Nell’ultimo caso, osservand0 i triangoli \(PBC\) e \(PDA\), e applicando il teorema dei seni, si ha: \[\frac{\overline{PC}}{\sin \left( \pi /6 \right)}=\frac{\overline{CB}}{\sin \left( x+\pi /6 \right)}\to \overline{PC}=\frac{a}{\sin \left( x+\pi /6 \right)}\]\[\frac{\overline{PB}}{\sin x}=\frac{\overline{CB}}{\sin \left( x+\pi /6 \right)}\to \overline{PB}=\frac{2a\sin x}{\sin \left( x+\pi /6 \right)}\]\[\frac{\overline{PD}}{\sin \left( x-\pi /6 \right)}=\frac{\overline{AD}}{\sin \left( x+\pi /6 \right)}\to \overline{PD}=\frac{a\sin \left( x-\pi /6 \right)}{\sin \left( x+\pi /6 \right)}\] per cui\[f\left( x \right)=\frac{\overline{PC}-\overline{PB}}{\overline{PD}}=\frac{\left( 1-2\sin x \right)}{\sin \left( x-\pi /6 \right)}=\frac{2\left( 1-2\sin x \right)}{\cos x-\sqrt{3}\sin x}\] e tale funzione è positiva per \[0\le x<\frac{\pi }{6}\vee \frac{\pi }{6}<x\le \frac{5}{6}\pi \vee \frac{7}{6}\pi <x\le 2\pi \quad .\]
Massimo Bergamini
