# Trigometry Identity

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?

Related Precalculus Mathematics Homework Help News on Phys.org
No. Youre right. There is another trigo identity that you might find useful:
cos2a + cos2b + cos2c=-1-4cosa cosb cosc.

VietDao29
Homework Helper
Have you covered the Sum to Product formulae?
$$\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)$$
$$\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)$$
$$\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)$$
$$\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)$$

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

Last edited:
Gib Z
Homework Helper
Heres a Trig Identity that you will find useful :

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

LOL

Thanks a lot :)