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a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1tanα+3tanα1=tan2α+1tan2α+3=cos2α1cos2α1+2=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169=817.
b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α+cos3α3cos3α+cos3α2sinαcos3αsinα−cos3αcosα=tan3α+3+2tanα(tan2α+1)tanα(tan2α+1)−(tan2α+1).
Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}B=22+3+22(2+1)2(2+1)−(2+1)=3+823(2−1).
vậy thì kết quả là
\(\sin2\alpha=-0.96\)
\(\)còn \(\cos\left(\alpha+\frac{\pi}{6}\right)\) thì đúng vì -(-0.8) mà sorry thiếu ngủ hôm qua -_-
a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(tan\alpha,cot\alpha>0\) và \(sin\alpha,cos\alpha< 0\).
\(\left\{{}\begin{matrix}tan\alpha-3cot\alpha=6\\tan\alpha cot\alpha=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\\left(6+3cot\alpha\right)cot\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\3cot^2\alpha+6cot\alpha-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\cot\alpha=\dfrac{-3+2\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=3+2\sqrt{3}\\cot\alpha=\dfrac{-3+2\sqrt{3}}{3}\end{matrix}\right.\).
Có \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{tan^2\alpha+1}\).
Có thể đề sai.
Ý bạn là \(\pi< a< \frac{3\pi}{2}\) và tìm \(cosa,tana,cota\)?
Khi đó \(cosa< 0\) \(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{12}{13}\)
\(tana=\frac{sina}{cosa}=\frac{5}{12}\)
\(cota=\frac{1}{tana}=\frac{12}{5}\)
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(tan\alpha< 0,cot\alpha< 0;cos\alpha< 0\).
Vì vậy: \(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{7}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{4}:\dfrac{-\sqrt{7}}{4}=\dfrac{-3}{\sqrt{7}}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-\sqrt{7}}{3}\).
\(A=\dfrac{2tan\alpha-3cot\alpha}{cos\alpha+tan\alpha}\)\(=\dfrac{2.\dfrac{-3}{\sqrt{7}}-3.\dfrac{-\sqrt{7}}{3}}{\dfrac{-\sqrt{7}}{4}+\dfrac{-3}{\sqrt{7}}}\)
\(=\dfrac{\dfrac{-6}{\sqrt{7}}+\sqrt{7}}{\dfrac{-7-12}{4\sqrt{7}}}\)\(=\dfrac{\dfrac{-6+7}{\sqrt{7}}.4\sqrt{7}}{-19}\)\(=\dfrac{\dfrac{1}{\sqrt{7}}.4\sqrt{7}}{-19}=-\dfrac{4}{19}\).
b) \(\dfrac{cos^2\alpha+cot^2\alpha}{tan\alpha-cot\alpha}=\dfrac{\left(-\dfrac{\sqrt{7}}{4}\right)^2+\left(\dfrac{-\sqrt{7}}{3}\right)^2}{\dfrac{-3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{\dfrac{7}{16}+\dfrac{7}{9}}{\dfrac{-9+7}{3\sqrt{7}}}=\dfrac{\dfrac{175}{144}}{\dfrac{-2}{3\sqrt{7}}}=\dfrac{-175}{96\sqrt{7}}\).
Do \(\pi< a< \frac{3\pi}{2}\Rightarrow cosa< 0\)
\(cosa=-\sqrt{1-sin^2a}=-\sqrt{1-0,6^2}=-\frac{4}{5}\)
\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)
\(cota=\frac{1}{tana}=-\frac{4}{3}\)
\(0< a< \frac{\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa>0\end{matrix}\right.\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cos^2a=\frac{1}{1+tan^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}\)
\(\Rightarrow cosa=\frac{1}{2}\Rightarrow sina=cosa.tana=\frac{\sqrt{3}}{2}\)
\(cos2a=2cos^2a-1=-\frac{1}{2}\)
\(sin2a=2sina.cosa=\frac{\sqrt{3}}{2}\)
\(\Rightarrow sin\left(2a-\frac{\pi}{3}\right)=sin2a.cos\frac{\pi}{3}-cos2a.sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)
\(tan\left(a+\frac{\pi}{4}\right)=\frac{tana+tan\frac{\pi}{4}}{1-tana.tan\frac{\pi}{4}}=-2-\sqrt{3}\)
a) Vì 90^{\circ}<\alpha<180^{\circ}90∘<α<180∘ nên \cos \alpha<0cosα<0 mặt khác \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 suy ra \cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\dfrac{1}{9}}=-\dfrac{2 \sqrt{2}}{3}cosα=−1−sin2α=−1−91=−322.
Do đó \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{1}{3}}{-\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}tanα=cosαsinα=−32231=−221.
b) Vì \sin ^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1 nên \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}sinα=1−cos2α=1−94=35 và \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}cotα=sinαcosα=35−32=−52.
c) Vì \tan \gamma=-2 \sqrt{2}<0 \Rightarrow \cos \alpha<0tanγ=−22<0⇒cosα<0 mặt khác \tan ^{2} \alpha+1=\dfrac{1}{\cos ^{2} \alpha}tan2α+1=cos2α1 nên \cos \alpha=-\sqrt{\dfrac{1}{\tan ^{2}+1}}=-\sqrt{\dfrac{1}{8+1}}=-\dfrac{1}{3}cosα=−tan2+11=−8+11=−31.
Ta có \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha=\tan \alpha \cdot \cos \alpha=-2 \sqrt{2} \cdot\left(-\dfrac{1}{3}\right)=\dfrac{2 \sqrt{2}}{3}tanα=cosαsinα⇒sinα=tanα⋅cosα=−22⋅(−31)=322 \Rightarrow \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{1}{3}}{\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}⇒cotα=sinαcosα=322−31=−221.
Sửa đề: cosa=3/5
3pi/2<a<2pi
=>sin a<0
\(sin^2a+cos^2a=1\)
=>\(sin^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)
mà sin a<0
nên sina =-4/5
tan a=-4/5:3/5=-4/3
cot a=1:(-4/3)=-3/4