Question

Se 3pi/2 < alfa< 2pi e cos alfa= 0,2, qual é o valor de tg alfa?

Answer 1

Olá Jonh.


Propriedades trigonométrica usadas:


$\boxed{\boxed{\mathsf{cos^2x+sen^2x=1}}}\\\\\\\\\boxed{\boxed{\mathsf{tg~x=\dfrac{sen~x}{cos~x}}}}$


Resolvendo.


$\mathsf{\dfrac{3\pi}{2} \ \textless \ \alpha\ \textless \ 2\pi}\\\\\\\mathsf{cos~\alpha=\dfrac{1}{5}}\\\\\\\mathsf{cos^2\alpha+sen^2\alpha=1}\\\\\mathsf{\Big(\dfrac{1}{5}\Big)^2+sen^2\alpha=1}\\\\\mathsf{\Big(\dfrac{1}{25}\Big)+sen^2\alpha=1}\\\\\mathsf{sen^2\alpha=1-\dfrac{1}{25}}\\\\\mathsf{sen~\alpha=\pm\sqrt{\dfrac{24}{25}}}\\\\\mathsf{sen~\alpha=-\dfrac{\sqrt{24}}{5}}$


Acima consideramos o seno sendo negativo, pois o intervalo de $\mathsf{\alpha}$ é do quarto quadrante, onde o seno é negativo.


$\mathsf{tg~\alpha=\dfrac{sen~\alpha}{cos~\alpha}}\\\\\\\mathsf{tg~\alpha=\dfrac{-\dfrac{\sqrt{24}}{5}}{\dfrac{1}{5}}}\\\\\\\mathsf{tg~\alpha=-\dfrac{\sqrt{24}}{\diagup\!\!\!\!5}\cdot \dfrac{\diagup\!\!\!\!5}{1}}\\\\\\\boxed{\mathsf{tg~\alpha=-\sqrt{24}}}$


Dúvidas? comente.

Answer 2

cosα =2/10 => cosα = 1/5 => secα = 5
α ∈ 4° quadrante, então tgα < 0.

tg² + 1 = sec²α

tg²α + 1 = 5²

tg²α = 25 -1

tg²α = 24

tgα = 2√6 (não serve) ou

tgα = -2√6