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Câu hỏi: Biến đổi thành tích
Lời giải chi tiết:
\(\begin{array}{l}
1 + \cos \left({\frac{\pi }{2} + 3\alpha } \right) - \sin \left({\frac{{3\pi }}{2} - 3\alpha } \right) + \cot \left({\frac{{5\pi }}{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha - \sin \left({\pi + \frac{\pi }{2} - 3\alpha } \right) + \cot \left({3\pi - \frac{\pi }{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha + \sin \left({\frac{\pi }{2} - 3\alpha } \right) + \cot \left({ - \frac{\pi }{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha + \cos 3\alpha - \cot \left({\frac{\pi }{2} - 3\alpha } \right)\\
= 1 - \sin 3\alpha + \cos 3\alpha - \tan 3\alpha \\
= 1 - \sin 3\alpha + \cos 3\alpha - \frac{{\sin 3\alpha }}{{\cos 3\alpha }}\\
= \frac{{\cos 3\alpha - \sin 3\alpha \cos 3\alpha + {{\cos }^2}3\alpha - \sin 3\alpha }}{{\cos 3\alpha }}\\
= \frac{{\left({\cos 3\alpha + {{\cos }^2}3\alpha } \right) - \left({\sin 3\alpha \cos 3\alpha + \sin 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{\cos 3\alpha \left({1 + \cos 3\alpha } \right) - \sin 3\alpha \left({1 + \cos 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{\left({1 + \cos 3\alpha } \right)\left({\cos 3\alpha - \sin 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{2{{\cos }^2}\frac{{3\alpha }}{2}.\sqrt 2 \cos \left({3\alpha + \frac{\pi }{4}} \right)}}{{\cos 3\alpha }}\\
= \frac{{2\sqrt 2 {{\cos }^2}\frac{{3\alpha }}{2}\cos \left({3\alpha + \frac{\pi }{4}} \right)}}{{\cos 3\alpha }}
\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}
\frac{{\cos 7\alpha - \cos 8\alpha - \cos 9\alpha + \cos 10\alpha }}{{\sin 7\alpha - \sin 8\alpha - \sin 9\alpha + \sin 10\alpha }}\\
= \frac{{\left({\cos 7\alpha - \cos 9\alpha } \right) - \left({\cos 8\alpha - \cos 10\alpha } \right)}}{{\left({\sin 7\alpha - \sin 9\alpha } \right) - \left({\sin 8\alpha - \sin 10\alpha } \right)}}\\
= \frac{{ - 2\sin 8\alpha \sin \left({ - \alpha } \right) + 2\sin 9\alpha \sin \left({ - \alpha } \right)}}{{2\cos 8\alpha \sin \left({ - \alpha } \right) - 2\cos 9\alpha \sin \left({ - \alpha } \right)}}\\
= \frac{{2\sin 8\alpha \sin \alpha - 2\sin 9\alpha \sin \alpha }}{{ - 2\cos 8\alpha \sin \alpha + 2\cos 9\alpha \sin \alpha }}\\
= \frac{{2\sin \alpha \left({\sin 8\alpha - \sin 9\alpha } \right)}}{{2\sin \alpha \left({\cos 9\alpha - \cos 8\alpha } \right)}}\\
= \frac{{\sin 8\alpha - \sin 9\alpha }}{{\cos 9\alpha - \cos 8\alpha }}\\
= \frac{{2\cos \frac{{17\alpha }}{2}\sin \left({ - \frac{\alpha }{2}} \right)}}{{ - 2\sin \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}\\
= \frac{{ - 2\cos \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}{{ - 2\sin \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}\\
= \frac{{\cos \frac{{17\alpha }}{2}}}{{\sin \frac{{17\alpha }}{2}}} = \cot \frac{{17\alpha }}{2}
\end{array}\)
Lời giải chi tiết:
$\begin{array}{l}
- \cos 5a.\cos 4a--\cos 4a.\cos 3a + 2{\cos ^2}2a.\cos a\\
= - \cos 4a\left( {\cos 5a + \cos 3a} \right) + 2{\cos ^2}2a.\cos a\\
= - 2\cos 4a.\cos 4a.\cos a + 2{\cos ^2}\;2a.\cos a\\
= 2\cos a({\cos ^2}2a - {\cos ^2}4a)\\
= 2\cos a\left( {\cos 2a + \cos 4a} \right)\left( {\cos 2a--\cos 4a} \right)\\
= 2\cos a.2\cos 3a.\cos a.2\sin 3a.\sin a\\
= 2\cos a\sin 2a\sin 6a.
