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Derive cos(A-B) from the law of cosines

  1. Jul 4, 2017 #1
    1. The problem statement, all variables and given/known data
    In the chapter about differentiation there is this question:
    Snap2.jpg
    2. Relevant equations
    Cosine rule:
    $$a^2=b^2+c^2-2bc\cos\alpha$$
    $$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$

    3. The attempt at a solution
    Snap1.jpg
    $$a^2=b^2+c^2-2bc\cos\alpha$$
    $$d^2=e^2+c^2-2ec\cos\beta$$
    $$(a-d)^2=b^2+e^2-2bc\cos(\alpha-\beta)$$
    It doesn't lead to the solution
     
  2. jcsd
  3. Jul 4, 2017 #2

    Mark44

    Staff: Mentor

    I wouldn't set it up the way you have in your drawing. Sketch the unit circle with one ray making an angle of ##\alpha## going out to the point ##(\cos \alpha, \sin \alpha)## on the unit circle. Draw another ray with a smaller angle of ##\beta## going out to the point ##(\cos \beta, \sin \beta)## on the unit circle.

    Use the Law of Cosines to find the distance between those two points on the unit circle. The result for ##\cos(\alpha - \beta)## is pretty easy after that.
     
  4. Jul 5, 2017 #3
    Snap1.jpg
    $$a^2=2(1-\cos\alpha),~~b^2=2(1-\cos\beta)$$
    $$x^2=2(1-\cos(\alpha-\beta))$$
    $$x\neq a-b$$
    Even if x were (a-b) it's complicated
     
  5. Jul 5, 2017 #4
    But you can compute what x is, because you know where the line segment x starts and ends. You only need pythagoras to do that.
    Then you use the law of cosines to find Cos(A-B) from x.
     
  6. Jul 5, 2017 #5
    $$x^2=(\sin\alpha-\sin\beta)^2+(\cos\beta-\cos\alpha)^2=2-2\cos(\alpha-\beta)$$
    $$\Rightarrow\cos(\alpha-\beta)=\sin\alpha\cdot\sin\beta+\cos\alpha\cdot \cos\beta$$
     
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