A partir das fórmulas
Funções seno, co-seno e tangente: Infinito 12 A - Parte 3 Pág. 34 Ex. 10
Enunciado
A partir das fórmulas correspondentes do seno e do co-seno, deduza uma fórmula para
- $\operatorname{tg} \left( {\alpha + \beta } \right)$
- $\operatorname{tg} \left( {\alpha – \beta } \right)$
Resolução
$$\operatorname{sen} \left( {\alpha + \beta } \right) = \operatorname{sen} \alpha \cos \beta + \cos \alpha \operatorname{sen} \beta $$
$$\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta – \operatorname{sen} \alpha \operatorname{sen} \beta $$
- Ora, $$\begin{array}{*{20}{l}}
{\operatorname{tg} \left( {\alpha + \beta } \right)}& = &{\frac{{\operatorname{sen} \left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}}} \\
{}& = &{\frac{{\operatorname{sen} \alpha \cos \beta + \cos \alpha \operatorname{sen} \beta }}{{\cos \alpha \cos \beta – \operatorname{sen} \alpha \operatorname{sen} \beta }}} \\
{}& = &{\frac{{\frac{{\operatorname{sen} \alpha \cos \beta }}{{\cos \alpha \cos \beta }} + \frac{{\cos \alpha \operatorname{sen} \beta }}{{\cos \alpha \cos \beta }}}}{{\frac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }} – \frac{{\operatorname{sen} \alpha \operatorname{sen} \beta }}{{\cos \alpha \cos \beta }}}}} \\
{}& = &{\frac{{\operatorname{tg} \alpha + \operatorname{tg} \beta }}{{1 – \operatorname{tg} \alpha \operatorname{tg} \beta }}}
\end{array}$$
Logo, $$\operatorname{tg} \left( {\alpha + \beta } \right) = \frac{{\operatorname{tg} \alpha + \operatorname{tg} \beta }}{{1 – \operatorname{tg} \alpha \operatorname{tg} \beta }}$$
- Tendo em consideração a relação obtida em 1, temos:
$$\begin{array}{*{20}{l}}
{\operatorname{tg} \left( {\alpha – \beta } \right)}& = &{\operatorname{tg} \left( {\alpha + ( – \beta )} \right)} \\
{}& = &{\frac{{\operatorname{tg} \alpha + \operatorname{tg} \left( { – \beta } \right)}}{{1 – \operatorname{tg} \alpha \operatorname{tg} \left( { – \beta } \right)}}} \\
{}& = &{\frac{{\operatorname{tg} \alpha – \operatorname{tg} \beta }}{{1 + \operatorname{tg} \alpha \operatorname{tg} \beta }}}
\end{array}$$
Logo, $$\operatorname{tg} \left( {\alpha – \beta } \right) = \frac{{\operatorname{tg} \alpha – \operatorname{tg} \beta }}{{1 + \operatorname{tg} \alpha \operatorname{tg} \beta }}$$
$$\operatorname{tg} \left( {\alpha + \beta } \right) = \frac{{\operatorname{tg} \alpha + \operatorname{tg} \beta }}{{1 – \operatorname{tg} \alpha \operatorname{tg} \beta }}$$
$$\operatorname{tg} \left( {\alpha – \beta } \right) = \frac{{\operatorname{tg} \alpha – \operatorname{tg} \beta }}{{1 + \operatorname{tg} \alpha \operatorname{tg} \beta }}$$
