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Integrate cos(x)-cos(x-c) from 0 to c/2

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Integrate cos(x)-cos(x-c) from 0 to c/2


    2. Relevant equations
    sin(x-c)=sinxcosc-cosxsinc


    3. The attempt at a solution
    sin(x)-sin(x-c) from 0 to c/2
    =sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
    =sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)

    Correct response:2sin(c/2)-sin(c)

    What am I doing wrong?
     
  2. jcsd
  3. Oct 16, 2012 #2

    SammyS

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    You're doing nothing wrong. They are equivalent.

    To get the answer in a more straight forward manner, simply plug the limits of integration into the anti-derivative, sin(x)-sin(x-c) .
     
  4. Oct 16, 2012 #3
    Sorry, I guess this is a question of trig then - how do I simplify my answer to the correct answer? It's probably something stupid but I'm having trouble figuring it out!
     
  5. Oct 16, 2012 #4
    $$sin(\frac{c}{2})-sin(\frac{c}{2})cos(c)+cos(\frac{c}{2})sin(c)-sin(c)$$
    $$sin(\frac{c}{2})-sin(c)+cos(\frac{c}{2})sin(c)-sin(\frac{c}{2})cos(c)$$

    You have to realize (from trig addition identity) that ##cos(\frac{c}{2})sin(c)-sin(\frac{c}{2})cos(c) = sin(c-\frac{c}{2}) = sin(\frac{c}{2})##

    Using the above, we have:

    $$sin(\frac{c}{2})-sin(c)+sin(\frac{c}{2})$$
    $$2sin(\frac{c}{2})-sin(c)$$
     
  6. Oct 16, 2012 #5
    Thanks!
     
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