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.

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

2.



















There you are for the second one :)

Thank you very much for help. You are right, the first one is not correct. Here is how it think it should be:

Code:

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\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}]

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I'm sorry if it doesn't look very good. I am trying to make it look like more mathematical.

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)

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...

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:

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

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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...

[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. :)

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 :(

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.