The sides of a triangle are in Arithmetic progression

Question

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.

Answer 1

Let the sides be $a-d,a,a+d$ (with $a>d)$ be the three sides of the triangle, so $X$ corresponds to the side with length $a-d$ and $Y$ that to with length $a+d$. Using cosine formula \begin{align*} \cos X & = \frac{(a+d)^2+a^2-(a-d)^2)}{2a(a+d)}=\frac{a+4d}{2(a+d)}\\ \cos Y & = \frac{(a-d)^2+a^2-(a+d)^2)}{2a(a-d)}=\frac{a-4d}{2(a-d)}\\ \end{align*} Then $$\cos X +\cos Y=\frac{a^2-4d^2}{a^2-d^2}=4 \frac{(a-2d)}{2(a+d)}\frac{(a+2d)}{2(a-d)}=4(1-\cos X)(1-\cos Y).$$

Answer 2

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.