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Simplify in sum and difference equation

  1. Feb 25, 2005 #1
    I was never taught how to do this I was just given problems with no solutions so I dont know what to do can someone plz help me I have two questions which are similar maybe if I get help with one I can solve the other one on my own.

    It says simplify the following
    [tex] \cos(\pi+x) + \cos (\pi-x) [/tex]
    for this one i expanded cos into the brackets and simplified and got 2cospi is this correct?

    and

    [tex] \cos (\frac {7\pi} {10}) \cos (\frac {\pi} {5}) +\sin (\frac {7\pi} {10}) \sin (\frac {\pi} {5}) [/tex]

    Help please quick! :yuck:
     
    Last edited: Feb 25, 2005
  2. jcsd
  3. Feb 25, 2005 #2

    xanthym

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    From standard trig identities:
    cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
    cos(-e) = cos(e)
    sin(-g) = -sin(g)

    cos(Pi + x) + cos(Pi - x) = cos(Pi)cos(x) - sin(Pi)sin(x) + cos(Pi)cos(-x) - sin(Pi)sin(-x) =
    = (-1)cos(x) - (0)sin(x) + (-1)cos(x) + (0)sin(x) =
    = (-2)cos(x)

    cos(7*Pi/10)cos(Pi/5) + sin(7*Pi/10)sin(Pi/5) = cos{(7*Pi/10) - (Pi/5)} =
    = cos{Pi/2} =
    = 0


    ~~
     
  4. Feb 25, 2005 #3
    Holy moly that looks really confusing I dont understand what is going on and that doesnt look very familiar from the unit ..... :cry:
     
  5. Feb 26, 2005 #4

    dextercioby

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    [tex] \cos(\pi+x)=\cos\pi\cos x-\sin\pi\sin x=-\cos x [/tex]

    [tex] \cos (\pi-x)=\cos\pi\cos x+\sin\pi\sin x=-\cos x [/tex]

    Add them and see what u get.

    As for the last,u have to use the identity:

    [tex] \cos a\cos b+\sin a\sin b=\cos(a-b) [/tex]

    See who's "a" & who's "b"...

    Daniel.

    EDIT:The typo was because of a missing space between \cos and b... o:)
     
    Last edited: Feb 26, 2005
  6. Feb 26, 2005 #5

    Curious3141

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    That, BTW, should be :

    [tex] \cos a\cos b+\sin a\sin b=\cos(a-b) [/tex]
     
  7. Feb 26, 2005 #6

    dextercioby

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    Check the code by clicking on the formula,it (the "cos b") was there...

    Daniel.
     
  8. Feb 26, 2005 #7

    Curious3141

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    Yes, I know it was, but you forgot the space.
     
  9. Feb 26, 2005 #8

    dextercioby

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    :blushing: :redface: :cry: :frown: You got me...Yes,as the Americans say :"S*** happens".Today i've had a lousy day...:yuck:

    daniel.
     
  10. Feb 26, 2005 #9
    [tex] \cos(\pi+x)=\cos\pi\cos x-\sin\pi\sin x=-\cos x [/tex]

    [tex] \cos (\pi-x)=\cos\pi\cos x+\sin\pi\sin x=-\cos x [/tex]

    What are u doing to these pies and xs to get -cosx? How do u get that?
     
  11. Feb 26, 2005 #10

    dextercioby

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    There's no "pie" there...:tongue2: I used the fact that
    [tex] \cos \pi =-1 [/tex]
    [tex] \sin \pi =0 [/tex]

    Daniel.
     
  12. Feb 26, 2005 #11

    See I dont know that :cry: wish I was as smart as you, ahhhhh so what do I do for the other question?? This is so confusing, I hate these sum and difference identities I dont know what to do with them :cry:
     
  13. Feb 26, 2005 #12

    dextercioby

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    What about the other one...?In post #2 it was solved...

    Daniel.
     
  14. Feb 26, 2005 #13
    [tex] \cos (\frac {7\pi} {10}) \cos (\frac {\pi} {5}) +\sin (\frac {7\pi} {10}) \sin (\frac {\pi} {5}) [/tex]

    here is the question in post 2 the person wrote [tex] cos [(\frac {7\pi}{10}) -(\frac {\pi} {5})] = \cos (\frac {\pi} {2}) = 0 [/tex]

    How come I dont know what is going on what is happening where are these numbers comming from how do I do this on my own?

    So the final simplified answer for this problem is 0? Can someone plz help me understand how to do these problems I dont have a clue I know how to evaluate sin(75 degrees) but thats about it because it uses special triangles but these problems are totally confusing for me . :frown: :yuck:
     
    Last edited: Feb 26, 2005
  15. Feb 26, 2005 #14

    dextercioby

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    So what...?That's perfectly correct...

    Daniel.
     
  16. Feb 26, 2005 #15
    :tongue2: ahhhhhhhh meany ur a smarty pant thats y u know how to do this if ud explain maybe i'd get it too :grumpy:
     
  17. Feb 26, 2005 #16

    dextercioby

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    Label
    [tex] \frac{7\pi}{10}=a [/tex]
    and
    [tex] \frac{\pi}{5}=b [/tex]
    And then compute
    [tex] \cos a\cos b+\sin a\sin b [/tex]

    using addition and subtraction formulas for cosine and sine.

    Daniel.
     
  18. Feb 26, 2005 #17
    addition and subtraction formulas for cos and sin?
     
  19. Feb 26, 2005 #18

    dextercioby

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    My dear god,the formulas:
    [tex] \cos (a-b),\cos(a+b),\sin(a-b),\sin(a+b) [/tex]

    Daniel.

    P.S.I'm going to bed now,so be a nice girl and don't make anything bad till i return.:wink:
     
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