Tính sin2x, cos2x, tan2x:  a) Biết sin x + cos x = 5/4 ( sin x > cosx) ​b) sin x – cos x = 2/3 , ( π/2 < x < π )

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1. $\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{a.\ sinx+cosx=\frac{5}{4} \Rightarrow sinx=\frac{5}{4} -cosx}\\ \mathrm{Ta\ có\ sin^{2} x+cos^{2} x=1}\\ \mathrm{\Leftrightarrow \left(\frac{5}{4} -cosx\right)^{2} +cos^{2} x=1}\\ \mathrm{\Leftrightarrow \ cosx=\frac{5+\sqrt{7}}{8} \Rightarrow sinx=\frac{5-\sqrt{7}}{8}}\\ \mathrm{\ \ \ \ \ hoặc:\ cosx=\frac{5-\sqrt{7}}{8} \Rightarrow sinx=\frac{5+\sqrt{7}}{8}( \ loại)}\\ \mathrm{sin2x=2sinxcosx=\frac{9}{16}}\\ \mathrm{cos2x=cos^{2} x-sin^{2} x=\frac{5\sqrt{7}}{16}}\\ \mathrm{tan2x=\frac{sin2x}{cos2x} =\frac{9\sqrt{7}}{35}}\\ \mathrm{b.\ sinx-cosx=\frac{2}{3} \Rightarrow sinx=\frac{2}{3} +cosx}\\ \mathrm{Do\ \frac{\pi }{2} < x< \pi \ \Rightarrow sinx< 0\ và\ cosx >0}\\ \mathrm{Ta\ có:\ sin^{2} x+cos^{2} x=1}\\ \mathrm{\Leftrightarrow \left(\frac{2}{3} +cosx\right)^{2} +cos^{2} x=1}\\ \mathrm{\Leftrightarrow cosx=\frac{-2+\sqrt{14}}{6} \ do\ cosx >0}\\ \mathrm{\Rightarrow sinx=\frac{2+\sqrt{14}}{6} \ không\ thoả\ mãn}\\ \end{array}$

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