sin a=12/13

cos^2a=1-(12/13)^2=25/169

=>cosa=5/13

tan a=12/13:5/13=12/5

cot a=1:12/5=5/12

sin b=căn 3/2

cos^2b=1-(căn 3/2)^2=1/4

=>cos b=1/2

tan b=căn 3/2:1/2=căn 3

cot b=1/căn 3

\(a,sin^2\alpha+cos^2\alpha=1\\ \Rightarrow cos\alpha=\pm\sqrt{1-sin^2\alpha}=\pm\sqrt{1-\left(\dfrac{\sqrt{3}}{3}\right)^2}=\pm\dfrac{\sqrt{6}}{3}\)

Vì \(0< \alpha< \dfrac{\pi}{2}\Rightarrow cos\alpha=\dfrac{\sqrt{6}}{3}\)

\(sin2\alpha=2sin\alpha cos\alpha=2\cdot\dfrac{\sqrt{3}}{3}\cdot\dfrac{\sqrt{6}}{3}=\dfrac{2\sqrt{2}}{3}\\ cos2\alpha=2cos^2\alpha-1=2\cdot\left(\dfrac{\sqrt{6}}{3}\right)^2-1=\dfrac{1}{3}\\ tan2\alpha=\dfrac{sin2\alpha}{cos2\alpha}=\dfrac{\dfrac{2\sqrt{2}}{3}}{\dfrac{1}{3}}=2\sqrt{2}\\ cot2\alpha=\dfrac{1}{tan2\alpha}=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{4}\)

\(b,sin^2\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}=1\\ \Rightarrow cos\dfrac{\alpha}{2}=\pm\sqrt{1-sin^2\dfrac{\alpha}{2}}=\pm\sqrt{1-\left(\dfrac{3}{4}\right)^2}=\pm\dfrac{\sqrt{7}}{4}\)

Vì \(\pi< \alpha< 2\pi\Rightarrow\dfrac{\pi}{2}< \dfrac{\alpha}{2}< \pi\Rightarrow cos\alpha=-\dfrac{\sqrt{7}}{4}\)

\(sin\alpha=2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}=2\cdot\dfrac{3}{4}\cdot\left(-\dfrac{\sqrt{7}}{4}\right)=-\dfrac{3\sqrt{7}}{8}\\ cos\alpha=2cos^2\dfrac{\alpha}{2}-1=2\cdot\left(-\dfrac{\sqrt{7}}{4}\right)^2-1=-\dfrac{1}{8}\\sin2\alpha=2sin\alpha cos\alpha=2\cdot\left(-\dfrac{3\sqrt{7}}{8}\right)\cdot\left(-\dfrac{1}{8}\right)=\dfrac{3\sqrt{7}}{32}\\ cos2\alpha=2cos^2\alpha-1=2\cdot\left(-\dfrac{1}{8}\right)^2-1=-\dfrac{31}{32}\\ tan2\alpha=\dfrac{sin2\alpha}{cos2\alpha}=\dfrac{\dfrac{3\sqrt{7}}{32}}{-\dfrac{31}{32}}=-\dfrac{3\sqrt{7}}{31}\\ cot2\alpha=\dfrac{1}{tan2\alpha}=\dfrac{1}{-\dfrac{3\sqrt{7}}{31}}=-\dfrac{31\sqrt{7}}{21}\)

có `cos α=1/2`

`=>cos^2 α=1/4`

Mà `cos^2 α +sin^2 α=1`

`=>1/4+sin^2 α=1`

`=>sin^2 α=1-1/4=3/4`

\(=>sin\alpha=\dfrac{\sqrt{3}}{2}\) (vì `sin α` >0)

ta có `sin α : cos α=tan α`

\(=>tan\alpha=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

ta có `tan α * cot α =1`

\(=>\sqrt{3}\cdot cot\alpha=1\\ =>cot\alpha=\dfrac{1}{\sqrt{3}}\)

tương tự ta có

\(\left\{{}\begin{matrix}sin\beta=\dfrac{\sqrt{2}}{2}\\cos\beta=1\\cot\beta=1\end{matrix}\right.\)

a) sin a=0,8

Ta có: \(\sin^2a+\cos^2a=1\)

\(\Rightarrow\cos^2a=1-\sin^2a=1-0,8^2=0,36\)

