Question

If A, B, C are the angles of a triangle and ${\sin }^3\theta =\sin (A-\theta )\sin (B-\theta )\sin (C-\theta )$, prove that $\cot \theta =\cot A+\cot B+\cot C$ and conversely.

Answer 1

The given equation can be written as
$\frac{\sin (A-\theta )}{\sin \theta }\cdot \frac{\sin (B-\theta )}{\sin \theta }\cdot \frac{\sin (C-\theta )}{\sin \theta }=1$
or $(\sin A\cot \theta -\cos A)(\sin B\cot \theta -\cos B)(\sin C\cot \theta -\cos C)=1$
$\sin A\sin B\sin C\left[\right(\cot \theta -\cot A\left)\right(\cot \theta -\cot B\left)\right(\cot \theta -\cot C\left)\right]=1$
Dividing by $\sin A\sin B\sin C$, we have
${\cot }^3\theta -{\cot }^2\theta S_1+\cot \theta S_2-S_3=\frac{1}{\sin A\sin B\sin C}$
where $S_2=\sum \cot A\cot B=1$
Also, $S_1=\cot A+\cot B+\cot C=\frac{\sum \cos A\sin B\sin C}{\sin A\sin B\sin C}$
$S_3=\cot A\cot B\cot C=\frac{\cos A\cos B\cos C}{\sin A\sin B\sin C}$
$\therefore \cot \theta (1+{\cot }^2\theta )=\frac{\sum \cos A\sin B\sin C}{\sin A\sin B\sin C}\cdot {\cot }^2\theta +\frac{1+\cos A\cos B\cos C}{\sin A\sin B\sin C}$
$=\frac{\sum \cos A\sin B\sin C}{\sin A\sin B\sin C}({\cot }^2\theta +1)$
[We have used $\cos (A+B+C)=\cos A\cos B\cos C-\sum \cos A\sin B\sin C=-1$]
Cancel $1+{\cot }^2\theta$
$\therefore \cot \theta =\frac{\sum \cos A\sin B\sin C}{sinA\sin B\sin C}=\cot A+\cot B+\cot C$.
Converse:
Let ${\sin }^3\theta =\sin (A-\theta )\sin (B-\theta )\sin (C-\theta )$
To prove: $\cot \theta =\cot A+\cot B+\cot C$
$\cot \theta -\cot A=\cot B+\cot C$
or $\frac{\sin (A-\theta )}{\sin \theta \sin A}=\frac{\sin (B+C)}{\sin B\sin C}=\frac{\sin A}{\sin B\sin C}$
$\therefore \sin (A-\theta )=\frac{\sin \theta {\sin }^{2}A}{\sin B\sin C}$ ..$\left(1\right)$
Similarly, $\sin (B-\theta )=\frac{\sin \theta {\sin }^{2}B}{\sin C\sin A}$ $\left(2\right)$
$\sin (C-\theta )=\frac{\sin \theta {\sin }^{2}C}{\sin C\sin B}$ .$\left(3\right)$
Multiplying $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$\sin (A-\theta )\sin (B-\theta )\sin (C-\theta )={\sin }^3\theta$.