Home › Forums › Math questions › Trigonometry questions

Conditional Identitity

Posted March 30th, 2010 by alpha
in

In a triangle, if $ cot A+ cot B + cot C=\sqrt{3} $ , prove that the triangle is equilateral.

‹ trigonometry equation Find the general solution ›

Comments

answer

On April 4th, 2010 Structure says:

For $ x,y\in (0,\pi) $ with $ 0<x+y<\pi $ we have

$$\cot x+\cot y=\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}=\frac{\sin(x+y)}{\sin x\sin y}=\frac{4\sin\frac{x+y}{2}\cos\frac{x+y}{2}}{\cos(x-y)-\cos(x+y)}\ge$$
$$\ge\frac{4\sin\frac{x+y}{2}\cos\frac{x+y}{2}}{1-\cos(x+y)}=\frac{4\sin\frac{x+y}{2}\cos\frac{x+y}{2}}{2\sin^2\frac{x+y}{2}}=2\cot\frac{x+y}{2}$$

so

$$\cot x+\cot y\ge2\cot\frac{x+y}{2}$$

with equality just for x=y.
So we have

$$\sqrt{3}+\frac{\sqrt3}{3}=\cot A+\cot B+\cot C+\cot\frac{\pi}{3}\ge 2\cot\frac{A+B}{2}+2\cot(\frac{C}{2}+\frac{\pi}{6})\ge 4\cot(\frac{A+B+C}{4}+\frac{\pi}{12})=4\cot\frac{\pi}{3}=\frac{4\sqrt 3}{3}$$

So $ A=B=C=\frac{\pi}{3} $

Back to top