Bài 2 trang 231 SBT đại số và giải tích 11

Biến đổi thành tích

LG 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:

\(\begin{array}{l}
1 + \cos \left( {\frac{\pi }{2} + 3\alpha } \right) - \sin \left( {\frac{{3\pi }}{2} - 3\alpha } \right) + \cot \left( {\frac{{5\pi }}{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha - \sin \left( {\pi + \frac{\pi }{2} - 3\alpha } \right) + \cot \left( {3\pi - \frac{\pi }{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha + \sin \left( {\frac{\pi }{2} - 3\alpha } \right) + \cot \left( { - \frac{\pi }{2} + 3\alpha } \right)\\
= 1 - \sin 3\alpha + \cos 3\alpha - \cot \left( {\frac{\pi }{2} - 3\alpha } \right)\\
= 1 - \sin 3\alpha + \cos 3\alpha - \tan 3\alpha \\
= 1 - \sin 3\alpha + \cos 3\alpha - \frac{{\sin 3\alpha }}{{\cos 3\alpha }}\\
= \frac{{\cos 3\alpha - \sin 3\alpha \cos 3\alpha + {{\cos }^2}3\alpha - \sin 3\alpha }}{{\cos 3\alpha }}\\
= \frac{{\left( {\cos 3\alpha + {{\cos }^2}3\alpha } \right) - \left( {\sin 3\alpha \cos 3\alpha + \sin 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{\cos 3\alpha \left( {1 + \cos 3\alpha } \right) - \sin 3\alpha \left( {1 + \cos 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{\left( {1 + \cos 3\alpha } \right)\left( {\cos 3\alpha - \sin 3\alpha } \right)}}{{\cos 3\alpha }}\\
= \frac{{2{{\cos }^2}\frac{{3\alpha }}{2}.\sqrt 2 \cos \left( {3\alpha + \frac{\pi }{4}} \right)}}{{\cos 3\alpha }}\\
= \frac{{2\sqrt 2 {{\cos }^2}\frac{{3\alpha }}{2}\cos \left( {3\alpha + \frac{\pi }{4}} \right)}}{{\cos 3\alpha }}
\end{array}\)

LG 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:

\(\begin{array}{l}
\frac{{\cos 7\alpha - \cos 8\alpha - \cos 9\alpha + \cos 10\alpha }}{{\sin 7\alpha - \sin 8\alpha - \sin 9\alpha + \sin 10\alpha }}\\
= \frac{{\left( {\cos 7\alpha - \cos 9\alpha } \right) - \left( {\cos 8\alpha - \cos 10\alpha } \right)}}{{\left( {\sin 7\alpha - \sin 9\alpha } \right) - \left( {\sin 8\alpha - \sin 10\alpha } \right)}}\\
= \frac{{ - 2\sin 8\alpha \sin \left( { - \alpha } \right) + 2\sin 9\alpha \sin \left( { - \alpha } \right)}}{{2\cos 8\alpha \sin \left( { - \alpha } \right) - 2\cos 9\alpha \sin \left( { - \alpha } \right)}}\\
= \frac{{2\sin 8\alpha \sin \alpha - 2\sin 9\alpha \sin \alpha }}{{ - 2\cos 8\alpha \sin \alpha + 2\cos 9\alpha \sin \alpha }}\\
= \frac{{2\sin \alpha \left( {\sin 8\alpha - \sin 9\alpha } \right)}}{{2\sin \alpha \left( {\cos 9\alpha - \cos 8\alpha } \right)}}\\
= \frac{{\sin 8\alpha - \sin 9\alpha }}{{\cos 9\alpha - \cos 8\alpha }}\\
= \frac{{2\cos \frac{{17\alpha }}{2}\sin \left( { - \frac{\alpha }{2}} \right)}}{{ - 2\sin \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}\\
= \frac{{ - 2\cos \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}{{ - 2\sin \frac{{17\alpha }}{2}\sin \frac{\alpha }{2}}}\\
= \frac{{\cos \frac{{17\alpha }}{2}}}{{\sin \frac{{17\alpha }}{2}}} = \cot \frac{{17\alpha }}{2}
\end{array}\)

LG c

-cos5a.cos4a – cos4a.cos3a + 2cos² 2a.cosa

Lời giải chi tiết:

-cos5a.cos4a – cos4a.cos3a + 2cos² 2a.cosa

= - cos4a(cos5a + cos3a) + 2cos² 2a.cosa

= - 2cos4a.cos4a.cosa + 2cos² 2a.cosa

= 2cosa(cos² 2a- cos² 4a)

= 2cosa(cos2a + cos4a)(cos2a – cos4a)

= 2cosa. 2cos3a.cosa. 2sin3a.sina

= 2cosa sin2a sin6a.