If the sides $a$, $b$ and $c$ of $\triangle ABC$ are in Arithmetic Progression

Question

If the sides $a$, $b$ and $c$ of $\triangle ABC$ are in Arithmetic Progression, then prove that: $$\cos (\dfrac {B-C}{2})=2\sin (\dfrac {A}{2})$$

My Attempt:

Since, $a,b,c$ are in AP $$2b=a+c$$ $$\sin A+\sin C=2\sin B$$ $$2\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2\sin B$$ $$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=\sin B$$ $$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2.\sin (\dfrac {A+C}{2}).\cos (\dfrac {A+C}{2})$$ $$2\cos (\dfrac {A+C}{2})=\cos (\dfrac {A-C}{2})$$

Answer 1

You can't, because it is not true. Consider a right-angle triangle with the sides $3,4,5$. Then: $$\cos \left(\frac{B-C}{2}\right)=\sqrt{\frac{1+\cos{(B-C)}}{2}}=\sqrt{\frac{1+\cos B\cos C+\sin B\sin C}{2}}=\frac{3}{\sqrt{10}},\\ 2\sin \frac A2=2\sqrt{\frac{1-\cos A}{2}}=\frac{2}{\sqrt{10}}.$$