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    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #1

    Sep 27, 2010, 12:24 PM
    2 questions trigonometry
    if A+B+C=\Pi:

    1) cosA + cosB*cosC + 1 = 4 cos(A/2)*cos(B/2)*cos(C/2)

    2) tgA + tgB + tgC = tgA*tgB*tgC

    I don't know what formulas to use. Trigonometry is kind of painful for me.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Sep 29, 2010, 04:08 AM

    Could you check the first one? I substituted some values and the proof is wrong.

    2.



















    There you are for the second one :)
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #3

    Sep 29, 2010, 05:54 AM
    Thank you very much for help. You are right, the first one is not correct. Here is how it think it should be:
    Code:
    \cosA+\cosB+\cosC+1= [2\cosfrac{A-B}{2}\cosfrac{A+B}{2}]+cosC= [2\cosfrac{C}{2}\cosfrac{A-B}{2}]+[2\cos^2frac{C}{2}-1]+1= [2\cosfrac{C}{2}2\cosfrac{{A-B+A+B}{2}}{2}\cosfrac{{A-B-A-B}{2}}{2}]= [4\cosfrac{C}{2}\cosfrac{A}{2}\cosfrac{B}{2}]= [4\cosfrac{A}{2}\cosfrac{B}{2}\cosfrac{C}{2}]
    I'm sorry if it doesn't look very good. I am trying to make it look like more mathematical.
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #4

    Sep 29, 2010, 06:01 AM
    Comment on lemon14's post
    cosA+cosB+cosC+1=2*cos[(A-B)/2]*cos[(A+B)/2]+cosC=2*cosC*cos[(A-B)/2]+2*cos^2(C/2) -1+1= 2*cos(C/2)*{cos[(A-B)/2]+2*cos(C/2)} = 2*cosC*{cos[(A-B)/2]+cos[(A+B)/2] } = 2*cos(C/2)*2cos{[(A-B+A+B)/2]/2}*cos{[(A-B-A-B)/2]/2} = cos(A/2)cos(B/2)cos(C/2)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Sep 29, 2010, 08:32 AM
    You can make use of the 'math' tabs :) It's practically the same. You insert '\' in front of frac only. And I use some for the brackets so that they become bigger. You can quote my answer to see how the code words. Except that the LaTeX has a little hiccup here, I don't know why. I had to put two brackets for the fractions and the bracket to work well, go figure...









    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #6

    Sep 29, 2010, 12:10 PM
    Here I tried to do something, but I'm not sure it is better. I haven't quoted your answer because I don't know how to do it (I'm sorry for being so annoying but I am a beginner and I don't know very much about this), but I tried to apply what you said.
    Code:
    cosA+cosB+cosC+1=4cos\({A}{2})cos\({B}{2})cos\({C}{2})
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #7

    Sep 29, 2010, 09:43 PM

    Ooh... I should have guessed earlier that you are in the 'Go Skin'. There are several interfaces of the site, the most recent one must be the one you are using and I must tell you that I don't like it that much... instead of inserting the [code], replace 'code' by 'math' and insert those between the tabs.

    cosA+cosB+cosC+1=4cos\(\(\frac{A}{2}\) cos\(\(\frac{B}{2}\) cos\(\(\frac{C}{2}\)

    to give:



    There is some error in the LaTeX again I don't know why... there normally should be a single "\(" but it doesn't appear when I put a single bracket... well...
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #8

    Sep 30, 2010, 08:15 AM
    [MATH]cosA+cosB+cosC+1=4cos\(\frac{A}{2}\) cos\(\frac{B}{2}\) cos\(\frac{C}{2}\)[\MATH]

    I hope I did it this time. Thank you for the tips. :)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #9

    Sep 30, 2010, 08:26 AM

    Maybe... have you tried typing manually in the answer box?

    I think that when you copy paste from where you are copying it, it types in a different symbol but which looks the same... I really don't know :(
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #10

    Sep 30, 2010, 08:59 AM
    No, I haven't typed it manually.

    Whatever, it's OK. Thank you for help. Maybe I'll figure it out one day. I just wanted to make it look better and easier to follow.

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