Cho góc tù α có sin α = $\frac{1}{3}$ .

a) Tính cos α, tan α, cot α

b) Tính giá trị biểu thức:

A = sinα . cot (180° – α) + cos(180° – α). cot (90°  – α)

B = $\frac{3\left(\sin \alpha +\sqrt{2}\cos \alpha \right)-2}{\sin \alpha -\sqrt{2}\cos \alpha }$ .

a) Vì α là góc tù nên cos α < 0

Ta có: sin²α + cos²α = 1

Suy ra: cos²α = 1 – sin²α = $1-\frac{1}{9}=\frac{8}{9}$

cos α = $\pm \sqrt{\frac{8}{9}}=\pm \frac{2\sqrt{2}}{3}$

Vì 90° < α < 180° nên cos α = $-\frac{2\sqrt{2}}{3}$

tan α = $\frac{\sin \alpha }{\cos \alpha }=\frac{1}{3}:\frac{-2\sqrt{2}}{3}=\frac{-1}{2\sqrt{2}}$

cot α =  $\frac{1}{\tan \alpha }=-2\sqrt{2}$

b) Ta có:

cot(180°– α) = – cotα

cos(180°– α) = – cosα

cot(90°– α) = tanα

Khi đó: A = sinα . cot (180° – α) + cos(180° – α). cot (90°  – α)

= sinα . (–cotα) + (–cosα). tanα = $\frac{1}{3}.2\sqrt{2}+\frac{2\sqrt{2}}{3}.\frac{-1}{2\sqrt{2}}=\frac{2\sqrt{2}}{3}-\frac{1}{3}=\frac{2\sqrt{2}-1}{3}$

B = $\frac{3\left(\sin \alpha +\sqrt{2}\cos \alpha \right)-2}{\sin \alpha -\sqrt{2}\cos \alpha }$

= $\frac{3.\left(\frac{1}{3}+\sqrt{2}.\frac{-2\sqrt{2}}{3}\right)-2}{\frac{1}{3}-\sqrt{2}.\frac{-2\sqrt{2}}{3}}=\frac{1-4-2}{\frac{1}{3}+\frac{4}{3}}=-3$.