<s>[INST] The sides of a triangle are in Arithmetic progression If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are $X$ and $Y$, then show that
$$4(1- \cos X)(1-\cos Y) = \cos X + \cos Y$$
I tried using sine rule but can't solve it.
 [/INST] The law of sines helps!
From the given we obtain
$$\sin X+\sin Y=2\sin(X+Y)$$ or
$$2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2}=4\sin\frac{X+Y}{2}\cos\frac{X+Y}{2}$$ or
$$\cos\frac{X-Y}{2}=2\cos\frac{X+Y}{2}$$ or
$$\cos\frac{X}{2}\cos\frac{Y}{2}=3\sin\frac{X}{2}\sin\frac{Y}{2}$$ or
$$\cos\frac{X-Y}{2}=4\sin\frac{X}{2}\sin\frac{Y}{2}$$ and
$$\cos\frac{X+Y}{2}=2\sin\frac{X}{2}\sin\frac{Y}{2}$$
We need to prove that
$$16\sin^2\frac{X}{2}\sin^2\frac{Y}{2}=2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2},$$ 
which is obvious now.
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