Câu hỏi của Thầy Tùng Dương

Question

a) Chosin $\alpha=\dfrac{1}{3}$ với $90^{\circ}<\alpha<180^{\circ}$. Tính $\cos \alpha$ và $	an \alpha$.

b) Cho $\cos \alpha=-\dfrac{2}{3}$. Tính $\sin \alpha$ và $\cot \alpha$.

c) Cho $	an \gamma...

Cao Thị Kim Ngân · Accepted Answer

a) Vì 90^{\circ}<\alpha<180^{\circ}90∘<α<180∘ nên \cos \alpha<0cosα<0 mặt khác \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 suy ra \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\dfrac{1}{9}}=-\dfrac{2 \sqrt{2}}{3}cosα=−1−sin2α​=−1−91​​=−322​​.

Do đó 	an \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{1}{3}}{-\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}tanα=cosαsinα​=−322​​31​​=−22​1​.

b) Vì \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 nên \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}sinα=1−cos2α​=1−94​​=35​​ và \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}cotα=sinαcosα​=35​​−32​​=−5​2​.

c) Vì 	an \gamma=-2 \sqrt{2}<0 \Rightarrow \cos \alpha<0tanγ=−22​<0⇒cosα<0 mặt khác 	an ^{2} \alpha+1=\dfrac{1}{\cos ^{2} \alpha}tan2α+1=cos2α1​ nên \cos \alpha=-\sqrt{\dfrac{1}{	an ^{2}+1}}=-\sqrt{\dfrac{1}{8+1}}=-\dfrac{1}{3}cosα=−tan2+11​​=−8+11​​=−31​.
Ta có 	an \alpha=\dfrac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha=	an \alpha \cdot \cos \alpha=-2 \sqrt{2} \cdot\left(-\dfrac{1}{3}ight)=\dfrac{2 \sqrt{2}}{3}tanα=cosαsinα​⇒sinα=tanα⋅cosα=−22​⋅(−31​)=322​​ \Rightarrow \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{1}{3}}{\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}⇒cotα=sinαcosα​=322​​−31​​=−22​1​.Câu trả lời: a) Vì 90^{\circ}<\alpha<180^{\circ}90∘<α<180∘ nên \cos \alpha<0cosα<0 mặt khác \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 suy ra \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\dfrac{1}{9}}=-\dfrac{2 \sqrt{2}}{3}cosα=−1−sin2α​=−1−91​​=−322​​.

Do đó 	an \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{1}{3}}{-\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}tanα=cosαsinα​=−322​​31​​=−22​1​.

b) Vì \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 nên \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}sinα=1−cos2α​=1−94​​=35​​ và \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}cotα=sinαcosα​=35​​−32​​=−5​2​.

c) Vì 	an \gamma=-2 \sqrt{2}<0 \Rightarrow \cos \alpha<0tanγ=−22​<0⇒cosα<0 mặt khác 	an ^{2} \alpha+1=\dfrac{1}{\cos ^{2} \alpha}tan2α+1=cos2α1​ nên \cos \alpha=-\sqrt{\dfrac{1}{	an ^{2}+1}}=-\sqrt{\dfrac{1}{8+1}}=-\dfrac{1}{3}cosα=−tan2+11​​=−8+11​​=−31​.
Ta có 	an \alpha=\dfrac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha=	an \alpha \cdot \cos \alpha=-2 \sqrt{2} \cdot\left(-\dfrac{1}{3}ight)=\dfrac{2 \sqrt{2}}{3}tanα=cosαsinα​⇒sinα=tanα⋅cosα=−22​⋅(−31​)=322​​ \Rightarrow \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{1}{3}}{\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}⇒cotα=sinαcosα​=322​​−31​​=−22​1​.

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