In a triangle $ABC$, if $\sin A$, $\sin B$ and $\sin C$ form an A.P, prove that $\cot (A/2)\cot(C/2) = 3$.

Question

I have tried using $2\sin B = \sin A +\sin C$. Then I substitute $B=180-(A+C)$, but still I''m stuck halfway. Someone please help !

Answer 1

from your formula we get $$2=\frac{\sin(C)}{\sin(B)}+\frac{\sin(A)}{\sin(B)}$$ and by the formula of sines we get $$2b=a+c$$ now we calculate $$\cot(A/2)=\frac{\sin(A)}{1-\cos(A)}$$ $$\cot(C/2)=\frac{\sin(C)}{1-\cos(C)}$$ and then we get $$\cot(A/2)\cot(C/2)=\frac{16A^2}{(2bc-b^2-c^2+a^2)(2ab-a^2-b^2+c^2)}$$ with $$b=\frac{a+c}{2}$$ and $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ we get $3$ as you stated.

Answer 2

$$4 \sin \frac{B}{2}\cos \frac{B}{2}=2 \sin B = \sin A + \sin C =$$

$$2 \sin \frac{A+C}{2}\cos \frac{A-C}{2}=2 \cos \frac{B}{2}\cos \frac{A-C}{2}$$

Simplifying we get

$$2 \sin \frac{B}{2}=\cos \frac{A-C}{2}$$

$$2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2}$$

$$2 \cos \frac{A}{2} \cos \frac{C}{2}-2\sin \frac{A}{2} \sin \frac{C}{2}= \cos \frac{A}{2} \cos \frac{C}{2}+\sin \frac{A}{2} \sin \frac{C}{2}$$

$$ \cos \frac{A}{2} \cos \frac{C}{2}=3\sin \frac{A}{2} \sin \frac{C}{2}$$

$$\cot \frac{A}{2} \cot \frac{C}{2}=3$$

Answer 3

As $A,B,C$ are in AP, $B=180^\circ-(A+C)$, i.e. $\sin B=\sin (A+C)$.

Also, given $\sin A, \sin B, \sin C$ are in AP. Hence.

$$\begin{align} 2\sin B&=\sin A+\sin C\\ 2\sin(A+C)&=\sin A+\sin C\\ 2\sin 2(\alpha+\gamma)&=\sin 2\alpha+\sin 2\gamma &&\scriptsize \left(\alpha=\frac A2, \gamma=\frac B2\right)\\ 2\cdot 2\sin (\alpha+\gamma)\cos (\alpha+\gamma)&=2\sin(\alpha+\gamma)\cos(\alpha-\gamma)\\ 2\cos(\alpha+\gamma)&=\cos(\alpha-\gamma)\\ 2(\cos\alpha\cos\gamma-\sin\alpha\sin\gamma)&=\cos\alpha\cos\gamma+\sin\alpha\sin\gamma\\ \div \sin \alpha\sin \gamma,\qquad\qquad 2(\cot\alpha\cot\gamma-1)&=\cot\alpha\cot\gamma+1\\ \cot\alpha\cot\gamma=\cot\left(\frac A2\right)\cot\left(\frac C2\right)&=3\;\color{red}\blacksquare \end{align}$$