$a)$ Biến đổi các tổng sau thành tích:
 $ 1/ 1+ cos{\alpha } + cos2{\alpha }$                               $2/ 1+  \frac{{1}}{{{1 + }\frac{{1}}{{{sin\alpha }}}}}$
$b$) Với giá trị nào của $k \in Z $ phương trình sau có nghiệm:
${si}{{n}^{2}}\frac{{x}}{{3}}{ + 
}\frac{{1}}{{2}}\left( {{1 + cos}\frac{{{2x}}}{{3}}} 
\right){ + 2si}{{n}^{2}}\left( {\frac{{\pi }}{{4}}{ - 
}\frac{{x}}{{2}}} \right){ = 2k}$. Tìm nghiệm đó.
$a) 1/$ Ta có:
$1+ cos\alpha  + cos2\alpha  = (1+ cos2\alpha )+ cos\alpha =2 cos2\alpha +  cos\alpha $
        $= 2 cos\alpha ( cos\alpha  + \frac{1}{2})=2 cos\alpha  (  cos\alpha +cos\frac{\pi }{3})$
        $=4cos\alpha cos(\frac{\alpha }{2} + \frac{\pi }{6})cos(\frac{\alpha  }{2} + \frac{\pi }{6}$)
$2/ 1+ \frac{1}{{1 + \frac{1}{{\sin \alpha }}}}$= 1+ $\frac{1}{{1 + \frac{{\sin \alpha  + 
1}}{{\sin \alpha }}}}$$ = 1 + \frac{{\sin \alpha }}{{\sin \alpha  + 1}} = \frac{{2\sin \alpha  + 
1}}{{\sin \alpha  + 1}}$
  $\begin{array}{l}
{ = }\frac{{{2}\left( {{sin\alpha  + }\frac{{1}}{{2}}} 
\right)}}{{{1 + cos(}\frac{{\pi }}{{2}}{ - \alpha )}}}{ = 
}\frac{{{2}\left[ {{sin\alpha  + sin}\frac{{\pi }}{{6}}} 
\right]}}{{{2co}{{s}^{2}}{(}\frac{{\pi }}{{4}}{ - 
}\frac{{\alpha }}{{2}}{)}}}\\
{ = }\frac{{{2sin(}\frac{{\alpha }}{{2}}{ + }\frac{{\pi 
}}{{{12}}}{)cos(}\frac{{\alpha }}{{2}}{ - }\frac{{\pi 
}}{{{12}}}{)}}}{{{co}{{s}^{2}}{(}\frac{{\pi 
}}{{4}}{ - }\frac{{\alpha }}{{2}}{)}}}
\end{array}$
$b$) Ta có: ${si}{{n}^{2}}\frac{{x}}{{3}}{ + 
}\frac{{1}}{{2}}\left( {{1 + cos}\frac{{{2x}}}{{3}}} 
\right){ + 2si}{{n}^{2}}\left( {\frac{{\pi }}{{4}}{ - 
}\frac{{x}}{{2}}} \right){ = 2k}$
$\begin{array}{l}
 \Leftrightarrow \frac{{{1 - cos}\frac{{{2x}}}{{3}}}}{{2}}{ + 
}\frac{{{1 + cos}\frac{{{2x}}}{{3}}}}{{2}}{ + 1 - 
cos(}\frac{{\pi }}{{2}}{ - x) = 2k}\\
 \Leftrightarrow {2 - sinx = 2k} \Leftrightarrow {sinx = 2(1 - k)}
\end{array}$
Để phương trình có nghiệm $ \Leftrightarrow {2}\left| {{1 - k}} \right| \le {1} 
\Leftrightarrow \left| {{1 - k}} \right| \le \frac{{1}}{{2}}$
    $\begin{array}{l}
 \Leftrightarrow { - }\frac{{1}}{{2}} \le {1 - k} \le 
\frac{{1}}{{2}} \Leftrightarrow \left. \begin{array}{l}
\frac{{1}}{{2}} \le {k} \le \frac{{3}}{{2}}\\
{k} \in {Z}
\end{array} \right\} \Rightarrow {k = 1}\\
 \Rightarrow {sinx = 0} \Leftrightarrow {x = k'\pi    (k'} \in {Z)}
\end{array}$
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