[ \begin{array}c a\\ b\\\end{array} \right]\] Now not to overload too much, just the purpose of this is that you will learn a lot about matrices and rotations and vectors and complex numbers the more you play with them by using what you do know as a rock to stand on. So I'll go ahead and give you more (think of this whole post as an exercise to understanding). This is a bit more complicated, but what if we had multiplied by \(e^{- i \theta}\), what matrix would we have gotten? What fundamental relationship(s) does it share with this rotation matrix in terms of matrix multiplication? What if we had instead used \(r e^{i \theta}\) from the beginning, we'd have also stretched! Can we represent the "r" matrix as something separate from the rotation matrix? Does this new "r" matrix commute with the rotation matrix, why or why not? Additionally, can you show that the length of the vector is unchanged by multiplication by these pure rotation matrices? These exercises will take a long time for you to complete, but if you do them and work hard, then you will understand that matrices, vectors, complex numbers, and algebra itself will come together in an intuitive and geometric way. At the end of the day, these are all just ways of getting algorithms to do very very very simple geometric manipulations that a child can do. Good luck! =D

  40. Astrophysics
    • one year ago
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    I don't understand rotation matrices, is there a visual representation for this? I'm sure it doesn't have to do with curl. @Empty

  41. Empty
    • one year ago
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    visual representation of rotation matrix:|dw:1436374843387:dw| The before vector is \[\left[ \begin{array}c 2\\ 0\\\end{array} \right]\] and then after multiplying by the rotation matrix with an angle of 45 degrees AKA \(\pi/4\). \[\left[ \begin{array}c \cos( \pi /4) & -\sin (\pi /4)\\\sin( \pi /4) & \cos( \pi /4)\\\end{array} \right]\left[ \begin{array}c 2\\ 0\\\end{array} \right] = \left[ \begin{array}c \sqrt{2}\\ \sqrt{2}\\\end{array} \right]\] So that's the new vector. Nothing weird or complicated to do with curl, just simply moving the hand on a clock, erm, except counterclockwise haha.

  42. Empty
    • one year ago
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    Check to make sure these vectors both have the same length, ala \(\sqrt{x^2+y^2}\). That's how you know it's just a rotation, right?

  43. Astrophysics
    • one year ago
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    OhOHohOHohOH that makes sense

  44. Astrophysics
    • one year ago
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    So the rotation is just a vector resultant sort off..@Empty

  45. Empty
    • one year ago
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    Forget rotations for a moment, let's think of the real number line. Let's look at the number 5. It's a good number. It starts at the origin and points to the place x=5 on the number line if we think of it as a vector. Let's look at 3. We can look at this also as being a vector, but let's not for a second. Let's instead think of it as just a scalar value. Now if we have 3*5=15 we can think of 3 being a transformation matrix stretching the vector 5 into a new vector, which starts at the origin and stretches to 15. So similarly we can think of \(e^{i \theta}\) as both a vector pointing from the origin rotated by the angle \(\theta\) of unit length and as a transformation that rotates an object by that angle when you multiply it. This is super convenient I think. It's like saying five sets of 3 apples is the same as three sets of 5 apples. Commutativity at its finest. Mmmm mmm. This is really quite amazing I think that we have this dual nature of how we can look at numbers. Similarly, permutation matrices work the same sort of intuitive way. It feels so nice and it's hard to explain unless you _get it_ and you will eventually. But what I mean is that permutation matrices ARE the permutations themselves. So in order to get a permutation matrix, you permute the columns of an identity matrix. And this matrix when you multiply it by something permutes it. So essentially if you want to get the permutation matrix you desire, you can multiply the identity matrix by the permutation matrix. This seems sort of cyclic (no pun intended... nevermind this joke probably doesn't make sense to you yet anyways because you are unaccustomed to cyclic permutations.. and this is a horrible joke ok I'm done) anyways, what the heck is a permutation matrix? \[\left[ \begin{array}c 1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{array} \right]\] Here's one. What does it do? Well, multiply it by a vector such as \[\left[ \begin{array}c a\\ b\\ c\\\end{array} \right]\] and we see that we get \[\left[ \begin{array}c 1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{array} \right]\left[ \begin{array}c a\\ b\\ c\\\end{array} \right]=\left[ \begin{array}c a\\ c\\ b\\\end{array} \right]\] Hey it just flipped them around. So maybe you can reread this section now that you know this. But remember, to get this permutation matrix which switches the second and third entries we had to switch the second and third columns of the identity matrix. In other words: \[\left[ \begin{array}c 1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{array} \right]\left[ \begin{array}c 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\\end{array} \right]=\left[ \begin{array}c 1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{array} \right]\] Kinda bizarre while also almost very natural at the same time. It might even be said to be obvious, but I don't know if that's true. I think it might be kind of subtle, I don't know. To me this sort of captures a very nice and what I personally believe to be a good indicator of mathematical truth, that it contains some circularity. For instance, doesn't this seem like a weird statement? \[i=e^{i \frac{\pi}{2}}\] We're saying that this is what i is, however at the same time it's more of an equation in which it satisfies. Now I will unforgivingly diverge hardcore and plug this equation into itself infinitely: \[\Large i=e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^{\frac{\pi}{2}\cdots}}}\] Oh so now we have this really weird infinite power tower of real numbers being equal to i. So apparently real numbers aren't even closed under exponentiation when done infinitely many times... Whatever that means. In fact we can... Nevermind I am finding myself wanting to talk about one of my favorite functions but I will spare you, one thing at a time haha I didn't get much sleep either. XD

  46. Astrophysics
    • one year ago
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    That was really nice, I think I'm starting to get it, except towards the end with the infinite power tower haha, that escalated pretty quickly. It sort of reminds me of the golden ratio though even if it's not entirely related, but I can see some use of the property..interesting.

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