If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are X and Y, then show that4(1−cos⁡X)(1−cos⁡Y)=cos⁡X+cos⁡YI tried using sine rule but can't solve it.

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If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are X and Y, then show that4(1−cos⁡X)(1−cos⁡Y)=cos⁡X+cos⁡YI tried using sine rule but can&#039;t solve it.

Aiden Cooper · Accepted Answer

Let the sides be a-d,a,a+d (with a &amp;gt;b) 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  cos⁡X=(a+d)2+a2−(a−d)2}{2a(a+d)}=a+4d2(a+d)    cos⁡Y=(a−d)2+a2−(a+d)2}{2a(a−d)}=a−4d2(a−d)  Then  cos⁡X+cos⁡Y=a2−4d2a2−d2=4(a−2d)2(a+d)(a+2d)2(a−d)=4(1−cos⁡X)(1−cos⁡Y)

Ronald Alvarez · Accepted Answer

The law of sines helps!   From the given we obtain  sin⁡X+sin⁡Y=2sin⁡(X+Y)  or  2sin⁡X+Y2cos⁡X−Y2=4sin⁡X+Y2cos⁡X+Y2  or  cos⁡X−Y2=2cos⁡X+Y2  or  cos⁡X2cos⁡Y2=3sin⁡X2sin⁡Y2  or  cos⁡X−Y2=4sin⁡X2sin⁡Y2  and  cos⁡X+Y2=2sin⁡X2sin⁡Y2  We need to prove that  16sin2X2sin2Y2=2cos⁡X+Y2cos⁡X−Y2  which is obvious now.

If the sides of a triangle are in Arithmetic progression

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