Question

If $\cos \left(A\right)+\cos \left(B\right)=4·{\sin }^2\left(\frac{C}{2}\right)$ then prove that the sides $a,c,b$ of a triangle $ABC$ is in A.P

Answer 1

Prove the given expression:

Given,

$$\begin{gathered} \cos \left(A\right)+\cos \left(B\right)=4·{\sin }^2\left(\frac{C}{2}\right) \\ \Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=4·{\sin }^2\left(\frac{C}{2}\right) \end{gathered}$$

Sum of angles of triangle $=A+B+C=180°$

So, $A+B=180°-C$

$$\begin{gathered} \Rightarrow 2\cos \left(\frac{180°-C}{2}\right)\cos \left(\frac{A-B}{2}\right)=4·{\sin }^2\left(\frac{C}{2}\right) \\ \Rightarrow 2\cos \left(90°-\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)=4·{\sin }^2\left(\frac{C}{2}\right) \\ \Rightarrow 2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)=4·{\sin }^2\left(\frac{C}{2}\right) \\ \Rightarrow \cos \left(\frac{A-B}{2}\right)=2·\sin \left(\frac{C}{2}\right) \end{gathered}$$

Multiply both sides by $\cos \left(\frac{C}{2}\right)$ we get

$$\begin{gathered} \Rightarrow \cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)=2·\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right) \\ \Rightarrow \cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)=\sin \left(C\right)\left[\because \sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)\right] \\ \Rightarrow \cos \left(\frac{180°-A-B}{2}\right)\cos \left(\frac{A-B}{2}\right)=\sin \left(C\right) \\ \Rightarrow \sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=\sin \left(C\right) \end{gathered}$$

Multiplying both side by $2$

$$\begin{gathered} \Rightarrow 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=2\sin \left(C\right) \\ \Rightarrow \sin \left(A\right)+\sin \left(B\right)=2\sin \left(C\right)....\left(1\right) \end{gathered}$$

We know that

$$\begin{gathered} \frac{\sin \left(A\right)}{a}=\frac{\sin \left(B\right)}{b}=\frac{\sin \left(C\right)}{c}=t[\mathrm{let}\text{'}\mathrm{s}\mathrm{say}] \\ \sin \left(A\right)=at,\sin \left(B\right)=bt\mathrm{and}\sin \left(\mathrm{C}\right)=\mathrm{ct} \\ \Rightarrow \mathrm{at}+\mathrm{bt}=2\mathrm{ct}(\mathrm{using}(1\left)\right) \\ \Rightarrow \mathrm{a}+\mathrm{b}=2\mathrm{c} \end{gathered}$$

So, the sides $a,c,b$ of a triangle $ABC$ are in A.P.

Hence proved.