Question

Proof!

Answer 1

\[\sqrt{A^2+B^2}\cos(wt-\arctan(\frac{B}{A}))=Acoswt+Bsinwt\]
This seems impossible to crack whichever way I go about it, but then all the ways were probably useless here.

Answer 2

By the way, I will only read any proofs most probably in 24 hours, after I've given it a proper go.

Answer 3

To get a grip on this - I would first of all draw the graphs. One is a circle with radius 2 a harmonic mean of A, B and the other an ellipse with main radii A & B

Answer 4

Obviously you have 4 solutions

Answer 5

http://www.wolframalpha.com/input/?i=%283cosx%2B2sinx%29
http://www.wolframalpha.com/input/?i=sqrt%284%2B9%29cos%28x-arctan%282%2F3%29%29

Answer 6

This is what you get

Answer 7

Why?

Answer 8

Are you assuming B is imaginary?

Answer 9

Because when I plot both, they come out sinusoidalesquely.

Answer 10

OK i thought it a 2-dim , but now I get your equality is pretty straightforward

Answer 11

I would understand what you're saying if A and B were unit vectors.

Answer 12

yes yes - it is different, but also easy, wait

Answer 13

I'll give the 2D one a go also. Thanks for your time in advance, I'll read your proof later.

Answer 14

OK SO the solution has stages :

Answer 15

1 Normalize by dividing and multiplying by \[\sqrt{A^2 + B^2}\]

Answer 16

Now consider the fact that the following numbers are components of SIN theta and COS theta \[\frac{ A }{ \sqrt{A^2 + B^2} } .......and... \frac{ B}{ \sqrt{A^2 + B^2} }\]

Answer 17

where \[\theta = \arctan \frac{B}{A}\]

Answer 18

now you have on the right hand side
\[\sqrt{A^2 + B^2}\left( \frac{ A }{ \sqrt{A^2 + B^2} }\cos wt + \frac{ B }{ \sqrt{A^2 + B^2} } \sin wt \right)\]

Answer 19

which is\[\sqrt{A^2 + B^2} \left( \cos \theta \cos wt + \sin \theta \sin wt\right)\]
and this is the well known addition formula

Answer 20

\[ \cos (wt - \theta)\]

Answer 21

Multiplied by\[\sqrt{A^2 + B^2}\]

Answer 22

thas-it ff-f-ff-oo-oo--oo-lks !

Answer 23

@henpen I solved it in less than 5 min -where did you go ?!

Answer 24

I did it a different way, but thanks for yours.

Answer 25

\[Acos(wt)+Bsin(wt)=Rcos(wt+ \alpha)\]From Euler's equation that was evident if alpha is a constant.
\[t=0\]
\[A=Rcos(\alpha)\] and\[t=\frac{\pi}{2w}\]
\[B=Rsin(\alpha)\]

Answer 26

I am prob. not familiar with this path . How so evident ?

Answer 27

\[R*e^{i(x+\alpha)}=R*e^{i(x)}*e^{i(\alpha)}=Rke^{ix}=Acoswt+Bsinwt\]

Answer 28

*A and B are not the same as e^ia 's real and im parts will be of a different magnitude

Answer 29

That should be wt, not x

Answer 30

well we are over-analyzing it. However I think that to be in any way evident one has at least to divide both sides with \[ \sqrt{A^2 + B^2} \]

Answer 31

So lets leave it at that

Answer 32

Yes. Let each leave with what solution works best for them (and, fo course, works). Thank you anyway.

Answer 33

YW (anyway)