Trig trig trig

1. Sep 29, 2005

mohlam12

hey everyone,
i have to show that in a triangle, there is : (A, B, C are the angles of that triangle)
$$\sin \left( 1/2\,B \right) \cos \left( 1/2\,C \right) +\sin \left( 1/2 \,C \right) \cos \left( 1/2\,B \right) =\cos \left( 1/2\,A \right)$$

for this one, here is what i got to...

$$1/4\,\sqrt {2-2\,{\it cosB}}\sqrt {2+2\,{\it cosC}}+1/4\,\sqrt {2-2\,{ \it cosC}}\sqrt {2+2\,{\it cosB}}$$

my question is, i dont have any A in this equation above, and i have to prove that it is equal to cos(a/2)!! i know that cos(a)=cos(pi-(b+c) ... please if someone can help me with that!

for the second one, we have : $$\left( \cos \right) \,\alpha={\frac {a}{b+c}}$$

and $$\left( \cos \right) \,\beta={\frac {b}{c+a}} and \left( \cos \right) \,\gamma={\frac {c}{a+b}}$$

and we have to show that:
$$1/2\,{\tan}^{2}\alpha+1/2\,{\tan}^{2}\beta+1/2\,{\tan}^{2}\gamma = 1$$

this one, i erally dont know what to do! if someone can help me out! or maybe give me hints...

2. Sep 29, 2005

TD

You can use the fact that

$$\tan ^2 \theta = \frac{{\sin ^2 \theta }} {{\cos ^2 \theta }} = \frac{{1 - \cos ^2 \theta }} {{\cos ^2 \theta }} = \frac{1} {{\cos ^2 \theta }} - 1$$

3. Sep 29, 2005

mohlam12

Sorry... for the second one, there is a mistake (I am still learning how to us LaTex so, yeah) here is the correct form: I am not gonna use LaTex, I can't get what I want...

tan²(α/2) + tan²(β/2) + tan²(γ/2) = 1

Thanks,

4. Sep 29, 2005

TD

Then you can use the fact that:

$$\tan \left( {\frac{\theta } {2}} \right) = \frac{{\sin \theta }} {{1 - \cos \theta }} \Leftrightarrow \tan ^2 \left( {\frac{\theta } {2}} \right) = \frac{{\sin ^2 \theta }} {{\left( {1 - \cos \theta } \right)^2 }} = \frac{{1 - \cos ^2 \theta }} {{\left( {1 - \cos \theta } \right)^2 }} = \frac{{\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}} {{\left( {1 - \cos \theta } \right)^2 }} = \frac{{1 + \cos \theta }} {{1 - \cos \theta }}$$

5. Sep 29, 2005

mohlam12

okay, so iam here:
$$\left( {\frac {a}{b+c}}+1 \right) \left( -{\frac {a}{b+c}}+1 \right) ^{-1}+ \left( 1+{\frac {b}{a+c}} \right) \left( 1-{\frac {b} {a+c}} \right) ^{-1}+ \left( 1+{\frac {c}{a+b}} \right) \left( 1-{ \frac {c}{a+b}} \right) ^{-1}$$

is there any way that this can be equal to 0 !!

PS: sorry about messing up the equation in latex, but i am sure u still can get what i did (i am using maple 10 to convert to latex, i think there is a problem there)

6. Sep 29, 2005

TD

I calculated it too and I don't think it simplifies to 1, although I can't find an error in my derivation for the tangents to cosines.

7. Sep 29, 2005

mohlam12

Okay, and for the first one, I noticed that what I had is a sine rule... it's equal tu sin((b+c)/2) but, where do i get it to get equal to cos (a/2) ... :huh:

8. Sep 29, 2005

TD

Very good, but we know that $a+b+c=180$ so:

$$\sin \left( {\frac{{a + b}} {2}} \right) = \sin \left( {\frac{{180 - c}} {2}} \right) = \sin \left( {90 - \frac{c} {2}} \right) = \cos \left( { - \frac{c} {2}} \right) = \cos \left( {\frac{c} {2}} \right)$$

9. Sep 29, 2005

mohlam12

yup, thanks!

10. Sep 29, 2005

TD

No problem

Never forget to use everythings that's given.

11. Sep 29, 2005

mohlam12

but anyone, can help me with the second problem?! pliiiiz

12. Sep 30, 2005

mohlam12

ok i got it. never mind everyone!