ITA 1963 ⇒ Questão 01 - ITA-1963 Tópico resolvido

[petrasMOD]

Calcular sen 105^o

Resposta

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Resposta:[tex3]\frac{\sqrt6+\sqrt2}{4}[/tex3]

[petrasMOD]

[tex3]\mathsf{
sen (A + B) = sen A cos B + cos Asen B\\
sen105^∘=sen(60^∘+45^∘)=sen60^∘cos45^∘+cos60^∘sen45^∘=\frac{\sqrt3}{2}.\frac{\sqrt2}{2}+\frac{1}{2}.\frac{\sqrt2}{2}\\
\therefore \boxed{sen105^o = \frac{\sqrt6+\sqrt2}{4}}


}
[/tex3]