✅ Após resolver os cálculos, concluímos que o valor do cosseno do ângulo "θ", a partir do seno valendo "1/3" é:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf \cos\theta = \pm\frac{2\sqrt{2}}{3}\:\:\:}}\end{gathered}$}[/tex]
Sejam os dados:
[tex]\LARGE\begin{cases}\tt \sin\theta = 1/3\\\tt\cos\theta = \:?\end{cases}[/tex]
Para resolver esta questão devemos nos lembrar da seguinte identidade trigonométrica fundamental, representada por:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\bf I \end{gathered}$}[/tex] [tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \sin^{2}\theta + \cos^{2}\theta = 1\end{gathered}$}[/tex]
Inserindo os dados na equação "I", simplificando e resolvendo, temos:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \bigg(\frac{1}{3}\bigg)^{2} + \cos^{2}\theta = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1}{9} + \cos^{2} \theta = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos^{2}\theta = 1 - \frac{1}{9}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\sqrt{1 - \frac{1}{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\sqrt{\frac{9 - 1}{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\sqrt{\frac{8}{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{\sqrt{8}}{\sqrt{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{\sqrt[\!\diagup\!]{2^{\!\diagup\!\!\!\!2}\cdot2}}{3}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
✅ Portanto, o valor do cosseno é:
[tex]\Large\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Prova:\:\:\:}}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
Temos:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \bigg(\frac{1}{3}\bigg)^{2} + \bigg(\frac{2\sqrt{2}}{3}\bigg)^{2} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1^{2}}{3^{2}} + \frac{(2\sqrt{2})^{2}}{3^{2}} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1}{9} + \frac{8}{9} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{9}{9} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt 1 = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = -\frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \bigg(\frac{1}{3}\bigg)^{2} + \bigg(-\frac{2\sqrt{2}}{3}\bigg)^{2} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1^{2}}{3^{2}} + \frac{(-2\sqrt{2})^{2}}{3^{2}} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}[/tex]
Saiba mais:
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✅ Após resolver os cálculos, concluímos que o valor do cosseno do ângulo "θ", a partir do seno valendo "1/3" é:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf \cos\theta = \pm\frac{2\sqrt{2}}{3}\:\:\:}}\end{gathered}$}[/tex]
Sejam os dados:
[tex]\LARGE\begin{cases}\tt \sin\theta = 1/3\\\tt\cos\theta = \:?\end{cases}[/tex]
Para resolver esta questão devemos nos lembrar da seguinte identidade trigonométrica fundamental, representada por:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\bf I \end{gathered}$}[/tex] [tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \sin^{2}\theta + \cos^{2}\theta = 1\end{gathered}$}[/tex]
Inserindo os dados na equação "I", simplificando e resolvendo, temos:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \bigg(\frac{1}{3}\bigg)^{2} + \cos^{2}\theta = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1}{9} + \cos^{2} \theta = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos^{2}\theta = 1 - \frac{1}{9}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\sqrt{1 - \frac{1}{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\sqrt{\frac{9 - 1}{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\sqrt{\frac{8}{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{\sqrt{8}}{\sqrt{9}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{\sqrt[\!\diagup\!]{2^{\!\diagup\!\!\!\!2}\cdot2}}{3}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
✅ Portanto, o valor do cosseno é:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \pm\frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Prova:\:\:\:}}}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = \frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
Temos:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \bigg(\frac{1}{3}\bigg)^{2} + \bigg(\frac{2\sqrt{2}}{3}\bigg)^{2} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1^{2}}{3^{2}} + \frac{(2\sqrt{2})^{2}}{3^{2}} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1}{9} + \frac{8}{9} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{9}{9} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt 1 = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \cos\theta = -\frac{2\sqrt{2}}{3}\end{gathered}$}[/tex]
Temos:
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \bigg(\frac{1}{3}\bigg)^{2} + \bigg(-\frac{2\sqrt{2}}{3}\bigg)^{2} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1^{2}}{3^{2}} + \frac{(-2\sqrt{2})^{2}}{3^{2}} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{1}{9} + \frac{8}{9} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt \frac{9}{9} = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered}\tt 1 = 1\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}[/tex]
Saiba mais: