`a) cos \alpha=-1/4; \pi<\alpha<(3\pi)/2`
Có: `sin^2\alpha+cos^2\alpha=1`
`<=> sin^2\alpha=1-cos^2\alpha`
`=> sin^2\alpha=1-(-1/4)^2=15/16`
`=> sin\alpha =-\sqrt{15}/4`
(do `\pi<\alpha<\(3\pi)/2=>sin\alpha<0`)
Khi đó $tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{-\sqrt{15}}{4}}{\dfrac{-1}{4}}=\sqrt{15}$
`-> cot \alpha=1/(tan\alpha)=1/\sqrt{15}`
Vậy `sin\alpha=-\sqrt{15}/4; tan\alpha=\sqrt{15};cot\alpha=1/\sqrt{15}`
`b) sin\alpha=1/3; 0<\alpha<\pi/2`
Có: `sin^2\alpha+cos^2\alpha=1`
`<=>cos^2\alpha=1-sin^2\alpha`
`=> cos^2\alpha=1-(1/3)^2=8/9`
`=> cos\alpha=(2\sqrt{2})/3`
(do `0<\alpha<\pi/2 => cos\alpha>0`)
Khi đó $tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{1}{3}}{\dfrac{2\sqrt{2}}{3}}=\dfrac{1}{2\sqrt{2}}$
`-> cot\alpha=1/(tan\alpha)=2\sqrt{2}`
Vậy `cos\alpha=(2\sqrt{2})/3;tan\alpha=1/(2\sqrt{2}); cot\alpha=2\sqrt{2}`
`c) tan\alpha=3/5; \pi/2<\alpha<\pi`
`=> cot\alpha=1/(tan\alpha)=5/3`
Lại có: `sin^2\alpha=1/(1+tan^2\alpha)=`$\dfrac{1}{1+(\dfrac{3}{5})^2}=\dfrac{1}{1+\dfrac{9}{25}}=\dfrac{1}{\dfrac{34}{25}}=\dfrac{25}{34}$
`=> sin\alpha=(5\sqrt{34})/34`
(do `\pi/2<\alpha<\pi=> sin\alpha>0`)
Khi đó `cos^2\alpha=1-sin^2\alpha=1-25/34=9/34`
`=> cos\alpha=(-3\sqrt{34})/34`
(do `\pi/2<\alpha<\pi=>cos\alpha<0`)
Vậy `sin\alpha=(5\sqrt{34})/34; cos\alpha=(-3\sqrt{34})/34; cot\alpha=5/3`
