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Análise o ciclo trigonométrico abaixo e determine, cos α + 3.sen α

Gu1me,

Círculo trigonométrico: raio = 1

[tex3]tg\alpha = \frac{1}{2} = \frac{sen\alpha}{cos\alpha} \therefore cos \alpha = 2sen\alpha\\
sen^2\alpha+cos^2\alpha =1 \implies sen^2 \alpha+(2sen\alpha)^2 = 1 \therefore sen\alpha = \frac{\sqrt5}{5}\\
\therefore cos \alpha+3sen \alpha = 2se\alpha+3sen\alpha = 5sen\alpha = 5\frac{\sqrt5}{5} = \boxed{\sqrt5} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]tg\alpha = \frac{1}{2} = \frac{sen\alpha}{cos\alpha} \therefore cos \alpha = 2sen\alpha\\
sen^2\alpha+cos^2\alpha =1 \implies sen^2 \alpha+(2sen\alpha)^2 = 1 \therefore sen\alpha = \frac{\sqrt5}{5}\\
\therefore cos \alpha+3sen \alpha = 2se\alpha+3sen\alpha = 5sen\alpha = 5\frac{\sqrt5}{5} = \boxed{\sqrt5} [/tex3]