\(VT=\frac{2\cos2\alpha.\cos\alpha}{2.\sin2\alpha\cos\alpha}.\frac{\sin2\alpha}{\cos2\alpha}-2\left(2\sin\alpha.\cos\alpha\right)^2\)

\(VT=1-2\left(\sin2\alpha\right)^2=\cos4\alpha\)

Ta có:

 \(\begin{array}{l}{\tan ^2}\alpha  + 1 = \frac{1}{{{{\cos }^2}\alpha }}\\ \Rightarrow {\left( {\frac{2}{3}} \right)^2} + 1 = \frac{1}{{{{\cos }^2}\alpha }}\\ \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = \frac{{13}}{9}\\ \Rightarrow \cos \alpha  =  \pm \frac{{3\sqrt {13} }}{{13}}\end{array}\)

Do \(\pi  < \alpha  < \frac{{3\pi }}{2} \Rightarrow \cos \alpha  =  - \frac{{3\sqrt {13} }}{{13}}\)

Ta có: \(\begin{array}{l}\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \frac{2}{3} = \sin \alpha :\left( { - \frac{{3\sqrt {13} }}{{13}}} \right)\\ \Rightarrow \sin \alpha  =  - \frac{{2\sqrt {13} }}{{13}}\end{array}\)

\(\cos \alpha  =  - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}}  =  - \frac{{12}}{{13}}\) (vì \(\pi  < \alpha  < \frac{{3\pi }}{2}\))

\(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)

\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha  + \sin \frac{\pi }{4}sin\alpha  = \frac{{ - 17\sqrt 2 }}{{26}}\)

Ta có:

a) \(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3  + 3\sqrt 2 }}{6}\)      

b) \(\cos \left( {\alpha  + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} =  - \frac{{3 + \sqrt 6 }}{6}\)

c) \(\sin \left( {\alpha  - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)

d) \(\cos \left( {\alpha  - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)

Này là kiến thức lớp 10 mà bạn...

a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)

\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha  + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha  =  \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}}  =  \pm \frac{{2\sqrt 2 }}{3}\end{array}\)

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

\(b)\;\,sin2\alpha  = 2sin\alpha .cos\alpha  = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} =  - \frac{{4\sqrt 2 }}{9}\)

\(c)\;cos(\alpha  + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6  + 1}}{6}\).

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 \(\cos \alpha  < 0 \Rightarrow \cos \alpha  =  - \sqrt {\frac{4}{5}} \)

Bài 1:

$\cos ^2a=1-\sin ^2a=1-\frac{1}{5^2}=\frac{24}{25}$

Vì $0< a< \frac{\pi}{2}$ nên $\cos a>0$

$\Rightarrow \cos a=\frac{\sqrt{24}}{5}$

\(\cos (a-\frac{\pi}{6})=\cos a.\cos \frac{\pi}{6}+\sin a\sin \frac{\pi}{6}=\cos a.\frac{\sqrt{3}}{2}+\sin a. \frac{1}{2}\)

\(=\frac{\sqrt{24}}{5}.\frac{\sqrt{3}}{2}+\frac{1}{5}.\frac{1}{2}=\frac{1+6\sqrt{2}}{10}\)

Bài 2:

$\cos x=\frac{-2}{3}\Rightarrow \sin ^2x=1-\cos ^2x=\frac{5}{9}$

Vì $x\in (\frac{\pi}{2}; \pi)$ nên $\sin x>0\Rightarrow \sin x=\frac{\sqrt{5}}{3}$

\(\Rightarrow \tan x=\frac{\sin x}{\cos x}=\frac{-\sqrt{5}}{2}\)

Do đó:

\(\tan (\frac{\pi}{4}+x)=\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4}.\tan x}=\frac{1+\tan x}{1-\tan x}=\frac{1-\frac{\sqrt{5}}{2}}{1+\frac{\sqrt{5}}{2}}=-9+4\sqrt{5}\)

Bài 3:

\(\tan a=\frac{-4}{7}=\frac{\sin a}{\cos a}\)

\(\Rightarrow \frac{\sin ^2a}{\cos ^2a}=\frac{16}{49}\Rightarrow \frac{1}{\cos ^2a}=\frac{65}{49}\) \(\Rightarrow \cos ^2a=\frac{49}{65}\)

Kết hợp điều kiện của $a$ suy ra $\cos a>0\Rightarrow \cos a=\frac{7}{\sqrt{65}}$

$\Rightarrow \sin a=\frac{-4}{7}\cos a=\frac{-4}{\sqrt{65}}$

Do đó:

\(\cos (2a-\frac{\pi}{2})=\cos 2a.\cos \frac{\pi}{2}+\sin 2a.\sin \frac{\pi}{2}\)

\(=(\cos ^2a-\sin ^2a).0+2\sin a\cos a.1=2\sin a\cos a=2.\frac{-4}{\sqrt{65}}.\frac{7}{\sqrt{65}}=\frac{56}{65}\)

Bài 4:

$\sin a=\frac{1}{2}$ và $0< a< \pi$ nên $a=\frac{\pi}{6}$ hoặc $a=\frac{5}{6}\pi$

Nếu $a=\frac{\pi}{6}$ thì $\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{-\sqrt{3}}{3}+\frac{1}{2}=\frac{3-2\sqrt{3}}{6}$

Nếu $a=\frac{5\pi}{6}$ thì:

\(\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{5\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{\sqrt{3}}{3}+\frac{1}{2}=\frac{3+2\sqrt{3}}{6}\)