TANGENTE DI UNA SOMMA
[math] C.E. \cos\alpha\cos\beta\ne 0[/math]
[math]\tan(\alpha+\beta) = {{\sin(\alpha+\beta)}\over {\cos(\alpha+\beta)}}= {{\sin\alpha\cos\beta +\cos\alpha\sin\beta }\over{\cos\alpha\cos\beta -\sin\alpha\sin\beta }[/math]
dividendo tutti i membri per
[math]\cos\alpha\cos\beta[/math]
e si ottiene:
[math]\tan(\alpha+\beta)[/math]
=
[math] {{\sin\alpha\cos\beta \over{cos\alpha\cos\beta}}+{{cos\alpha\sin\beta}\over{cos\alpha\cos\beta}}}\over{{cos\alpha\cos\beta/cos\alpha\cos\beta-{{\sin\alpha\sin\beta}/{\cos\alpha\cos\beta}}}[/math]
semplifico e avrò dimostrato che:
[math]\tan(\alpha+\beta) = {{\tan\alpha + \tan\beta}\over{1-\tan\alpha\tan\beta}}[/math]
TANGENTE DI UNA DIFFERENZA
[math]\tan(\alpha-\beta) = {{\sin(\alpha-\beta)}\over {\cos(\alpha-\beta)}}= {{\sin\alpha\cos\beta -\cos\alpha\sin\beta }\over{\cos\alpha\cos\beta +\sin\alpha\sin\beta }[/math]
dividendo tutti i membri per
[math]\cos\alpha\cos\beta[/math]
si ottiene:
[math]\tan(\alpha-\beta)[/math]
=
[math] {{\sin\alpha\cos\beta \over{cos\alpha\cos\beta}}-{{cos\alpha\sin\beta}\over{cos\alpha\cos\beta}}}\over{{cos\alpha\cos\beta/cos\alpha\cos\beta+{{\sin\alpha\sin\beta}/{\cos\alpha\cos\beta}}}[/math]
semplificando si ottiene che:
[math]\tan(\alpha-\beta) = {{\tan\alpha - \tan\beta}\over{1+\tan\alpha\tan\beta}}[/math]
COTANGENTE DI UNA SOMMA
[math] cotg(\alpha+\beta) = {{cos(\alpha+\beta)} \over \sin (\alpha+\beta)}= {{\cos\alpha\cos\beta - \sin\alpha\sin\beta}\over \sin\alpha\cos\beta+\cos\alpha\sin\beta}=[/math]
divido tutto per
[math] \sin\alpha\sin\beta[/math]
e avrò:
[math] cotg (\alpha+\beta)[/math]
=
[math] {{cos\alpha\cos\beta \over{sin\alpha\sin\beta}}-{{\sin\alpha\sin\beta}\over{\sin\alpha\sin\beta}}}\over{{\sin\alpha\cos\beta/sin\alpha\sin\beta+{{\cos\alpha\sin\beta}/{\sin\alpha\sin\beta}}}[/math]
=
[math]cotg(\alpha+\beta)[/math]
=
[math]{cotg\alpha cotg\beta -1}\over {cotg\beta+cotg\alpha}[/math]
COTANGENTE DI UNA DIFFERENZA
[math] cotg(\alpha-\beta) = {{cos(\alpha-\beta)} \over \sin (\alpha-\beta)}= {{\cos\alpha\cos\beta + \sin\alpha\sin\beta}\over \sin\alpha\cos\beta-\cos\alpha\sin\beta}=[/math]
divido tutti i membri per
[math] \sin\alpha\sin\beta[/math]
e avrò:
[math] cotg (\alpha-\beta)[/math]
=
[math] {{cos\alpha\cos\beta \over{sin\alpha\sin\beta}}+{{\sin\alpha\sin\beta}\over{\sin\alpha\sin\beta}}}\over{{\sin\alpha\cos\beta/sin\alpha\sin\beta-{{\cos\alpha\sin\beta}/{\sin\alpha\sin\beta}}}[/math]
=
[math]cotg(\alpha-\beta)[/math]
=
[math]{cotg\alpha cotg\beta +1}\over {cotg\beta-cotg\alpha}[/math]