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\(\sin^2x=\sqrt{1-\left(-\dfrac{4}{5}\right)^2}=\dfrac{9}{25}\)
mà \(\sin x>0\)
nên \(\sin x=\dfrac{3}{5}\)
=>\(\tan x=-\dfrac{3}{4}\)
\(\Leftrightarrow\cot x=-\dfrac{4}{3}\)
\(\cos^2x=\sqrt{1-\dfrac{9}{25}}=\dfrac{16}{25}\)
mà \(\cos x< 0\)
nên \(\cos x=-\dfrac{4}{5}\)
=>\(\tan x=-\dfrac{3}{4};\cot x=-\dfrac{4}{3}\)
a.Ta có : \(x\in\left(\pi;\dfrac{3}{2}\pi\right)\Rightarrow cosx< 0\)
\(cosx=-\sqrt{1-sin^2x}=-\sqrt{1-0,8^2}=-0,6\)
\(tanx=\dfrac{4}{3};cotx=\dfrac{3}{4}\)
b. cos 2x = \(cos^2x-sin^2x=0,6^2-0,8^2=-0,28\)
\(P=2.cos2x=-0,56\)
\(Q=tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{tan2x+tan\dfrac{\pi}{3}}{1-tan2x.tan\dfrac{\pi}{3}}=\dfrac{tan2x+\sqrt{3}}{1-tan2x.\sqrt{3}}\)
tan 2x = \(\dfrac{2tanx}{1-tan^2x}=\dfrac{\dfrac{2.4}{3}}{1-\left(\dfrac{4}{3}\right)^2}=\dfrac{-24}{7}\)
\(Q=\dfrac{-\dfrac{24}{7}+\sqrt{3}}{1+\dfrac{24}{7}.\sqrt{3}}\) \(=\dfrac{-24+7\sqrt{3}}{7+24\sqrt{3}}\)
b)\(P=cos2a-cos(\dfrac{\pi}{3}-a) \\=2cos^2a-1-cos\dfrac{\pi}{3}cosa-sin\dfrac{\pi}{3}sina \\=2.(\dfrac{-2}{5})^2-1-\dfrac{1}{2}.\dfrac{-2}{5}-\dfrac{\sqrt3}{2}.\dfrac{-\sqrt{21}}{5} \\=\dfrac{-24+15\sqrt7}{50}\)
a, Vì : \(\pi< a< \dfrac{3\pi}{2}\) nên \(cos\alpha< 0\) mà \(cos^2\alpha=1-sin^2\alpha=1-\dfrac{4}{25}=\dfrac{21}{25},\)
do đó : \(cos\alpha=-\dfrac{\sqrt{21}}{5}\)
từ đó suy ra : \(tan\alpha=\dfrac{2}{\sqrt{21}},cot\alpha=\dfrac{\sqrt{21}}{2}\)
Em 2k8 ms học nên k chắc
Vì 0 < \(\alpha< \dfrac{\pi}{2}\) => sin \(\alpha>0\)
Cos \(\alpha=\dfrac{1}{3}\) \(\Rightarrow sin\alpha=\sqrt{1-\dfrac{1}{9}}=\dfrac{2\sqrt{2}}{3}\)
tan \(\alpha=2\sqrt{2}\) ; cot \(\alpha=\dfrac{1}{2\sqrt{2}}\)
\(0< a< \dfrac{\pi}{2}\Rightarrow0< \dfrac{a}{2}< \dfrac{\pi}{4}\Rightarrow sin\dfrac{a}{2}>0\)
\(\Rightarrow sin\dfrac{a}{2}=\sqrt{1-cos^2\dfrac{a}{2}}=\dfrac{3}{5}\)
\(sina=2sin\dfrac{a}{2}cos\dfrac{a}{2}=2.\left(\dfrac{4}{5}\right)\left(\dfrac{3}{5}\right)=\dfrac{24}{25}\)
\(cosa=\pm\sqrt{1-sin^2a}=\pm\dfrac{7}{25}\)
\(tana=\dfrac{sina}{cosa}=\pm\dfrac{24}{7}\)
\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)
\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)
\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
\(\left\{{}\begin{matrix}tan\alpha=-\dfrac{7}{3}\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=-\dfrac{7}{3}\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\sin^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\\dfrac{49}{9}cos^2\alpha+cos^2\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\cos^2\alpha=\dfrac{9}{58}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{3}cos\alpha\\cos\alpha=\dfrac{3}{\sqrt{58}}\end{matrix}\right.\) (Vì \(\dfrac{3\pi}{2}< \alpha< 2\pi\Rightarrow cos\alpha>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=-\dfrac{7}{\sqrt{58}}\\cos\alpha=\dfrac{3}{\sqrt{58}}\end{matrix}\right.\)
\(cot\alpha=\dfrac{1}{tan\alpha}=-\dfrac{3}{7}\)
\(\dfrac{\pi}{2}< x< \pi\Rightarrow cosx< 0\)
\(\Rightarrow cosx=-\sqrt{1-sin^2x}=-\dfrac{20}{29}\)
\(tanx=\dfrac{sinx}{cosx}=-\dfrac{21}{20}\)
\(cotx=\dfrac{1}{tanx}=-\dfrac{20}{21}\)