Assegnato il triangolo rettangolo BAC, si posiziona il vertice A nell'origine degli assi cartesiani e il vertice B sul semiasse positivo delle ascisse.[br]Disegnata la circonferenza goniometrica, è noto che: sen([math]\alpha[/math]) = [color=#ff00ff]DE[/color]; cos([math]\alpha[/math]) = [color=#6aa84f]AE[/color]; tg([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) = [color=#00ff00]FH[/color]; ctg([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) = [color=#ff0000]GI[/color].[br]- Dalla similitudine dei triangoli EAD e BAC si ha: [br]1) DE : CB = AD : AC e, sostituendo: sen([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) : CB = 1 : AC, quindi: sen([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) = CB/AB (cateto opposto all'angolo diviso ipotenusa);[br]2) AE : AB = AD : AC e, sostituendo: cos([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) : AB = 1 : AC, quindi: cos([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) = AB/AC (cateto adiacente all'angolo diviso ipotenusa).[br]- Dalla similitudine dei triangoli HAF e BAC si ha: HA : AB = FH : CB e, sostituendo: = 1 : AB = tg([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) : CB, [br] quindi: tg([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) = CB/AB (cateto opposto all'angolo diviso cateto adiacente).[br]- Dalla similitudine dei triangoli GAI e BAC si ha: IG : AB = AI : BC e, sostituendo: ctg([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) : AB = 1 : BC,[br] quindi: ctg([img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAA0AAAALCAYAAACksgdhAAAAAXNSR0IArs4c6QAAALdJREFUKFOd0a+KQmEQhvHfuQCbySBewEZhs8l/sBgUxCR4AWIwq80g3oLJvmmVLRvFsFWsJu9AjHLkEw6Ho8FJw8s8M+/MRN6I6AlTRB5HXNI1WdAJMTRDCz0cQv4dN0hDfXRRC90nuGKOMv7TUBVbNPEToA6W+MXgYTM5aYw2PhM7NLAK1nZZUD3YGiagL4xQSR4jvdMaORTwgTNKiF0ssiY9tCn22AQhdvAXDnKXnv3p5cvfgm6cxhoMOJglIQAAAABJRU5ErkJggg==[/img]) = AB/BC (cateto adiacente all'angolo diviso cateto opposto).[br]( analogo ragionamento si può ripetere per l'angolo [math]\beta[/math])[br][url=https://www.geogebra.org/m/ar9w6fcp#material/k9k9xdkj]seno e coseno[/url] ; [url=https://www.geogebra.org/m/ar9w6fcp#material/pksskxf8]tangente e cotangente[/url][br]