# Proving Triangle ABC Equality

• ciubba
In summary, the conversation discusses a proof for the equation cos(A)+cos(B)+cos(C)=1+4sin(\frac{1}{2}A)sin(\frac{1}{2}B)sin(\frac{1}{2}C) for any triangle ABC. The conversation suggests using product to sum formulas and potentially substituting C with 180° - (A+B) or C =##\pi## - (A+B) to simplify the equation. The person mentions trying to use double angle formulas but is unsure how to proceed. They also suggest doing a Google search for triangle "sin A + sin B + sin C" for further help.

#### ciubba

Prove for any triangle ABC that:
$$cos(A)+cos(B)+cos(C)=1+4sin(\frac{1}{2}A)sin(\frac{1}{2}B)sin(\frac{1}{2}C)$$

I tried using the half-angle formula on the right side to get:

$$1+4 \left(\frac{1-cos(A)}{2}\frac{1-cos(b)}{2}sin(\frac{1}{2}C)\right) {}$$

which simplifies to

$$1+\bigg(1-cos(A)-cos(B)+cos(A)cos(B) \bigg)*sin(\frac{1}{2}C)$$

by the product to sum rule, this simplifies to

$$1+\bigg(1-cos(A)-cos(B)+\frac{1}{2}\Big[cos(A+B)+cos(A-B)\Big]\bigg)*sin(\frac{1}{2}C)$$

I tried expanding the sin 1/2 C, but that just made things even more complicated. How should I approach this problem?

Last edited:
Use product to sum formulas.

I remember their being a shortcut for this problem where u rewrite the terms on the left allowing you to use double angle formulas (dont quote me on this). I'll try to give a shot.

Forget the last part I said about product to sum. It just gives you back the original statement. I'll try to work on it later after work. Just had a 15 min break.

Don't forget at some stage you are going to substitute C with 180° - (A+B)
or C = ##\pi## - (A+B)

whichever you prefer.

If stymied, it's often worth trying a google search, such as: triangle "sin A + sin B + sin C"