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