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Trigonometry proof

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data
    show that

    (a cos A - b cos B) / (b cos A - a cos B)= cos C

    if ABC is triangle

    2. Relevant equations
    trigonometry identity



    3. The attempt at a solution
    A + B + C = 180o
    A = 180o - (B+C)
    cos A = - cos (B+C)

    substitute cos A to the LHS: (- a cos (B+C) - b cos B) / (b cos A - a cos B) = (-a cos B cos C + a sin B sin C - b cos B) / (b cos A - a cos B)

    How to continue? Thank you very much
     
  2. jcsd
  3. Sep 17, 2010 #2

    ehild

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    What are a and b?

    ehild
     
  4. Sep 18, 2010 #3
    the question doesn't explain what a and b are, but I think they are the length of side BC and AC respectively
     
  5. Sep 18, 2010 #4

    Mentallic

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    For your RHS:

    [tex]cos(C)=-cos(A+B)[/tex]

    [tex]=-cosAcosB+sinAsinB[/tex]

    If you multiply through by the denominator of the fraction on the LHS:

    [tex]acosA-bcosB=bcosAsinAsinB-bcos^2AcosB-acosBsinAsinB+acosAcos^2B[/tex]

    Now use the sine rule: [tex]\frac{sinA}{a}=\frac{sinB}{b}[/tex]

    and the rule [tex]cos^2x=1-sin^2x[/tex] to convert your squared terms on the RHS and together with the sine rule, things will cancel nicely to equal the LHS.
     
  6. Sep 18, 2010 #5
    I guess I am not allowed to use sine rule because the teacher hasn't covered that yet. I also confused whether your method is valid or not because as far as I know, proofing must be done only by doing LHS.
    Let's make it clear first. When proofing, is it valid if:
    a. we do the RHS only
    b. we do both side simultaneously (like if we change LHS to form X and we also change RHS to form X, then we conclude it is proven)
    c. like your method, we multiply the fraction of one side to the other side

    In my opinion, proofing is valid if we do the LHS only, so option (a) to (c) all are not valid methods for proofing.

    Thank you very much
     
  7. Sep 18, 2010 #6

    Mentallic

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    Since this is an equality, whenever we apply the standard rules of operation to both sides, we keep both sides equal so the equality is still valid. If we work on both ends and come to the conclusion that x=x thus 0=0 then yes, we have proven it.
    However, I don't know if it is customary to do proofs like these by only working on one side (even though the way I did it is still a valid proof) because my teacher used to make me do it also. She would never give a reason why, other than "just do it".

    Ok so can you please post which trigonometric rules you are allowed to use? It seems kind of weird that you haven't been shown the sine rule yet because as far as my curriculum worked, the sine rule came about 2 years prior to doing proofs like these.
     
  8. Sep 18, 2010 #7
    I can use all the rules here
    http://www.themathpage.com/atrig/trigonometric-identities.htm

    Our teachers have different thought. Because this is a proof, we still don't know whether LHS is the same as RHS. If we multiply the fraction of LHS to RHS it means that the equality is already proven true. That's the reason why I am not allowed to do so; because we still don't know whether the proof is true or not.
     
  9. Sep 18, 2010 #8

    Mentallic

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    Yes we don't know, but what we do is assume they are equal and then we apply the usual rules of algebra and finally convert the equality into 0=0, an equality that we are certain about and don't need any more convincing of. If however the equality is not true, then when we apply the rules we will end up with something like 1=0, a form that we are certain about again which is that 1[itex]\neq[/itex]0 and thus the original equality is not true.

    Say for example, we want to prove [tex]\frac{1-cos^2x}{cos^2x}=tan^2x[/tex]

    If you multiply through you have [tex]1-cos^2x=(cosx.tanx)^2[/tex]

    and then apply the rules, [tex]sin^2x=sin^2x[/tex]

    You can go further and take sin2x from each side to end up with 0=0 but this obviously isn't necessary.

    Now rather than multiplying through and dealing with both sides, we can just manipulate things on the LHS in a different manner to end up with the result on the right.
    Since in the first case we multiplied through and we ended up with cos2xtan2x on the RHS, we can instead multiply top and bottom of the LHS by tan2x to get that same effect we did in the first case

    [tex]\frac{(1-cos^2x)tan^2x}{tan^2xcos^2x}=\frac{sin^2xtan^2x}{sin^2x}=tan^2x=RHS[/tex]

    There really is no difference to the first method since you are basically doing the same thing, but honestly, it is harder to try and realize what you need to multiply by and such when dealing with just 1 side. This is probably why the teachers force you to do it this way.

    So if you can do it one way, you can do it the other way.


    Oh and by the way, without the use of the sine rule I'm completely stumped as to how to solve it...
     
  10. Sep 30, 2010 #9
    I also still stuck on this question until now and my teacher hasn't given the solution yet. Thank you very much for your help and explanation
     
  11. Oct 1, 2010 #10

    Mentallic

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    If your teacher gives a solution by following all the rules that you've stated, would you mind posting it up here so I can take a look at it? :smile:
     
  12. Oct 4, 2010 #11
    Of course I will. That is one way to repay your help :smile:
     
  13. Oct 5, 2010 #12

    Mentallic

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    lol this is a summary of where my 'help' lead us...

     
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