\(\frac{\pi}{2}< a< \frac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{3}}{2}\)

\(A=cosa.cos\frac{4\pi}{3}+sina.sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}.\left(-\frac{1}{2}\right)+\frac{1}{2}.\left(-\frac{\sqrt{3}}{2}\right)=0\)

\(B=cos\left(2a+2019.2\pi\right)=cos2a=1-2sin^2a=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}\)

a)     Đồ thị hàm số:

-        Với mỗi \(m \in \left[ { - 1;1} \right]\) chỉ có 1 giá trị \(\alpha  \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right]\) sao cho \(\sin \alpha  = m\)

b)     Đồ thị hàm số:

-        Với mỗi \(m \in \left[ { - 1;1} \right]\) có 1 giá trị \(\alpha  \in \left[ {0;\pi } \right]\) sao cho \(\cos \alpha  = m\)

c)     Đồ thị hàm số:

-        Với mỗi \(m \in \mathbb{R}\), có 2 giá trị \(\alpha  \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right]\) sao cho \(\tan \alpha  = m\)

d)     Đồ thị hàm số:

-        Với mỗi \(m \in \mathbb{R}\), có 2 giá trị \(\alpha  \in \left[ {0;\pi } \right]\) sao cho \(\cot \alpha  = m\)

Này là kiến thức lớp 10 mà bạn...

\(\frac{a}{2}\in\left(\frac{\pi}{2};\frac{3\pi}{4}\right)\Rightarrow tan\frac{a}{2}< 0\) ; \(sin\frac{a}{2}>0;cos\frac{a}{2}< 0\)

Đặt \(tan\frac{a}{2}=x< 0\)

\(\frac{2x}{1-x^2}=3\Leftrightarrow3x^2+2x-3=0\Rightarrow tan\frac{a}{2}=x=\frac{-1-\sqrt{10}}{3}\)

\(tan2a=\frac{2tana}{1-tan^2a}=\frac{6}{1-9}=-\frac{3}{4}\)

\(tan4a=\frac{2tan2a}{1-tan^22a}=-\frac{24}{7}\)

\(cos\frac{a}{2}=-\frac{1}{\sqrt{1+tan^2\frac{a}{2}}}=\) số thật kinh khủng

\(sin\frac{a}{2}=\sqrt{1-cos^2\frac{a}{2}}=...\)

\(sin\left(\frac{a}{2}+\frac{\pi}{2}\right)=\sqrt{2}\left(sin\frac{a}{2}+cos\frac{a}{2}\right)=...\)

Rốt cuộc là \(a\in\left(0;\frac{\pi}{2}\right)\) hay \(a\in\left(\frac{\pi}{2};\pi\right)\) bạn?

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

Ta có:

 \(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi  + \alpha } \right) =  \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2}  + \alpha } \right) - \cos \left( {12\pi  + \pi + \alpha } \right) =  \sin \left( {-8\pi  + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha  \right) = \cos \left( \alpha  \right) + \cos \left( \alpha  \right) = 2\cos \left( \alpha  \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)

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

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

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

\(tanx=tan\alpha\Rightarrow x=\alpha+k\pi\)