What else could we generate from $A+B+C=\pi$, when $A,\, B,\,C$ are three angles from a triangle?

Following previous blog post, we have shown that if

If $A,\, B,\,C$ are three angles from a triangle, i.e. $A+B+C=\pi$, then we should have known the following equality holds.

$\displaystyle \tan A+\tan B+\tan C=\tan A\tan B\tan C$

On this blog post, we now try to generate another relation between $A,\,B$ and $C$ for cotangent functions:

1.

$A+B+C=\pi$

$A+B=\pi-C$

Taking the sine on both sides of the equation:

$\sin (A+B)=\sin (\pi-C)$

$\sin A \cos B+\cos A\sin B=\sin\pi \cos C-\cos \pi \sin C$

$\sin A \cos B+\cos A\sin B=(0)( \cos C)-(-1)(\sin C)$

$\sin A \cos B+\cos A\sin B=\sin C$---(1)

Taking the cosine on both sides of the equation:

$\cos (A+B)=\cos (\pi-C)$

$\cos A \cos B-\sin A\sin B=\cos \pi \cos C+\sin \pi \sin C$

$\cos A \cos B-\sin A\sin B=(-1)( \cos C)+(0)(\sin C)$

$\cos A \cos B-\sin A\sin B=-\cos C$---(2)

Dividing (2) by (1), we see that we have:

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

Dividing the numerator and denominator on the left by $\sin A\sin B$, we get:

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

$\displaystyle \frac{\cot A \cot B-1}{\cot B+\cot A}=-\cot C$

$\displaystyle \cot A \cot B-1=(-\cot C)(\cot B+\cot A)$

$\displaystyle \cot A \cot B-1=-\cot C\cot B-\cot C \cot A)$

$\displaystyle \color{yellow}\bbox[5px,green]{\cot A \cot B+\cot B\cot C+\cot A\cot C=1}$

2.


$A+B+C=\pi$

$\displaystyle \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$

Taking the sine on both sides of the equation:

$\displaystyle \sin \left(\frac{A}{2}+\frac{B}{2}\right)=\sin \left(\frac{\pi}{2}-\frac{C}{2}\right)$

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

$\displaystyle \sin \frac{A}{2} \cos \frac{B}{2}+\cos \frac{A}{2}\sin \frac{B}{2}=(1)\cos \frac{C}{2}-(0)\sin \frac{C}{2}$

$\displaystyle \sin \frac{A}{2} \cos \frac{B}{2}+\cos \frac{A}{2}\sin \frac{B}{2}=\cos \frac{C}{2}$---(3)

Taking the cosine on both sides of the equation:

$\displaystyle \cos \left(\frac{A}{2}+\frac{B}{2}\right)=\cos \left(\frac{\pi}{2}-\frac{C}{2}\right)$

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

$\displaystyle \cos \frac{A}{2} \cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2}=(0) \cos \frac{C}{2}+(1) \sin \frac{C}{2}$

$\displaystyle \cos \frac{A}{2} \cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2}= \sin \frac{C}{2}$---(4)

Dividing (3) by (4), we see that we have:

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

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

Dividing the numerator and denominator on the left by $\sin \frac{A}{2}\sin \frac{B}{2}$, we get:

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

$\displaystyle \frac{\cot \frac{B}{2}+\cot \frac{A}{2}}{\cot \frac{A}{2} \cot \frac{B}{2}-1}=\cot \frac{C}{2}$

$\displaystyle \cot \frac{B}{2}+\cot \frac{A}{2}=\cot \frac{A}{2} \cot \frac{B}{2}\cot \frac{C}{2} -\cot \frac{C}{2}$

$\displaystyle \color{yellow}\bbox[5px,purple]{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2}\cot \frac{C}{2} }$