Đáp án:
Giải thích các bước giải:
a) `cos\ \alpha=5/13`
Ta có: `sin^2 \alpha+cos^2 \alpha=1`
`⇒ sin\ \alpha=\sqrt{1-(5/13)^2}=\frac{12}{13}`
`tan\ \alpha=\frac{sin\ \alpha}{cos\ \alpha}=\frac{12}{13}:\frac{5}{13}=12/5`
`cot\ \alpha=\frac{cos\ \alpha}{sin\ \alpha}=\frac{5}{13}:\frac{12}{13}=5/12`
b) `sin\ \alpha=1/2`
Ta có: `sin^2 \alpha+cos^2 \alpha=1`
`⇒ cos\ \alpha=\sqrt{1-(1/2)^2}=\frac{\sqrt{3}}{2}`
`tan\ \alpha=\frac{sin\ \alpha}{cos\ \alpha}=\frac{1}{2}:\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}`
`cot\ \alpha=\frac{cos\ \alpha}{sin\ \alpha}=\frac{\sqrt{3}}{2}:\frac{1}{2}=\sqrt{3}`
c) `tan\ \alpha=3`
Ta có: `cot\ \alpha=\frac{1}{tan\ \alpha}`
`⇒ cot\ \alpha=1/3`
`1+tan^2 \ alpha=\frac{1}{cos^2 \alpha}`
`⇔ 1+9=\frac{1}{cos^2 \alpha}`
`⇒ cos\ \alpha=\sqrt{\frac{1}{10}}=\frac{\sqrt{10}}{10}`
`sin^2 \alpha+cos^2 \alpha=1`
`⇒ sin\ \alpha=\sqrt{1-cos^2 \alpha}=\frac{3\sqrt{10}}{10}`
