Prove Trig Identity: A+B+C = 180 → 1+4 cosAcosBcosC

[Harmony]

Question Statement
If A+B+C = 180, prove that cos (A+B-C) + cos (B+C-A) + cos (C+A-B) = 1+4 cosAcosBcosC

My Attempt
If A+B+C=180,
Then A+B-C=180-2C
cos (A+B-c)=cos(180-2C)

(After some substitution and caculation)

cos (A+B-C) = -cos 2C

Similarily, I obtain the same expression for cos (B+C-A), cos (C+A-B).

But that does not get me to the desire result.

Is my attempt wrong from the beginning?

 

[chaoseverlasting]

No. Youre right. There is another trigo identity that you might find useful:
cos2a + cos2b + cos2c=-1-4cosa cosb cosc.

 

[VietDao29]

Have you covered the Sum to Product formulae?
[tex]\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex]
[tex]\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]
[tex]\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex]
[tex]\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]

So you've shown the LHS to be equal to:
- (cos (2A) + cos (2B) + cos(2C))
Using the identites above, we have:
[tex]- (\cos (2A) + \cos (2B) + \cos(2C)) = - (2 \cos(A + B) \cos(A - B) + \cos (2C)) = -(2 \cos(\pi - C) \cos (A - B) + 2 \cos ^ 2 C - 1)[/tex]
[tex]= 1 - (- 2 \cos C \cos (A - B) + 2 \cos ^ 2 C) = ...[/tex]
Can you go from here? :)

 

[Gib Z]

Heres a Trig Identity that you will find useful :

[tex]\mbox{If A+B+C = 180, then} \cos (A+B-C) +\cos (B+C-A) + \cos (C+A-B) = 1 + 4 \cos A \cos B \cos C[/tex].

LOL

 

[Harmony]

Thanks a lot :)