A partir da fórmula
Funções seno, co-seno e tangente: Infinito 12 A - Parte 3 Pág. 34 Ex. 9
Enunciado
A partir da fórmula $$\operatorname{sen} \left( {\alpha + \beta } \right) = \operatorname{sen} \alpha \cos \beta + \cos \alpha \operatorname{sen} \beta $$ encontre uma fórmula para:
- $\operatorname{sen} \left( {\alpha – \beta } \right)$
- $\operatorname{sen} \left( {2\alpha } \right)$
Resolução
$$\operatorname{sen} \left( {\alpha + \beta } \right) = \operatorname{sen} \alpha \cos \beta + \cos \alpha \operatorname{sen} \beta $$
- Ora, $$\begin{array}{*{20}{l}}
{\operatorname{sen} \left( {\alpha – \beta } \right)}& = &{\operatorname{sen} \left( {\alpha + ( – \beta )} \right)} \\
{}& = &{\operatorname{sen} \alpha \cos \left( { – \beta } \right) + \cos \alpha \operatorname{sen} \left( { – \beta } \right)} \\
{}& = &{\operatorname{sen} \alpha \cos \beta – \cos \alpha \operatorname{sen} \beta }
\end{array}$$
Logo, $\operatorname{sen} \left( {\alpha – \beta } \right) = \operatorname{sen} \alpha \cos \beta – \cos \alpha \operatorname{sen} \beta $.
- Ora, $$\begin{array}{*{20}{l}}
{\operatorname{sen} \left( {2\alpha } \right)}& = &{\operatorname{sen} \left( {\alpha + \alpha } \right)} \\
{}& = &{\operatorname{sen} \alpha \cos \alpha + \cos \alpha \operatorname{sen} \alpha } \\
{}& = &{2\operatorname{sen} \alpha \cos \alpha }
\end{array}$$
Logo, $\operatorname{sen} \left( {2\alpha } \right) = 2\operatorname{sen} \alpha \cos \alpha $.
$$\operatorname{sen} \left( {\alpha – \beta } \right) = \operatorname{sen} \alpha \cos \beta – \cos \alpha \operatorname{sen} \beta $$
$$\operatorname{sen} \left( {2\alpha } \right) = 2\operatorname{sen} \alpha \cos \alpha $$
