Como eu fiz:
Sabemos:
[tex3]\cos(\alpha + \beta) = 0 \\
\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sen(\alpha)\sen(\beta) \\
\cos(\alpha) \cos(\beta) - \sen(\alpha)\sen(\beta) = 0 \implies \boxed{{\color{blue}\cos(\alpha) \cos(\beta)} = \sen(\alpha)\sen(\beta)}[/tex3]
O requerido é isso:
[tex3]\sen(\alpha + 2 \beta) = ? \\
\sen(\alpha + 2 \beta) = \sen(\alpha) \cos(2 \beta) + \sen(2 \beta) \cos(\alpha) \\
\sen(\alpha + 2 \beta) = \sen(\alpha) [\cos^2(\beta) - \sen^2(\beta)] + 2 \sen(\beta) {\color{blue}\cos(\beta) \cos(\alpha)} \\
\sen(\alpha + 2 \beta) = \sen(\alpha) [\cos^2(\beta) - \sen^2(\beta)] + 2 \sen(\beta) {\color{blue}\sen(\alpha) \sen(\beta)} \\
\sen(\alpha + 2 \beta) = \sen(\alpha) \cdot [\sen^2(\beta) + \cos^2(\beta)] \\
\sen(\alpha + 2 \beta) = \sen(\alpha)[/tex3]
Poliedro — Se cos(a + b) = 0, então sen(a + 2b) pode ser igual a:
(a) sen a * cos b
(b) cos b
(c) cos a
(d) sen b
(e) sen a
Resposta
b
tags: alfa beta