- LG a
- LG b
- LG c
Rút gọn
LG a
\[\dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - 4{{\sin }^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha }}{{4 - {{\sin }^2}2\alpha - 4{{\sin }^2}\alpha }}\];
Lời giải chi tiết:
\[\dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - 4{{\sin }^2}\alpha {{\cos }^2}\alpha }}{{4 - {{\sin }^2}2\alpha - 4{{\sin }^2}\alpha }} \] \[= \dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - {{\left[ {2\sin \alpha \cos \alpha } \right]}^2}}}{{4\left[ {1 - {{\sin }^2}\alpha } \right] -{{{\left[ {2\sin \alpha \cos \alpha } \right]}^2}}}}\]
\[= \dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - {{\sin }^2}2\alpha }}{{4{{\cos }^2}a - 4{{\sin }^2}\alpha {{\cos }^2}\alpha }}\]
=\[\dfrac{{4{{\sin }^4}\alpha }}{{4\cos ^2\alpha [1 - {{\sin }^2}\alpha ]}}\] \[= \dfrac{{{{\sin }^4}\alpha }}{{{{\cos }^4}\alpha }}\] \[ = {\tan ^4}\alpha \].
LG b
\[3 - 4\cos 2a + \cos 4a\];
Lời giải chi tiết:
\[3 - 4\cos 2a + \cos 4a \] \[= 3 - 4[1 - 2{\sin ^2}a] + [1 - 2{\sin ^2}2a]\]
\[= 3 - 4 + 8{\sin ^2}\alpha + 1 - 2{\left[ {2\sin \alpha \cos \alpha } \right]^2}\]
\[=8{\sin ^2}a - 8{\sin ^2}a{\cos ^2}a \]
\[= 8{\sin ^2}a[1 - {\cos ^2}a]\]
\[=8{\sin ^4}a\]
LG c
\[\dfrac{{{\mathop{\rm cota}\nolimits} + \tan a}}{{1 + \tan 2a\tan a}}\]
Lời giải chi tiết:
\[\dfrac{{\cot a + \tan a}}{{1 + \tan 2a\tan a}} \] \[= \dfrac{{\dfrac{{\cos a}}{{\sin a}} + \dfrac{{\sin a}}{{\cos a}}}}{{1 + \dfrac{{\sin 2a\sin a}}{{\cos 2a\cos a}}}}\]
=\[\dfrac{1}{{\sin a\cos a}}.\dfrac{{\cos acos2a}}{{\cos 2a\cos a + \sin 2a\sin a}}\]
=\[\dfrac{2}{{\sin 2a}}.\dfrac{{\cos acos2a}}{{\cos [2a - a]}} = 2\cot 2a\]