\(\Rightarrow\orbr{\begin{cases}\cos a=0,6\\\cos a=-0,6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\tan a=\frac{4}{3}\\\tan a=\frac{-4}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\cot a=\frac{3}{4}\\\cot a=\frac{-3}{4}\end{cases}}\)

\(\sin a=0,8\)

\(\sin^2a=1-\sin^2a=1\)

\(\cos^2a=1-\sin^2a=1-0,8^2=0,36\)

\(\Rightarrow\hept{\begin{cases}\cos a=0,6\\\cos a=-0,6\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\tan a=\frac{4}{3}\\\tan a=\frac{-4}{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\cot a=\frac{3}{4}\\\cot a=\frac{-3}{4}\end{cases}}\)

Code : Breacker

a) Vì \(0<\alpha <\frac{\pi }{2} \) nên \(\sin \alpha  > 0\). Mặt khác, từ \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) suy ra

\(\sin \alpha  = \sqrt {1 - {{\cos }^2}a}  = \sqrt {1 - \frac{1}{{25}}}  = \frac{{2\sqrt 6 }}{5}\)

Do đó, \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{{2\sqrt 6 }}{5}}}{{\frac{1}{5}}} = 2\sqrt 6 \) và \(\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{\frac{1}{5}}}{{\frac{{2\sqrt 6 }}{5}}} = \frac{{\sqrt 6 }}{{12}}\)

b) Vì \(\frac{\pi }{2} < \alpha  < \pi\) nên \(\cos \alpha  < 0\). Mặt khác, từ \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) suy ra

       \(\cos \alpha  = \sqrt {1 - {{\sin }^2}a}  = \sqrt {1 - \frac{4}{9}}  = -\frac{{\sqrt 5 }}{3}\)

Do đó, \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{2}{3}}}{{-\frac{{\sqrt 5 }}{3}}} = -\frac{{2\sqrt 5 }}{5}\) và \(\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{-\frac{{\sqrt 5 }}{3}}}{{\frac{2}{3}}} = -\frac{{\sqrt 5 }}{2}\)

c) Ta có: \(\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{1}{{\sqrt 5 }}\)

Ta có: \({\tan ^2}\alpha  + 1 = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\cos ^2}\alpha  = \frac{1}{{{{\tan }^2}\alpha  + 1}} = \frac{1}{6} \Rightarrow \cos \alpha  =  \pm \frac{1}{{\sqrt 6 }}\)

Vì \(\pi  < \alpha  < \frac{{3\pi }}{2} \Rightarrow \sin \alpha  < 0\;\) và \(\,\,\cos \alpha  < 0 \Rightarrow \cos \alpha  = -\frac{1}{{\sqrt 6 }}\)

Ta có: \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \sin \alpha  = \tan \alpha .\cos \alpha  = \sqrt 5 .(-\frac{1}{{\sqrt 6 }}) = -\sqrt {\frac{5}{6}} \)

d) Vì \(\cot \alpha  =  - \frac{1}{{\sqrt 2 }}\;\,\) nên \(\,\,\tan \alpha  = \frac{1}{{\cot \alpha }} =  - \sqrt 2 \)

Ta có: \({\cot ^2}\alpha  + 1 = \frac{1}{{{{\sin }^2}\alpha }} \Rightarrow {\sin ^2}\alpha  = \frac{1}{{{{\cot }^2}\alpha  + 1}} = \frac{2}{3} \Rightarrow \sin \alpha  =  \pm \sqrt {\frac{2}{3}} \)

Vì \(\frac{{3\pi }}{2} < \alpha  < 2\pi  \Rightarrow \sin \alpha  < 0 \Rightarrow \sin \alpha  =  - \sqrt {\frac{2}{3}} \)

Ta có: \(\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }} \Rightarrow \cos \alpha  = \cot \alpha .\sin \alpha  = \left( { - \frac{1}{{\sqrt 2 }}} \right).\left( { - \sqrt {\frac{2}{3}} } \right) = \frac{{\sqrt 3 }}{3}\)

