Câu hỏi của Shinobu Kochou

Question

Nguyễn Huy Tú · Accepted Answer

a, $sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\Leftrightarrow cosx=\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=-\dfrac{3}{5}:\left(\dfrac{4}{5}\right)=-\dfrac{3}{4}$$cotx=-\dfrac{4}{3}$c, $sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\Leftrightarrow sinx=\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}$$cotx=\dfrac{3}{4}$b, $cos^2x+sin^2x=1\Leftrightarrow sin^2x=1-\dfrac{1}{16}=\dfrac{15}{16}\Leftrightarrow sinx=\dfrac{\sqrt{15}}{4}$$tanx=\dfrac{\sqrt{15}}{4}:\dfrac{1}{4}=\sqrt{15}$$cotx=\dfrac{1}{\sqrt{15}}$d, $sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{25}{169}=\dfrac{144}{169}\Leftrightarrow sinx=\dfrac{12}{13}$$tanx=\dfrac{12}{13}:\left(-\dfrac{5}{13}\right)=-\dfrac{12}{5}$$cotx=-\dfrac{5}{12}$Câu trả lời: a, $sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\Leftrightarrow cosx=\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=-\dfrac{3}{5}:\left(\dfrac{4}{5}\right)=-\dfrac{3}{4}$$cotx=-\dfrac{4}{3}$c, $sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\Leftrightarrow sinx=\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}$$cotx=\dfrac{3}{4}$b, $cos^2x+sin^2x=1\Leftrightarrow sin^2x=1-\dfrac{1}{16}=\dfrac{15}{16}\Leftrightarrow sinx=\dfrac{\sqrt{15}}{4}$$tanx=\dfrac{\sqrt{15}}{4}:\dfrac{1}{4}=\sqrt{15}$$cotx=\dfrac{1}{\sqrt{15}}$d, $sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{25}{169}=\dfrac{144}{169}\Leftrightarrow sinx=\dfrac{12}{13}$$tanx=\dfrac{12}{13}:\left(-\dfrac{5}{13}\right)=-\dfrac{12}{5}$$cotx=-\dfrac{5}{12}$

Nguyễn Lê Phước Thịnh · Answer

a: $\Omega< x< \dfrac{3}{2}\Omega$=>cosx<0Ta có: $sin^2x+cos^2x=1$=>$cos^2x=1-sin^2x=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}$mà cosx<0nên $cosx=-\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}$$cotx=\dfrac{1}{tanx}=\dfrac{4}{3}$b: $0< x< \dfrac{\Omega}{2}$=>sin x>0$sin^2x+cos^2x=1$=>$sin^2x=1-\left(\dfrac{1}{4}\right)^2=\dfrac{15}{16}$mà sin x>0nên $sinx=\dfrac{\sqrt{15}}{4}$$tanx=\dfrac{sinx}{cosx}=\dfrac{\sqrt{15}}{4}:\dfrac{1}{4}=\sqrt{15}$$cotx=\dfrac{1}{tanx}=\dfrac{1}{\sqrt{15}}=\dfrac{\sqrt{15}}{15}$c: 0sin x>0$sin^2x+cos^2x=1$=>$sin^2x=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}=\left(\dfrac{4}{5}\right)^2$mà sin x>0nên $sinx=\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}$$cotx=1:\dfrac{4}{3}=\dfrac{3}{4}$d: $180^0< x< 270^0$=>sin x<0$sin^2x+cos^2x=1$=>$sin^2x=1-\left(-\dfrac{5}{13}\right)^2=1-\dfrac{25}{169}=\dfrac{144}{169}$mà sin x<0nên $sinx=-\dfrac{12}{13}$$tanx=\dfrac{sinx}{cosx}=\dfrac{-12}{13}:\dfrac{-5}{13}=\dfrac{12}{5}$$cotx=\dfrac{1}{tanx}=\dfrac{5}{12}$Câu trả lời: a: $\Omega< x< \dfrac{3}{2}\Omega$=>cosx<0Ta có: $sin^2x+cos^2x=1$=>$cos^2x=1-sin^2x=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}$mà cosx<0nên $cosx=-\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}$$cotx=\dfrac{1}{tanx}=\dfrac{4}{3}$b: $0< x< \dfrac{\Omega}{2}$=>sin x>0$sin^2x+cos^2x=1$=>$sin^2x=1-\left(\dfrac{1}{4}\right)^2=\dfrac{15}{16}$mà sin x>0nên $sinx=\dfrac{\sqrt{15}}{4}$$tanx=\dfrac{sinx}{cosx}=\dfrac{\sqrt{15}}{4}:\dfrac{1}{4}=\sqrt{15}$$cotx=\dfrac{1}{tanx}=\dfrac{1}{\sqrt{15}}=\dfrac{\sqrt{15}}{15}$c: 0sin x>0$sin^2x+cos^2x=1$=>$sin^2x=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}=\left(\dfrac{4}{5}\right)^2$mà sin x>0nên $sinx=\dfrac{4}{5}$$tanx=\dfrac{sinx}{cosx}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}$$cotx=1:\dfrac{4}{3}=\dfrac{3}{4}$d: $180^0< x< 270^0$=>sin x<0$sin^2x+cos^2x=1$=>$sin^2x=1-\left(-\dfrac{5}{13}\right)^2=1-\dfrac{25}{169}=\dfrac{144}{169}$mà sin x<0nên $sinx=-\dfrac{12}{13}$$tanx=\dfrac{sinx}{cosx}=\dfrac{-12}{13}:\dfrac{-5}{13}=\dfrac{12}{5}$$cotx=\dfrac{1}{tanx}=\dfrac{5}{12}$

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