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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
\(B = \left( {\cos \frac{\pi }{9} + \cos \frac{{5\pi }}{9}} \right) + \cos \frac{{11\pi }}{9} = \left( {2\cos \frac{{\frac{\pi }{9} + \frac{{5\pi }}{9}}}{2}\cos \frac{{\frac{\pi }{9} - \frac{{5\pi }}{9}}}{2}} \right) + \cos \frac{{11\pi }}{9} = 2\cos \frac{\pi }{3}\cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9}\)
\( = \cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9} = 2\cos \frac{{\frac{{2\pi }}{9} + \frac{{11\pi }}{9}}}{2}\cos \frac{{\frac{{2\pi }}{9} - \frac{{11\pi }}{9}}}{2} = 2\cos \frac{{13\pi }}{{18}}\cos \frac{\pi }{2} = 0\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\). Do đó \(\cos a = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \frac{1}{3}} = - \frac{{\sqrt 6 }}{3}\)
Ta có: \(\cos \left( {a + \frac{\pi }{6}} \right) = \cos a\cos \frac{\pi }{6} - \sin a\sin \frac{\pi }{6} = - \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 3 }}.\frac{1}{2} = - \frac{{\sqrt 3 + 3\sqrt 2 }}{6}\)
b) Vì \(\pi < a < \frac{{3\pi }}{2}\) nên \(\sin a < 0\). Do đó \(\sin a = \sqrt {1 - {{\cos }^2}a} = \sqrt {1 - \frac{1}{9}} = - \frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a\; = \frac{{\sin a}}{{\cos a}} = \frac{{ - \frac{{2\sqrt 2 }}{3}}}{{ - \frac{1}{3}}} = 2\sqrt 2 \)
Ta có: \(\tan \left( {a - \frac{\pi }{4}} \right) = \frac{{\tan a - \tan \frac{\pi }{4}}}{{1 + \tan a\tan \frac{\pi }{4}}} = \frac{{\frac{{\sin a}}{{\cos a}} - 1}}{{1 + \frac{{\sin a}}{{\cos a}}}} = \frac{{2\sqrt 2 - 1}}{{1 + 2\sqrt 2 }} = \frac{{9 - 4\sqrt 2 }}{7}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)
Ta có: \({\sin ^2}a + {\cos ^2}a = 1\)
\(\Leftrightarrow \frac{1}{9} + {\cos ^2}a = 1\)
\(\Leftrightarrow {\cos ^2}a = 1 - \frac{1}{9}= \frac{8}{9}\)
\(\Leftrightarrow \cos a =\pm\sqrt { \frac{8}{9}} = \pm \frac{{2\sqrt 2 }}{3}\)
Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)
Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} = - \frac{{\sqrt 2 }}{4}\)
Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) = - \frac{{4\sqrt 2 }}{9}\)
\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)
\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} = - \frac{{4\sqrt 2 }}{7}\)
b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)
\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)
Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 = - \frac{3}{4}\)
Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)
\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)
\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)
\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)
\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)
\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 = - \frac{{\sqrt 7 }}{4}\)
\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
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) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)
\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
\(\begin{array}{l}A = \sin \left( {a - 17^\circ } \right)\cos \left( {a + 13^\circ } \right) - \sin \left( {a + 13^\circ } \right)\cos \left( {a - 17^\circ } \right)\\A = \sin \left( {a - 17^\circ - a - 13^\circ } \right) = \sin \left( { - 30^\circ } \right) = - \frac{1}{2}\end{array}\)
\(\begin{array}{l}B = \cos \left( {b + \frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{6} - b} \right) - \sin \left( {b + \frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{6} - b} \right)\\B = \cos \left( {b + \frac{\pi }{3} + \frac{\pi }{6} - b} \right) = \cos \frac{\pi }{2} = 0\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) Ta có: VT = \(\cos \left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \cos \frac{\pi }{{6}} = \frac{{\sqrt 3 }}{2}\)
\(VP = \cos \frac{\pi }{3}\cos \frac{\pi }{6} + \sin \frac{\pi }{3}\sin \frac{\pi }{6} = \frac{{1 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2}.\frac{1}{2} = \frac{{\sqrt 3 }}{2} = VT\)
Vậy \(\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\)
b) Ta có: \(\cos \left( {a + b} \right) = \cos (a--b) = \cos a\cos \left( { - b} \right) + \sin a\sin \left( { - b} \right) = \cos a\cos b - \sin a\sin b\)
c) Ta có: \(\sin \left( {a - b} \right) = \cos \left[ {\frac{\pi }{2} - \left( {a - b} \right)} \right] = \cos \left[ {\left( {\frac{\pi }{2} - a} \right) + b} \right] = \cos \left( {\frac{\pi }{2} - a} \right)\cos b + \sin \left( {\frac{\pi }{2} - a} \right)\sin b\)
\( = \left( {\cos \frac{\pi }{2}\cos a + \sin \frac{\pi }{2}\sin a} \right)\cos b + \sin \left( {\frac{\pi }{2} - a} \right)\sin b = \sin a\cos b + \cos a\sin b\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
Ta có:
\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)