<s>[INST] If the sides $a$, $b$ and $c$ of $\triangle ABC$ are in Arithmetic Progression 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
$$2b=a+c$$
$$\sin A+\sin C=2\sin B$$
$$2\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2\sin B$$
$$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=\sin B$$
$$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2.\sin (\dfrac {A+C}{2}).\cos (\dfrac {A+C}{2})$$
$$2\cos (\dfrac {A+C}{2})=\cos (\dfrac {A-C}{2})$$
 [/INST] You can't, because it is not true. Consider a right-angle triangle with the sides $3,4,5$. Then:
$$\cos \left(\frac{B-C}{2}\right)=\sqrt{\frac{1+\cos{(B-C)}}{2}}=\sqrt{\frac{1+\cos B\cos C+\sin B\sin C}{2}}=\frac{3}{\sqrt{10}},\\
2\sin \frac A2=2\sqrt{\frac{1-\cos A}{2}}=\frac{2}{\sqrt{10}}.$$
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