Question

(EPUSP-63) Provar que se π/4 < a < π/2 e π/4 < b < π/2 então

$\mathsf{sen(a+b)\ \textless \ sen~a+\frac{4}{5}\cdot sen~b}$

Answer 1

Olá Vinicius.



Somando os ângulos, temos:


$\mathsf{\Big(\dfrac{\pi}{4}\ \textless \ a\ \textless \ \dfrac{\pi}{2}\Big)~~+~~\Big(\dfrac{\pi}{4}\ \textless \ b\ \textless \ \dfrac{\pi}{2}\Big) ~\Rightarrow~~\Big(\dfrac{\pi}{2}\ \textless \ a+b\ \textless \ \pi\Big)}$


A soma dos ângulos a e b, será um ângulo do segundo quadrante. Se tirarmos o seno na inequação de (a + b), precisamos inverter os sinais. Já que depois que o ângulo passa de 90º até 180º, ele vai decrescendo.


$\mathsf{\Big(~sen\Big(\dfrac{\pi}{2}\Big)\ \textgreater \ sen(a+b)\ \textgreater \ sen(\pi)~\Big)~\Rightarrow~~\Big(1\ \textgreater \ sen(a+b)\ \textgreater \ 0\Big)}$


Checando o intervalo do lado direito da inequação:


$\mathsf{\Big(\dfrac{\pi}{4}\ \textless \ b\ \textless \ \dfrac{\pi}{2}\Big)~\Rightarrow~\Big(~sen\Big(\dfrac{\pi}{4}\Big)\ \textless \ sen~b\ \textless \ sen\Big(\dfrac{\pi}{2}\Big)~\Big)}\\\\=\\\\\mathsf{\Big(\dfrac{\sqrt{2}}{2}\ \textless \ sen~b\ \textless \ 1\Big)\cdot\Big(\dfrac{4}{5}\Big)\Rightarrow ~\Big(\dfrac{2\sqrt{2}}{5}\ \textless \ \dfrac{4}{5}\cdot sen~b~\ \textless \ \dfrac{4}{5}\Big)}}\\\\=\\\\\mathsf{\Big(\dfrac{2\sqrt{2}}{5}\ \textless \ \dfrac{4}{5}\cdot sen~b~\ \textless \ \dfrac{4}{5}\Big)+\Big(\dfrac{\sqrt{2}}{2}\ \textless \ sen~a\ \textless \ 1\Big)}\\\\=$


$\mathsf{\Big(\dfrac{4\sqrt{2}+5\sqrt{2}}{10}\ \textless \ sen~a+\dfrac{4}{5}\cdot sen~b\ \textless \ \dfrac{5+4}{5}\Big)}\\\\=\\\\\mathsf{\Big(\dfrac{9\sqrt{2}}{10}\ \textless \ sen~a+\dfrac{4}{5}\cdot sen~b\ \textless \ \dfrac{9}{5}\Big)~~ou~~\Big(1,27\ \textless \ sen~a+\dfrac{4}{5}\cdot sen~b\ \textless \ 1,8\Big)}$


$\mathsf{Nota:}~\textsl{Perceba~que~nesse~caso~os~sinais~n\~ao~foram~invertidos.}\\\textsl{~~~~~~~~~Isso~porque~os~intervalos~est\~ao~no~primeiro~quadrante,}\\\textsl{~~~~~~~~~onde~ele~inicia~no~0~e~termina~no~1.}$


Comparando os intervalos::


$\mathsf{\Big(0\ \textless \ sen(a+b)\ \textless \ 1\ \textless \ 1,27\ \textless \ sen~a+\dfrac{4}{5}\cdot sen~b\ \textless \ 1,8\Big)}}\\\\\\=\\\\\\\boxed{\mathsf{sen(a+b)\ \textless \ sen~a+\dfrac{4}{5}\cdot sen~b}}~~~\checkmark$


Dúvidas? comente