Question

(Fuvest - sp) o valor de (sen 22°30¹ + cos 22°30¹)² é

Answer 1

$\boxed{\boxed{1 \ grau \ = \ 60 \ minutos}} \\ \\ \boxed{\boxed{1 \ minuto \ = \ 60 \ segundos}}$


$Nota\c{c}\~ao \ angular \ : \ g^\circ \ m' \ s''\ (g \ : \ grau; \ m \ : \ minuto; \ s \ : \ segundo)$

$Sendo \ o \ \^angulo \ \alpha \ = \ \underbrace{22^\circ}_{graus} \ \underbrace{30'}_{minutos} \ e \ seja \ \beta \ = \ 2 \ \cdot \ \alpha \ \rightarrow \\ \\ \beta \ = \ 2 \ \cdot \ 22^\circ \ 30' \ \rightarrow \\ \\ \beta \ = \ 22^\circ \ 30' \ + \ 22^\circ \ 30' \ \rightarrow \\ \\ Vamos \ somar \ angularmente \ : \ grau \ com \ grau, \etc \ \dots \ \righatrrow \\ \\ \beta \ = \ 44^\circ \ + \ \underbrace{60'}_{= \ 1^\circ} \ \rightarrow \\ \\ \beta \ = \ 44^\circ \ + \ 1^\circ \ \rightarrow$

$\boxed{\boxed{\beta \ = \ 45^\circ}}$

$\beta \ \'e \ o \ \bold{arco \ duplo} \ de \ \alpha \ e \ tem \ as \ seguintes \ propriedades \ : \\ \\ sen(\beta) \ \rightarrow \ sen(\alpha \ + \ \alpha) \ = \ \boxed{2 \ \cdot \ sen(\alpha) \ \cdot \ cos(\alpha)} \\ \\ cos(\beta) \ \rightarrow \ cos(\alpha \ + \ \alpha) \ = \ \boxed{cos^2(\alpha) \ - \ sen^2(\alpha)}$

$sen(\beta) \ = \ \underbrace{sen(45^\circ)}_{\^angulo \ not\'avel} \ = \ \boxed{\frac{\sqrt{2}}{2}}$

$Observe \ \rightarrow \\ \\ sen(\beta) \ = \ sen(2 \ \cdot \ \alpha) \ \rightarrow \\ \\ sen(\beta) \ = \ sen(\alpha \ + \ \alpha) \ \rightarrow \\ \\ sen(\beta) \ = \ 2 \ \cdot \ sen(\alpha) \ \cdot \ cos(\alpha) \ \rightarrow \\ \\ \boxed{\boxed{2 \ \cdot \ sen(\alpha) \ \cdot \ cos(\alpha) \ = \ \frac{\sqrt{2}}{2}}}$

$Agora, \ desenvolvendo \ por \ \bold{produto \ not\'avel} \ \rightarrow \\ \\ (sen(\alpha) \ + \ cos(\alpha))^2 \ \rightarrow \\ \\ sen^2(\alpha) \ + \ 2 \ \cdot \ sen(\alpha) \ \cdot \ cos(\alpha) \ + \ cos^2(\alpha) \ \rightarrow \\ \\ \boxed{\boxed{Identidade \ Fundamental \ \longrightarrow \ sen^2(\alpha) \ + \ cos^2(\alpha) \ = \ 1)}} \\ \\ \\ 1 \ + \ \underbrace{2 \ \cdot \ sen(\alpha) \ \cdot \ cos(\alpha)}_{= \ \frac{\sqrt{2}}{2}} \ \rightarrow \\ \\ \\ 1 \ + \ \frac{\sqrt{2}}{2} \ \rightarrow$

$\frac{2}{2} \ + \ \frac{\sqrt{2}}{2} \ \rightarrow \\ \\ \frac{2 \ + \ \sqrt{2}}{2} \ \rightarrow \\ \\ \frac{\sqrt{2} \ \cdot \ \sqrt{2} \ + \ \sqrt{2}}{2} \ \rightarrow \\ \\ \boxed{\boxed{\frac{\sqrt{2} \ \cdot \ (\sqrt{2} \ + \ 1)}{2}{}}}$