Alguém para ajudar?
Anexos:
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Soluções para a tarefa
Respondido por
1
Se
, então
é um arco do 2º quadrante.

a)

Se
, então
é um arco do 2º quadrante, e seu seno é positivo. Portanto,

b)

a)
Se
b)
Micax:
Muito obrigada Lukyo *-*
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