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Law of cosines being an immediate consequence of

  1. Sep 1, 2008 #1
    I have a problem with this question where I need to show the law of cosines (c^2=a^2+b^2-2ab*cos(angle) is an immediate consequence of the identity: (a+b)^2=a^2+b^2+2ab. I've played around with it for awhile and come somewhat close but can't get quite all the way there. It's basically all an algebraic problem but I'm not too good at that when deriving other equations. I drew a triangle with sides a,b, and c, to try to re-write the 2ab part as 2(c*cos(angle))^2 but don't know how to get rid of one to have 2abCos(angle). Any bit of help or just a hint could probably get me through this.
     
  2. jcsd
  3. Sep 1, 2008 #2

    Doc Al

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    Are you familiar with the scalar product of two vectors?

    [tex]\vec{a}\cdot\vec{b} = \rm{??}[/tex]
     
  4. Sep 1, 2008 #3
    yes i know when you multiply two vectors you get a scalar product, that comes into play with the cosine part?
     
  5. Sep 1, 2008 #4

    Doc Al

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    Sure. Just treat that identity as a scalar product.
     
  6. Sep 3, 2008 #5
    still can't come up with anything, it doesnt make sense since the identity is the same thing on both sides how can you transform it to the law of cosines? The question states to show how it's an immediate consequence so I would think its something fairly quick and not too exhaustive of a problem but I've been working on it for past 2 days and can't come up with an answer.
     
  7. Sep 3, 2008 #6

    Doc Al

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    If the identity were exactly the same on both sides it would be a tautology and of little interest.

    My suggestion is to treat this as a vector problem. Draw the corresponding vector diagram (it will be your triangle):

    [tex]\vec{c} = \vec{a} + \vec{b}[/tex]

    Now apply the identity to the right hand side, treating it as a problem in vector multiplication:

    [tex](\vec{a} + \vec{b})^2 = (\vec{a} + \vec{b})\cdot (\vec{a} + \vec{b})[/tex]

    See what happens.
     
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