Question

In a triangle $ABC$ prove that $\cos \left(\frac{\mathrm{A}}{2}\right)+\cos \left(\frac{\mathrm{B}}{2}\right)+\cos \left(\frac{\mathrm{C}}{2}\right)=4.\cos \left(\frac{\mathrm{\pi }-\mathrm{A}}{4}\right).\cos \left(\frac{\mathrm{\pi }-\mathrm{B}}{4}\right).\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right)$.

Answer 1

Prove the given equation.

In the question, it is given that $ABC$ is a triangle.

We know that, according to the angle sum property $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=\mathrm{\pi }$.

Simplify the Right-hand side.

$$\begin{gathered} 4·\cos \left(\frac{\mathrm{\pi }-\mathrm{A}}{4}\right)·\cos \left(\frac{\mathrm{\pi }-\mathrm{B}}{4}\right)·\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right) \\ \Rightarrow 2·\left(2\cos \left(\frac{\mathrm{\pi }-\mathrm{A}}{4}\right)·\cos \left(\frac{\mathrm{\pi }-\mathrm{B}}{4}\right)\right)·\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right)\left[\because \cos \left(\mathrm{C}\right)+\cos \left(\mathrm{D}\right)=2\cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\right] \\ \Rightarrow 2·\left(\cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{A}+\mathrm{B}}{4}\right)+\cos \left(\frac{\mathrm{B}-\mathrm{A}}{4}\right)\right)·\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right)\left[\because \cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{A}+\mathrm{B}}{4}\right)=\sin \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)\right] \\ \Rightarrow 2·\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right)\sin \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)+2·\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right)\cos \left(\frac{\mathrm{B}-\mathrm{A}}{4}\right)\left[\because \angle \mathrm{C}=\mathrm{\pi }-\left(\angle \mathrm{A}+\angle \mathrm{B}\right)\right] \\ \Rightarrow 2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)\sin \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)+2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)\cos \left(\frac{\mathrm{B}-\mathrm{A}}{4}\right)\left[\because \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)\sin \left(\frac{\mathrm{A}+\mathrm{B}}{4}\right)\right] \\ \Rightarrow \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)+\cos \left(\frac{\mathrm{B}}{2}\right)+\cos \left(\frac{-\mathrm{A}}{2}\right)\left[\because \angle \mathrm{A}+\angle \mathrm{B}=\mathrm{\pi }-\angle \mathrm{C},\cos \left(-\mathrm{x}\right)=\cos \left(\mathrm{x}\right)\right] \\ \Rightarrow \sin \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{C}}{2}\right)+\cos \left(\frac{\mathrm{B}}{2}\right)+\cos \left(\frac{\mathrm{A}}{2}\right)\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)=\cos \left(\mathrm{x}\right)\right] \\ \Rightarrow \cos \left(\frac{\mathrm{C}}{2}\right)+\cos \left(\frac{\mathrm{B}}{2}\right)+\cos \left(\frac{\mathrm{A}}{2}\right)=\mathrm{LHS} \end{gathered}$$

Hence proved $\cos \left(\frac{\mathrm{A}}{2}\right)+\cos \left(\frac{\mathrm{B}}{2}\right)+\cos \left(\frac{\mathrm{C}}{2}\right)=4.\cos \left(\frac{\mathrm{\pi }-\mathrm{A}}{4}\right).\cos \left(\frac{\mathrm{\pi }-\mathrm{B}}{4}\right).\cos \left(\frac{\mathrm{\pi }-\mathrm{C}}{4}\right)$.