Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing $4\sin(A)\sin(B)\sin(C)$ to $$4\sin(B+C)\sin(A+C)\sin(A+B)$$ by making substitutions by reorganizing $A+B+C=\pi$. I then did the same thing to the other side to get $$-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))$$ and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. I am pretty sure there is some simpler way of proving the equality, but I can't seem to figure it out. Maybe there is a geometric interpretation or maybe it can be done using just algebra and trig. Any hint's would be appreciated (I would prefer an algebraic approach, but it would be nice to see some geometric proofs as well)
Best Answer
Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities \begin{align} \sin(\pi-\theta) & = \sin(\theta)\\ \cos(\pi-\theta) & = -\cos(\theta)\\ \sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\ \sin(2\theta) & = 2\sin(\theta) \cos(\theta)\\ \cos(\theta - \phi) - \cos(\theta + \phi) & = 2 \sin(\theta) \sin(\phi) \end{align} We have that \begin{align} \sin(2A) + \sin(2B) & = 2 \sin(A+B) \cos(A-B)\\ & = 2 \sin(\pi-C) \cos(A-B)\\ & = 2 \sin(C) \cos(A-B) \end{align} Hence, \begin{align} \sin(2A) + \sin(2B) + \sin(2C) & = 2 \sin(C) \cos(A-B) + 2 \sin(C) \cos(C)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(C) \right)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(\pi-(A+B)) \right)\\ & = 2 \sin(C) \left(\cos(A-B) - \cos(A+B) \right)\\ & = 2 \sin(C) \times 2 \sin(A) \sin(B)\\ & = 4 \sin(A) \sin(B) \sin(C) \end{align}