\(A = \cos {75^0}\cos {15^0} = \frac{1}{2}\left[ {\cos \left( {{{75}^0} - {{15}^0}} \right) + \cos \left( {{{75}^0} + {{15}^0}} \right)} \right] \\= \frac{1}{2}.\cos {60^0}.\cos {90^0} = 0\)

\(B = \sin \frac{{5\pi }}{{12}}\cos \frac{{7\pi }}{{12}} = \frac{1}{2}\left[ {\sin \left( {\frac{{5\pi }}{{12}} - \frac{{7\pi }}{{12}}} \right) + \sin \left( {\frac{{5\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right)} \right] \\= \frac{1}{2}\sin \left( { - \frac{{2\pi }}{{12}}} \right).\sin \left( {\frac{{12\pi }}{{12}}} \right) =  - \frac{1}{2}\sin \frac{\pi }{6}\sin \pi  = 0\)

a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}} = \frac{{\sin \left( {\frac{\pi }{{15}} + \frac{\pi }{{10}}} \right)}}{{\cos \left( {\frac{{2\pi }}{{15}} + \frac{\pi }{5}} \right)}} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{3}}} = 1\)

b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)

Ta có:

\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)

Ta có:

\(\begin{array}{l}\sin \frac{\pi }{{24}}\cos \frac{{5\pi }}{{24}} = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{{24}} + \frac{{5\pi }}{{24}}} \right) + \sin \left( {\frac{\pi }{{24}} - \frac{{5\pi }}{{24}}} \right)} \right]\\ = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{4}} \right) + \sin \left( { - \frac{\pi }{6}} \right)} \right]\\ = \frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} - \frac{1}{2}} \right] = \frac{{\sqrt 2  - 1}}{4}\end{array}\)

Ta có:

\(\begin{array}{l}\sin \frac{{7\pi }}{8}\sin \frac{{5\pi }}{8} = \frac{1}{2}\left[ {\cos \left( {\frac{{7\pi }}{8} - \frac{{5\pi }}{8}} \right) - \cos \left( {\frac{{7\pi }}{8} + \frac{{5\pi }}{8}} \right)} \right]\\ = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{{3\pi }}{2}} \right)} \right]\\ = \frac{1}{2}.\left( {\frac{{\sqrt 2 }}{2} + 0} \right) = \frac{{\sqrt 2 }}{4}\end{array}\)

\(cosx+cos\left(x+\frac{\pi}{5}\right)+cos\left(x+\frac{9\pi}{5}\right)+cos\left(x+\frac{2\pi}{5}\right)+cos\left(x+\frac{8\pi}{5}\right)+...+cos\left(x+\frac{5\pi}{5}\right)\)

\(=cosx-2cosx.cos\frac{4\pi}{5}-2cosx.cos\frac{3\pi}{5}-2cosx.cos\frac{2\pi}{5}-2cosx.cos\frac{\pi}{5}-cosx\)

\(=-2cosx\left(cos\frac{\pi}{5}+cos\frac{4\pi}{5}+cos\frac{2\pi}{5}+cos\frac{3\pi}{5}\right)\)

\(=-2cosx\left(2cos\frac{\pi}{2}.cos\frac{3\pi}{10}+2cos\frac{\pi}{2}cos\frac{\pi}{10}\right)\)

\(=0\) (do \(cos\frac{\pi}{2}=0\))

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

\(\begin{array}{l}Q = {\tan ^2}\frac{\pi }{3} + {\sin ^2}\frac{\pi }{4} + \cot \frac{\pi }{4} + \cos \frac{\pi }{2}\\\,\,\,\,\, = \,{\left( {\sqrt 3 } \right)^2} + {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} + 1 + 0 = \frac{7}{2}\end{array}\)