\end{array}$
Câu a
\(1 + \cos \left( {\frac{\pi }{2} + 3\alpha } \right) - \sin \left({\frac{{3\pi }}{2} - 3\alpha } \right) + \cot \left({\frac{{5\pi }}{2} + 3\alpha } \right)\)Lời giải chi tiết:
\(\begin{array}{l}
1 + \cos \left({\frac{\pi }{2} + 3\alpha } \right) - \sin \left({\frac{{3\pi }}{2} - 3\alpha } \right) + \cot \left({\frac{{5\pi }}{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha - \sin \left({\pi + \frac{\pi }{2} - 3\alpha } \right) + \cot \left({3\pi - \frac{\pi }{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha + \sin \left({\frac{\pi }{2} - 3\alpha } \right) + \cot \left({ - \frac{\pi }{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha + \cos 3\alpha - \cot \left({\frac{\pi }{2} - 3\alpha } \right)\\
= 1 - \sin 3\alpha + \cos 3\alpha - \tan 3\alpha \\
= 1 - \sin 3\alpha + \cos 3\alpha - \frac{{\sin 3\alpha }}{{\cos 3\alpha }}\\
= \frac{{\cos 3\alpha - \sin 3\alpha \cos 3\alpha + {{\cos }^2}3\alpha - \sin 3\alpha }}{{\cos 3\alpha }}\\
= \frac{{\left({\cos 3\alpha + {{\cos }^2}3\alpha } \right) - \left({\sin 3\alpha \cos 3\alpha + \sin 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{\cos 3\alpha \left({1 + \cos 3\alpha } \right) - \sin 3\alpha \left({1 + \cos 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{\left({1 + \cos 3\alpha } \right)\left({\cos 3\alpha - \sin 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{2{{\cos }^2}\frac{{3\alpha }}{2}.\sqrt 2 \cos \left({3\alpha + \frac{\pi }{4}} \right)}}{{\cos 3\alpha }}\\
= \frac{{2\sqrt 2 {{\cos }^2}\frac{{3\alpha }}{2}\cos \left({3\alpha + \frac{\pi }{4}} \right)}}{{\cos 3\alpha }}
\end{array}\)
Câu b
\(\frac{{\cos 7\alpha - \cos 8\alpha - \cos 9\alpha + \cos 10\alpha }}{{\sin 7\alpha - \sin 8\alpha - \sin 9\alpha + \sin 10\alpha }}\)Lời giải chi tiết:
\(\begin{array}{l}
\frac{{\cos 7\alpha - \cos 8\alpha - \cos 9\alpha + \cos 10\alpha }}{{\sin 7\alpha - \sin 8\alpha - \sin 9\alpha + \sin 10\alpha }}\\
= \frac{{\left({\cos 7\alpha - \cos 9\alpha } \right) - \left({\cos 8\alpha - \cos 10\alpha } \right)}}{{\left({\sin 7\alpha - \sin 9\alpha } \right) - \left({\sin 8\alpha - \sin 10\alpha } \right)}}\\
= \frac{{ - 2\sin 8\alpha \sin \left({ - \alpha } \right) + 2\sin 9\alpha \sin \left({ - \alpha } \right)}}{{2\cos 8\alpha \sin \left({ - \alpha } \right) - 2\cos 9\alpha \sin \left({ - \alpha } \right)}}\\
= \frac{{2\sin 8\alpha \sin \alpha - 2\sin 9\alpha \sin \alpha }}{{ - 2\cos 8\alpha \sin \alpha + 2\cos 9\alpha \sin \alpha }}\\
= \frac{{2\sin \alpha \left({\sin 8\alpha - \sin 9\alpha } \right)}}{{2\sin \alpha \left({\cos 9\alpha - \cos 8\alpha } \right)}}\\
= \frac{{\sin 8\alpha - \sin 9\alpha }}{{\cos 9\alpha - \cos 8\alpha }}\\
= \frac{{2\cos \frac{{17\alpha }}{2}\sin \left({ - \frac{\alpha }{2}} \right)}}{{ - 2\sin \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}\\
= \frac{{ - 2\cos \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}{{ - 2\sin \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}\\
= \frac{{\cos \frac{{17\alpha }}{2}}}{{\sin \frac{{17\alpha }}{2}}} = \cot \frac{{17\alpha }}{2}
\end{array}\)
Câu c
$ - \cos 5a.\cos 4a--\cos 4a.\cos 3a + 2{\cos ^2}2a.\cos a$Lời giải chi tiết:
$\begin{array}{l}
- \cos 5a.\cos 4a--\cos 4a.\cos 3a + 2{\cos ^2}2a.\cos a\\
= - \cos 4a\left( {\cos 5a + \cos 3a} \right) + 2{\cos ^2}2a.\cos a\\
= - 2\cos 4a.\cos 4a.\cos a + 2{\cos ^2}\;2a.\cos a\\
= 2\cos a({\cos ^2}2a - {\cos ^2}4a)\\
= 2\cos a\left( {\cos 2a + \cos 4a} \right)\left( {\cos 2a--\cos 4a} \right)\\
= 2\cos a.2\cos 3a.\cos a.2\sin 3a.\sin a\\
= 2\cos a\sin 2a\sin 6a.
\end{array}$
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