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.

A + B + C = \pi

A + B = \pi- C

tan(A + B)= tan(\pi- C)

\frac{tanA + tanB}{1-tanAtanB}= \frac{tan\pi- tanC}{1+tan\pi tanC}

\frac{tanA + tanB}{1-tanAtanB}= \frac{0- tanC}{1+0}

\frac{tanA + tanB}{1-tanAtanB}= - tanC

tanA + tanB= - tanC(1-tanAtanB)

tanA + tanB= - tanC+tanAtanBtanC

tanA + tanB+ tanC=tanAtanBtanC

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:

\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.

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...

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

= 2cos \(\( \frac{C}{2}\) cos\( \( \frac{A-B}{2}\) + 2cos^2\(\frac{C}{2}\) + 1-1

= 2cos \(\( \frac{C}{2}\) (2cos\( \( \frac{(A-B+A+B)/2}{2}\)cos\(\(\frac{(A-B-A-B)/2}{2}\) \)

= 2cos \(\( \frac{C}{2}\) (2cos\( \( \frac{A}{2}\)cos\(\(\frac{-B}{2}\) \)

= 4cos \(\( \frac{A}{2}\)cos\( \( \frac{B}{2}\)cos\(\(\frac{C}{2}\)

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.

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

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:

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

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.