Toán Lớp 9: Bài 4: a) Biết sin ∝= $\frac{2}{3}$ . Tính A=2.sin^2 ∝+5.cos^2 ∝; B= tan^2 ∝-2cot^2 ∝ b) Biết cot B= $\frac{1}{3}$ . Tính P

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1. Giải đáp:

   $4)\\ a)A=\dfrac{11}{3}\\ B=-\dfrac{17}{10}\\ b)P=\dfrac{11}{2}.$

   Lời giải và giải thích chi tiết:

   $4)\\ a)\sin \alpha=\dfrac{2}{3}\\ A=2\sin^2 \alpha+5\cos^2 \alpha\\ =5\sin^2 \alpha+5\cos^2 \alpha-3\sin^2 \alpha\\ =5\left(\sin \alpha+\cos \alpha\right)-3\sin^2 \alpha\\ =5-3\sin^2 \alpha\\ =5-3 \left(\dfrac{2}{3}\right)^2\\ =\dfrac{11}{3}\\ B=\tan^2\alpha-2\cot^2\alpha\\ =\dfrac{\sin^2\alpha}{\cos^2\alpha}-2\dfrac{\cos^2\alpha}{\sin^2\alpha}\\ =\dfrac{\sin^2\alpha}{1-\sin^2\alpha}-2\dfrac{1-\sin^2\alpha}{\sin^2\alpha}\\ =\dfrac{\left(\dfrac{2}{3}\right)^2}{1-\left(\dfrac{2}{3}\right)^2}-2\dfrac{1-\left(\dfrac{2}{3}\right)^2}{\left(\dfrac{2}{3}\right)^2}\\ =-\dfrac{17}{10}\\ b)P=\dfrac{3\sin \widehat{B}+2\cos \widehat{B}}{\sin \widehat{B} – \cos \widehat{B}}\\ =\dfrac{3+2\dfrac{\cos \widehat{B}}{\sin \widehat{B}}}{1 – \dfrac{\cos \widehat{B}}{\sin \widehat{B}}}\\ =\dfrac{3+2\cot \widehat{B}}{1 – \cot \widehat{B}}\\ =\dfrac{3+2.\dfrac{1}{3}}{1 – \dfrac{1}{3}}\\ =\dfrac{11}{2}.$

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