Question

How do you derive this using vector algebra

Answer 1

\[\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\] and \[\sin( \alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\]

Answer 2

try

\[\cos(\alpha-\beta)=\dfrac{<\cos\alpha, \sin\alpha> \cdot <\cos\beta, \sin \beta>}{\|<\cos\alpha, \sin\alpha>\|\| <\cos\beta, \sin \beta>\|}\]

Answer 3

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

Answer 4

Ok let me try xD

Answer 5

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 : \(\alpha-\beta\)

Answer 6

* looks i had the angles wrong \(\beta-\alpha\)

Answer 7

That was much simpler than I thought it would be

Answer 8

Thanks haha, and a way I gooooo.....

Answer 9

Looks we can also use law of cosines

Answer 10

Oh forgot we derive that from law of cosines xD

Answer 11

there are different ways to go about these, my fav is always to use complex numbers

Answer 12

Yeah, I'm not very good with complex numbers, never really learnt them, so I have to use khan academy haha

Answer 13

btw you can derive sin(a-b) similarly by replacing a by pi/2-a and b by -b

Answer 14

Ooh nice

Answer 15

as you can see, vectors proof is much nicer and simpler than using trig

Answer 16

similarly using complex numbers is much simpler than vectors

Answer 17

Yeah, I would've thought it was harder, but these are easier and funner

Answer 18

you just need to see it done ocne

Answer 19

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!

Answer 20

Does that have anything to do with Euler's formula by any chance

Answer 21

Oh yes it does, I see!

Answer 22

it is using euler formula all the way yeah

Answer 23

Yeah I just realized it haha, that's ncie

Answer 24

Wow that's pretty awesome with this we can get both plus minus just by considering them odd/ even functions

Answer 25

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

Answer 26

Ah ok I get it, I was just thinking of \[\cos(\alpha - \beta)\]

Answer 27

\[\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}\]

Answer 28

Haha, great now I know I can derive it with complex, vector algebra, and trig!

Answer 29

each in increasing order of pain

Answer 30

XDDDDD

Answer 31

Thanks haha, that was fun

Answer 32

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 \(\phi=\arctan(b/a)\)

Answer 33

you may use trig/vectors/complex to prove it

Answer 34

Looks like fun, I will give it a try tomorrow