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Đề bài

Cho $\alpha$ thoả mãn $\sin \alpha =\frac{3}{5}$. Tính cos$\alpha$, tan$\alpha$, cot$\alpha$, sin(90° - $\alpha$), cos(90° - $\alpha$), sin(180° – $\alpha$),

cos(180° – $\alpha$) trong các trường hợp sau:

a) 0° < $\alpha$ < 90°

b) 90° < $\alpha$ < 180°

Lời giải chi tiết

a) Theo giả thiết, 0° < $\alpha$ < 90° $\Rightarrow \cos \alpha >0,\tan \alpha >0,\cot \alpha >0$

+ Ta có: ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\cos }^2\alpha =1-{\sin }^2\alpha =1-{\left(\frac{3}{5}\right)}^2=\frac{16}{25}$ $\Rightarrow \cos \alpha =\frac{4}{5}$

+ $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{3}{4};\cot \alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{4}{3}$

+ $\sin ({90}^0-\alpha )=\cos \alpha =\frac{4}{5};\cos ({90}^0-\alpha )=\sin \alpha =\frac{3}{5}$

+ $\sin \left({180}^0-\alpha \right)=\sin \alpha =\frac{3}{5};\cos \left({180}^0-\alpha \right)=-\cos \alpha =-\frac{4}{5}$

b) Theo giả thiết, 90° < $\alpha$ < 180° $\Rightarrow \cos \alpha <0,\tan \alpha <0,\cot \alpha <0$

+ Ta có: ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\cos }^2\alpha =1-{\sin }^2\alpha =1-{\left(\frac{3}{5}\right)}^2=\frac{16}{25}$ $\Rightarrow \cos \alpha =-\frac{4}{5}$

+ $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=-\frac{3}{4};\cot \alpha =\frac{\cos \alpha }{\sin \alpha }=-\frac{4}{3}$

+ $\sin ({90}^0-\alpha )=\cos \alpha =-\frac{4}{5};\cos ({90}^0-\alpha )=\sin \alpha =\frac{3}{5}$

+ $\sin \left({180}^0-\alpha \right)=\sin \alpha =\frac{3}{5};\cos \left({180}^0-\alpha \right)=-\cos \alpha =\frac{4}{5}$