# 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!