Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle

Solution 1:

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}

Solution 2:

Here's another way to do it. Since $A+B+C=\pi$, we have the three angles of a triangle. Inscribe that triangle in a circle of unit diameter (not unit radius). Then $\sin A$, $\sin B$, and $\sin C$ are actually the lengths of the sides opposite that three angles---that's essentially the law of sines. The area of any triangle is the product of the lengths of two sides times the sine of the angle between them, divided by $2$. So the area in this case is $(\sin A\sin B\sin C)/2$. That's the right side of the identity, except that you've got $4$ rather than $1/2$ as the coefficient. So it is enough to show that the area is $(\sin(2A) + \sin(2B) + \sin(2C))/8$. If the center of the circle is inside the triangle, you can draw lines from the center to each of the three vertices, thus breaking the triangle into three smaller triangles. It is then enough to show that $\sin(2A)/8$ is the area of one of those (and the other two are shown "similarly"). All you need is that two lengths of sides are each $1/2$ and the angle between them is $2A$.

If the center of the circle is not inside the triangle, then one of the three angles at the center---say $2A$---is more than $180^\circ$ and so $\sin(2A)$ is negative. You get a signed area. Two of those three parts add up to more than the whole triangle, and then you subtract the third part and get the right amount.

This also generalizes to other polygons inscribed in a circle, telling us, for example, that $$ \text{if }A+B+C+D+E=\pi $$ $$ \begin{align} & \text{then }\sin(2A)+\sin(2B)+\sin(2C)+\sin(2D)+\sin(2E) \\[8pt] & = \underbrace{4\,\overbrace{\sin A\sin B\sin C}^{\text{three sines}}\,\, \overbrace{\cos D\cos E}^{\text{two cosines}} + \cdots\cdots}_{\text{10 terms}} - 8\sin A\sin B\sin C\sin D\sin E. \end{align} $$ (The five angles here are the angles between adjacent diagonals of the inscribed pentagon.)

Solution 3:

The other two answers are great, but if you're ever not feeling clever enough to come up with them you can always try the sledgehammer method (using Euler's formula and complex exponentials): $$4\sin(A)\sin(B)\sin(C)=4\left(\frac{e^{iA}-e^{-iA}}{2i}\right)\left(\frac{e^{iB}-e^{-iB}}{2i}\right)\left(\frac{e^{iC}-e^{-iC}}{2i}\right) \, $$ Now, multiply out this product and you'll get eight terms. The two where all the signs are the same will be $e^{i\pi}/(-2i)$ and $-e^{-i\pi}/(-2i)$ since $A+B+C=\pi$, so they cancel. The other six will collect into three terms that look like $$\frac{-e^{i(A+B-C)}+e^{i(-A-B+C)}}{-2i}=\sin(A+B-C)$$ (possibly with the variables permuted). But $A+B-C=\pi-2C$, which means $$\sin(A+B-C)=\sin(\pi-2C)=\sin(2C) \, ,$$ and similarly with the other two terms. So their sum is the left-hand side of your identity.