Mostra que
Trigonometria: Matematicamente Falando 9 - Parte 2 Pág. 62 Ex. 10
Enunciado
Mostra que:
- \(2{{\mathop{\rm sen}\nolimits} ^2}\alpha – 1 = \left( {{\mathop{\rm sen}\nolimits} \alpha – \cos \alpha } \right)\left( {{\mathop{\rm sen}\nolimits} \alpha + \cos \alpha } \right)\)
- \(1 + \frac{1}{{{{{\mathop{\rm tg}\nolimits} }^2}\theta }} = \frac{1}{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}\)
Resolução
- Tendo em consideração a Fórmula Fundamental da Trigonometria e o caso notável “diferença de dois quadrados”, temos:
\[\begin{array}{*{20}{l}}{2{{{\mathop{\rm sen}\nolimits} }^2}\alpha – 1}& = &{2{{{\mathop{\rm sen}\nolimits} }^2}\alpha – \left( {{{{\mathop{\rm sen}\nolimits} }^2}\alpha + {{\cos }^2}\alpha } \right)}\\{}& = &{2{{{\mathop{\rm sen}\nolimits} }^2}\alpha – {{{\mathop{\rm sen}\nolimits} }^2}\alpha – {{\cos }^2}\alpha }\\{}& = &{{{{\mathop{\rm sen}\nolimits} }^2}\alpha – {{\cos }^2}\alpha }\\{}& = &{\left( {{\mathop{\rm sen}\nolimits} \alpha – \cos \alpha } \right)\left( {{\mathop{\rm sen}\nolimits} \alpha + \cos \alpha } \right)}\end{array}\]
- Tendo em consideração \({\mathop{\rm tg}\nolimits} \alpha = \frac{{{\mathop{\rm sen}\nolimits} \alpha }}{{\cos \alpha }}\) e a Formula Fundamental da Trigonometria, vem:
\[\begin{array}{*{20}{l}}{1 + \frac{1}{{{{{\mathop{\rm tg}\nolimits} }^2}\theta }}}& = &{1 + \frac{1}{{\frac{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}{{{{\cos }^2}\theta }}}}}\\{}& = &{1 + \frac{{{{\cos }^2}\theta }}{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}}\\{}& = &{\frac{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }} + \frac{{{{\cos }^2}\theta }}{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}}\\{}& = &{\frac{{{{{\mathop{\rm sen}\nolimits} }^2}\theta + {{\cos }^2}\theta }}{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}}\\{}& = &{\frac{1}{{{{{\mathop{\rm sen}\nolimits} }^2}\theta }}}\end{array}\]
