Question

If $A+B+C=\pi$, the prove that $\sin A+\sin B+\sin C=4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$.

Answer 1

Hint : We use the theorem of sum of two trigonometric ratios along with submultiple formula of $\sin X=2\sin \dfrac{X}{2}\cos \dfrac{X}{2}$. We also change the angles and their ratios according to the need. We take the ratio and convert the sum into multiple forms.

Complete step-by-step answer :
We try to use the theorem of sum of two trigonometric ratios.
So, $\sin A+\sin B+\sin C=\left( \sin A+\sin B \right)+\sin C$.
We use the formula of $\sin X+\sin Y=2\sin \dfrac{X+Y}{2}\cos \dfrac{X-Y}{2}$. We also use the submultiple formula of $\sin X=2\sin \dfrac{X}{2}\cos \dfrac{X}{2}$.
Therefore, $\sin A+\sin B+\sin C=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}+2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$.
Now we use the given property of $A+B+C=\pi$. We get $A+B=\pi -C$ and $C=\pi -\left( A+B \right)$.

& \sin A+\sin B+\sin C=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}+2\sin \dfrac{C}{2}\cos \dfrac{C}{2} \\\ & \Rightarrow \sin A+\sin B+\sin C=2\sin \dfrac{\pi -C}{2}\cos \dfrac{A-B}{2}+2\sin \dfrac{\pi -\left( A+B \right)}{2}\cos \dfrac{C}{2} \\\ \end{aligned}$$ For general form of $\sin \left( x \right)$, we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha $ from that multiple of $\dfrac{\pi }{2}$ to make it equal to x. Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha $, $k\in \mathbb{Z}$. Here we took the addition of $\alpha $. We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$. Now we take the value of k. If it’s even then keep the ratio as sin and if it’s odd then the ratio changes to cos ratio from sin. Then we find the position of the given angle as a quadrant value measured in counter clockwise movement from the origin and the positive side of the X-axis. If the angel falls in the first or second quadrant then the sign remains positive but if it falls in the third or fourth quadrant then the sign becomes negative. The final form becomes $\sin \dfrac{\pi -C}{2}=\sin \left( \dfrac{\pi }{2}-\dfrac{C}{2} \right)=\cos \dfrac{C}{2}$ and $\sin \dfrac{\pi -\left( A+B \right)}{2}=\cos \dfrac{A+B}{2}$. $$\begin{aligned} & 2\sin \dfrac{\pi -C}{2}\cos \dfrac{A-B}{2}+2\sin \dfrac{C}{2}\cos \dfrac{C}{2} \\\ & =2\cos \dfrac{C}{2}\cos \dfrac{A-B}{2}+2\cos \dfrac{A+B}{2}\cos \dfrac{C}{2} \\\ & =2\cos \dfrac{C}{2}\left( \cos \dfrac{A-B}{2}+\cos \dfrac{A+B}{2} \right) \\\ \end{aligned}$$ We again use the formula of $$\cos X+\cos Y=2\cos \dfrac{X+Y}{2}\cos \dfrac{X-Y}{2}$$. $$\begin{aligned} & 2\cos \dfrac{C}{2}\left( \cos \dfrac{A-B}{2}+\cos \dfrac{A+B}{2} \right) \\\ & =2\cos \dfrac{C}{2}\left( 2\cos \dfrac{A-B+A+B}{4}. \cos \dfrac{A-B-A-B}{4} \right) \\\ & =4\cos \dfrac{C}{2}\cos \dfrac{A}{2}\cos \left( -\dfrac{B}{2} \right) \\\ & =4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} \\\ \end{aligned}$$ Thus proved $\sin A+\sin B+\sin C=4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$. **Note** : We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of $\pi $ instead of $\dfrac{\pi }{2}$. It makes the multiplied number always even. In that case we don’t have to change the ratio. If $x=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha $. Value of $2k$ is always even.