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Solve Cos(a+b)

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    cos(a+b)

    2. Relevant equations

    find formulas for above

    3. The attempt at a solution

    cos (a+b)

    = cos (a-(b))

    =cos(a)cos(-b) + sin(a)sin(-b)

    =cos(a)cos(b) - sin(a)sin(b)




    I been scratching my head at this for hours now, and I still can't figure out how this works.
     
  2. jcsd
  3. Sep 14, 2008 #2

    Dick

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    Re: cos(a+b)

    cos(a+b) is not equal cos(a-b). Do you have a formula for one that you are trying to transform into a formula for the other?
     
  4. Sep 14, 2008 #3
    Re: cos(a+b)

    @dick - no, he has written a - b, but i think he meant a - (-b), as he has followed it up with the relevant calculations.

    What i can't understand is what the question is. You seem to have solved and found solution already.
     
  5. Sep 14, 2008 #4

    Dick

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    Re: cos(a+b)

    I agree, my question was trying to get at what the question is. I'm guessing it's given cos(a+b) find cos(a-b). And it is done, except for general sloppiness in the whole statement and stating the parts that make it work like cos(-b)=cos(b) and sin(-b)=-sin(b). I was just running around looking for unanswered questions.
     
  6. Sep 15, 2008 #5
    Re: cos(a+b)

    opps, sorry guys..... typo

    cos(a+b)= cos (a-(-b))

    =cosAcos(-b) + sin(a)sin(-b)

    =cosAcosB - sinAsinB



    I don't understand how the second line comes in. Where does sin come from?

    The question asks to find formulas for cos(a+b). The answer is copied right out of my prof's notes, I just don't understand how it works.
     
  7. Sep 15, 2008 #6

    Dick

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    Re: cos(a+b)

    You're professor must have ALREADY derived cos(a-b)=cos(a)cos(b)+sin(a)sin(b). You now want a formula for cos(a+b) without repeating something like the previous derivation. If you write a+b=a-(-b) then you've written the sum as a difference and you can use the difference formula you've already derived.
     
  8. Sep 15, 2008 #7
    Re: cos(a+b)

    thnkx, I see what the notes mean now. Yes, cos(a-b) was derived on a page earlier.....

    so we set cos(a+b) = cos(a-(-b)) so it fits the cos(a-b) derivation, so we don't have to do it all over again.

    got it!! thnkx...
     
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