Đáp án:
$a)cosx=-\frac{\sqrt{10}}{10}\\
sinx=\frac{3\sqrt{10}}{10}\\
cotx=\frac{-1}{3}\\
b) tanx=5\\
sinx=- \frac{5\sqrt{26}}{26}\\
cosx=\frac{\sqrt{26}}{26}\\
c) cosx=\frac{\sqrt{65}}{9}\\
tanx=\frac{4\sqrt{65}}{65}\\
cotx=\frac{\sqrt{65}}{4}\\
d) sinx=\frac{2\sqrt{6}}{7}\\
tanx=\frac{-2\sqrt{6}}{5}\\
cotx=\frac{-5\sqrt{6}}{12}\\$
Giải thích các bước giải:
$a) 1+tan^2x=\frac{1}{cos^2x}\\
\Rightarrow \frac{1}{cos^2x}=1+(-3)^2=10\\
\Rightarrow cos^2x=\frac{1}{10}\\
\Rightarrow cosx=\pm \frac{\sqrt{10}}{10}$
Vì $\frac{\pi}{2}<\alpha <\pi$ nên $cos=-\frac{\sqrt{10}}{10}$
$tanx=\frac{sinx}{cosx}=-3\\
\Rightarrow sinx=-3.\frac{-\sqrt{10}}{10}=\frac{3\sqrt{10}}{10}\\
cotx=\frac{1}{tanx}=\frac{-1}{3}\\
b) tanx=\frac{1}{cotx}=\frac{1}{\frac{1}{5}} =5\\
cot^2x+1=\frac{1}{sin^2x}\\
\Rightarrow \frac{1}{sin^2x}=1+(\frac{1}{5})^2=\frac{26}{25}\\
\Rightarrow sin^2x=\frac{25}{26}\\
\Rightarrow sinx=\pm \sqrt{\frac{25}{26}}=\pm \frac{5\sqrt{26}}{26}$
Vì $\pi < \alpha >\frac{3\pi}{2}$
nên $sinx=-\frac{5\sqrt{26}}{16}\\
cotx=\frac{cosx}{sinx}\Rightarrow cosx=cotx.sinx=\frac{1}{5}.\frac{5\sqrt{26}}{26}=\frac{\sqrt{26}}{26}\\
c) sin^2x+cos^2x=1\Rightarrow cos^2x=1-sin^2x=1-\frac{4}{9}^2=\frac{65}{81}\\
\Rightarrow cosx=\pm \frac{\sqrt{65}}{9}$
Vì $0<\alpha <\frac{\pi}{2}$ nên $cosx=\frac{\sqrt{65}}{9}$
$tanx=\frac{sinx}{cosx}=\frac{\frac{4}{9}}{\frac{\sqrt{65}}{9}}=\frac{4\sqrt{65}}{65}\\
cotx=\frac{1}{tanx}=\frac{\sqrt{65}}{4}\\
d) sin^2x+cos^2x=1\Rightarrow sin^2x=1-cos^2x=1-\left (\frac{-5}{7} \right )^2=\frac{24}{49}\\
\Rightarrow sinx=\pm \frac{2\sqrt{6}}{7}$
Vì $\frac{\pi}{2}<\alpha <\pi$ nên $sinx=\frac{2\sqrt{6}}{7}$
$tanx=\frac{sinx}{cosx}=\frac{\frac{2\sqrt{6}}{7}}{\frac{-5}{7}}=\frac{-2\sqrt{6}}{5}\\
cotx=\frac{1}{tanx}=\frac{-5\sqrt{6}}{12}\\$
