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Astrophysics
 one year ago
How do you derive this using vector algebra
Astrophysics
 one year ago
How do you derive this using vector algebra

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\cos(\alpha  \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\] and \[\sin( \alpha  \beta) = \sin \alpha \cos \beta  \cos \alpha \sin \beta\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4try \[\cos(\alpha\beta)=\dfrac{<\cos\alpha, \sin\alpha> \cdot <\cos\beta, \sin \beta>}{\<\cos\alpha, \sin\alpha>\\ <\cos\beta, \sin \beta>\}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I know for second one we can do \[\sin^2(\alpha  \beta) = 1\cos^2(\alpha  \beta)\] I think this may work

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1436340353271:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4let 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 : \(\alpha\beta\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4* looks i had the angles wrong \(\beta\alpha\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1That was much simpler than I thought it would be

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Thanks haha, and a way I gooooo.....dw:1436340890359:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Looks we can also use law of cosines

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh forgot we derive that from law of cosines xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4there are different ways to go about these, my fav is always to use complex numbers

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, I'm not very good with complex numbers, never really learnt them, so I have to use khan academy haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4btw you can derive sin(ab) similarly by replacing a by pi/2a and b by b

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4as you can see, vectors proof is much nicer and simpler than using trig

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4similarly using complex numbers is much simpler than vectors

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, I would've thought it was harder, but these are easier and funner

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you just need to see it done ocne

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Here 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!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Does that have anything to do with Euler's formula by any chance

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh yes it does, I see!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4it is using euler formula all the way yeah

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah I just realized it haha, that's ncie

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Wow that's pretty awesome with this we can get both plus minus just by considering them odd/ even functions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Hmm 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ah ok I get it, I was just thinking of \[\cos(\alpha  \beta)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\begin{align}&\color{blue}{\cos(\alpha\pm\beta) }+ i\color{purple}{\sin(\alpha\pm\beta)} \\~\\ &= e^{i(\alpha\pm\beta)} \\~\\ &= e^{i\alpha}e^{\pm i\beta} \\~\\ &=[\cos(\alpha)+i\sin(\alpha)][\cos(\pm\beta)+i\sin(\pm \beta)] \\~\\ &=\color{blue}{cos(\alpha)\cos(\pm \beta) \sin(\alpha)\sin(\pm\beta)} + i[\color{purple}{\sin(\alpha)\cos(\pm\beta)+ \cos( \alpha)\sin(\pm\beta)}] \end{align}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, great now I know I can derive it with complex, vector algebra, and trig!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4each in increasing order of pain

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Thanks haha, that was fun

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Here 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 \(\phi=\arctan(b/a)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you may use trig/vectors/complex to prove it

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Looks like fun, I will give it a try tomorrow

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4ok remember, trig is for highschoolers vectors is for undergrads complex is for real men ;p

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok, sounds good, I will try all though!

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Hahahaha I am loving this whole question ty ganeshie

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Another 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: \[\sqrt{z^*z} = \sqrt{(xiy)(x+iy)} = \sqrt{x^2+y^2}\] \[\sqrt{(e^{i \theta})( e^{i \theta})^* } =\sqrt{e^{i \theta}e^{i \theta} }= \sqrt{e^{i \theta  i \theta}} = \sqrt{e^0} = 1 \] 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I don't understand rotation matrices, is there a visual representation for this? I'm sure it doesn't have to do with curl. @Empty

Empty
 one year ago
Best ResponseYou've already chosen the best response.1visual 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Check 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?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1OhOHohOHohOH that makes sense

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So the rotation is just a vector resultant sort off..@Empty

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Forget 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1That 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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