\(\tan\alpha=\dfrac{1}{3}\Leftrightarrow1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}\\ \Leftrightarrow\cos\alpha=\dfrac{1}{\sqrt{1+\tan^2\alpha}}=\dfrac{1}{\sqrt{\dfrac{9}{8}}}=\dfrac{2\sqrt{2}}{3}\\ \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\dfrac{8}{9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\\ \tan\alpha=\dfrac{1}{3}\Leftrightarrow\cot\alpha=3\)

a)  Ta có \({\cos ^2}\alpha  + {\sin ^2}\alpha \,\,\, = \,1\)

mà \(\sin \alpha  = \frac{{\sqrt {15} }}{4}\) nên \({\cos ^2}\alpha  + {\left( {\frac{{\sqrt {15} }}{4}} \right)^2}\,\,\, = \,1 \Rightarrow {\cos ^2}\alpha  = \frac{1}{{16}}\)

Lại có \(\frac{\pi }{2} < \alpha  < \pi \) nên \(\cos \alpha  < 0 \Rightarrow \cos \alpha  =  - \frac{1}{4}\)

Khi đó \(\tan \alpha  = \frac{{\sin \alpha }}{{co{\mathop{\rm s}\nolimits} \alpha }} =  - \sqrt {15} ;\cot \alpha  = \frac{1}{{\tan \alpha }} =  - \frac{1}{{\sqrt {15} }}\)

b)

Ta có \({\cos ^2}\alpha  + {\sin ^2}\alpha \,\,\, = \,1\)

mà \(\cos \alpha  =  - \frac{2}{3}\) nên \({\sin ^2}\alpha  + {\left( {\frac{{ - 2}}{3}} \right)^2}\,\,\, = \,1 \Rightarrow {\sin ^2}\alpha  = \frac{5}{9}\)

Lại có \( - \pi  < \alpha  < 0\) nên \(\sin \alpha  < 0 \Rightarrow \sin \alpha  =  - \frac{{\sqrt 5 }}{3}\)

Khi đó \(\tan \alpha  = \frac{{\sin \alpha }}{{co{\mathop{\rm s}\nolimits} \alpha }} = \frac{{\sqrt 5 }}{2};\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{2}{{\sqrt 5 }}\)

c)

Ta có \(\tan \alpha  = 3\) nên

\(\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{1}{3}\)

\(\frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha \,\,\, = \,1 + {3^2} = 10\,\, \Rightarrow {\cos ^2}\alpha  = \frac{1}{{10}}\)

Mà \({\cos ^2}\alpha  + {\sin ^2}\alpha \,\,\, = \,1 \Rightarrow {\sin ^2}\alpha  = \frac{9}{{10}}\)

Với \( - \pi  < \alpha  < 0\) thì \(\sin \alpha  < 0 \Rightarrow \sin \alpha  =  - \sqrt {\frac{9}{{10}}} \)

Với \( - \pi  < \alpha  <  - \frac{\pi }{2}\) thì \(\cos \alpha  < 0 \Rightarrow \cos \alpha  =  - \sqrt {\frac{1}{{10}}} \)

và  \( - \frac{\pi }{2} \le \alpha  < 0\) thì \(\cos \alpha  > 0 \Rightarrow \cos \alpha  = \sqrt {\frac{1}{{10}}} \)

d)

Ta có \(\cot \alpha  =  - 2\) nên

\(\tan \alpha  = \frac{1}{{\cot \alpha }} =  - \frac{1}{2}\)

\(\frac{1}{{{{\sin }^2}\alpha }} = 1 + co{{\mathop{\rm t}\nolimits} ^2}\alpha \,\,\, = \,1 + {( - 2)^2} = 5\,\, \Rightarrow {\sin ^2}\alpha  = \frac{1}{5}\)

Mà \({\cos ^2}\alpha  + {\sin ^2}\alpha \,\,\, = \,1 \Rightarrow {\cos ^2}\alpha  = \frac{4}{5}\)

Với \(0 < \alpha  < \pi \) thì \(\sin \alpha  > 0 \Rightarrow \sin \alpha  = \sqrt {\frac{1}{5}} \)

Với \(0 < \alpha  < \frac{\pi }{2}\) thì \(\cos \alpha  > 0 \Rightarrow \cos \alpha  = \sqrt {\frac{4}{5}} \)

và  \(\frac{\pi }{2} \le \alpha  < \pi \) thì \(\cos \alpha  < 0 \Rightarrow \cos \alpha  =  - \sqrt {\frac{4}{5}} \)