Bài tập 1 trang 14 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) Ta có ${\text{cos}}^2\alpha =1-{\text{sin}}^2\alpha =1-{\left(-\frac{4}{5}\right)}^2=\frac{9}{25}$.

Vì $\pi <\alpha <\frac{3\pi }{2};$ nên cosα < 0.

Do đó $\text{cos}\alpha =-\frac{3}{5}.$

Suy ra $\text{tan}\alpha =\frac{\text{sin}\alpha }{\text{cos}\alpha }=\frac{-\frac{4}{5}}{-\frac{3}{5}}=\frac{4}{3}$ và $\text{cot}\alpha =\frac{1}{\text{tan}\alpha }=\frac{1}{\frac{4}{3}}=\frac{3}{4}.$

b) Ta có ${\text{sin}}^2\alpha =1-{\cos }^2\alpha =1-{\left(\frac{11}{61}\right)}^2={\left(\frac{60}{61}\right)}^2$.

Vì $0<\alpha <\frac{\pi }{2};$ nên sinα > 0.

Do đó $\text{sin}\alpha =\frac{60}{61}.$

Suy ra $\text{tan}\alpha =\frac{\text{sin}\alpha }{\text{cos}\alpha }=\frac{\frac{60}{61}}{\frac{11}{61}}=\frac{60}{11}$ và $\text{cot}\alpha =\frac{1}{\text{tan}\alpha }=\frac{1}{\frac{60}{11}}=\frac{11}{60}.$

c) Ta có $\text{cot}\alpha =\frac{1}{\text{tan}\alpha }=\frac{1}{-\frac{15}{8}}=-\frac{8}{15};\frac{1}{{\text{cos}}^2\alpha }=1+{\text{tan}}^2\alpha =1+{\left(-\frac{15}{8}\right)}^2=\frac{289}{64}$

Suy ra ${\text{cos}}^2\alpha =\frac{64}{289}$.

Vì $-90°<\alpha <90°;$ nên $\text{cos}\alpha >0$.

Do đó $\text{cos}\alpha =\frac{8}{17}.$

Suy ra $\text{sin}\alpha =\text{tan}\alpha \text{cos}\alpha =-\frac{15}{8}\cdot \frac{8}{17}=-\frac{15}{17}.$

d) Ta có $\text{tan}\alpha =\frac{1}{\text{cot}\alpha }=\frac{1}{-\mathrm{2,4}}=-\frac{5}{12};\frac{1}{{\text{sin}}^2\alpha }=1+{\text{cot}}^2\alpha =1+{\left(-\mathrm{2,4}\right)}^2=\frac{676}{100}$

Suy ra ${\text{sin}}^2\alpha =\frac{100}{676}$.

Vì ‒180° < α < 0° nên $\text{sin}\alpha <0$.

Do đó $\text{sin}\alpha =-\frac{5}{13}.$

Suy ra $\text{cos}\alpha =\text{cot}\alpha \text{sin}\alpha =-\mathrm{2,4}\cdot \left(-\frac{5}{13}\right)=\frac{12}{13}.$

-- Mod Toán 11 HỌC247