Câu hỏi của Nguyễn Phương Linh

Question

Giúp mình câu này nhé mọi người <3 Kamsamita

Lê Thị Thục Hiền · Accepted Answer

Vì $\pi< \alpha< \dfrac{3\pi}{2}$ $\Rightarrow\dfrac{\pi}{2}< \dfrac{\alpha}{2}< \dfrac{3\pi}{4}$$\Rightarrow sin\dfrac{\alpha}{2}>0;cos\dfrac{\alpha}{2}< 0$$\pi< \alpha< \dfrac{3\pi}{2}\Rightarrow cos\alpha< 0$$\Rightarrow cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{3}{5}$Có $sin^2\dfrac{\alpha}{2}=\dfrac{1-cosa}{2}=\dfrac{4}{5}\Rightarrow sin\dfrac{\alpha}{2}=\sqrt{\dfrac{4}{5}}=\dfrac{2\sqrt{5}}{5}$$cos^2\dfrac{\alpha}{2}=\dfrac{1+cosa}{2}=\dfrac{1}{5}\Rightarrow cos\dfrac{\alpha}{2}=-\sqrt{\dfrac{1}{5}}=-\dfrac{\sqrt{5}}{5}$$tan\dfrac{\alpha}{2}=\dfrac{sin\dfrac{\alpha}{2}}{cos\dfrac{\alpha}{2}}=-2$$cot\dfrac{\alpha}{2}=-\dfrac{1}{2}$ Câu trả lời: Vì $\pi< \alpha< \dfrac{3\pi}{2}$ $\Rightarrow\dfrac{\pi}{2}< \dfrac{\alpha}{2}< \dfrac{3\pi}{4}$$\Rightarrow sin\dfrac{\alpha}{2}>0;cos\dfrac{\alpha}{2}< 0$$\pi< \alpha< \dfrac{3\pi}{2}\Rightarrow cos\alpha< 0$$\Rightarrow cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{3}{5}$Có $sin^2\dfrac{\alpha}{2}=\dfrac{1-cosa}{2}=\dfrac{4}{5}\Rightarrow sin\dfrac{\alpha}{2}=\sqrt{\dfrac{4}{5}}=\dfrac{2\sqrt{5}}{5}$$cos^2\dfrac{\alpha}{2}=\dfrac{1+cosa}{2}=\dfrac{1}{5}\Rightarrow cos\dfrac{\alpha}{2}=-\sqrt{\dfrac{1}{5}}=-\dfrac{\sqrt{5}}{5}$$tan\dfrac{\alpha}{2}=\dfrac{sin\dfrac{\alpha}{2}}{cos\dfrac{\alpha}{2}}=-2$$cot\dfrac{\alpha}{2}=-\dfrac{1}{2}$

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