Questão 01 - ITA-1963
Enviado: 01 Jul 2024, 12:26
Calcular sen 105o
Resposta:[tex3]\frac{\sqrt6+\sqrt2}{4}[/tex3]
Resposta
Resposta:[tex3]\frac{\sqrt6+\sqrt2}{4}[/tex3]
[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]