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\(A = \cos {75^0}\cos {15^0} = \frac{1}{2}\left[ {\cos \left( {{{75}^0} - {{15}^0}} \right) + \cos \left( {{{75}^0} + {{15}^0}} \right)} \right] \\= \frac{1}{2}.\cos {60^0}.\cos {90^0} = 0\)
\(B = \sin \frac{{5\pi }}{{12}}\cos \frac{{7\pi }}{{12}} = \frac{1}{2}\left[ {\sin \left( {\frac{{5\pi }}{{12}} - \frac{{7\pi }}{{12}}} \right) + \sin \left( {\frac{{5\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right)} \right] \\= \frac{1}{2}\sin \left( { - \frac{{2\pi }}{{12}}} \right).\sin \left( {\frac{{12\pi }}{{12}}} \right) = - \frac{1}{2}\sin \frac{\pi }{6}\sin \pi = 0\)
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:
\(\begin{array}{l}\sin \frac{{7\pi }}{8}\sin \frac{{5\pi }}{8} = \frac{1}{2}\left[ {\cos \left( {\frac{{7\pi }}{8} - \frac{{5\pi }}{8}} \right) - \cos \left( {\frac{{7\pi }}{8} + \frac{{5\pi }}{8}} \right)} \right]\\ = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{{3\pi }}{2}} \right)} \right]\\ = \frac{1}{2}.\left( {\frac{{\sqrt 2 }}{2} + 0} \right) = \frac{{\sqrt 2 }}{4}\end{array}\)
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) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
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 \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {\frac{4}{5}} \)
Ta có
\(\begin{array}{l}Q = {\tan ^2}\frac{\pi }{3} + {\sin ^2}\frac{\pi }{4} + \cot \frac{\pi }{4} + \cos \frac{\pi }{2}\\\,\,\,\,\, = \,{\left( {\sqrt 3 } \right)^2} + {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} + 1 + 0 = \frac{7}{2}\end{array}\)
Ta có:
\(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi + \alpha } \right) = \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2} + \alpha } \right) - \cos \left( {12\pi + \pi + \alpha } \right) = \sin \left( {-8\pi + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha \right) = \cos \left( \alpha \right) + \cos \left( \alpha \right) = 2\cos \left( \alpha \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)
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) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)