- LG a
- LG b
- LG c
Rút gọn
LG a
\[4{\cos ^4}a - 2\cos 2a - \dfrac{1}{2}\cos 4a\];
Lời giải chi tiết:
\[4{\cos ^4}a - 2\cos 2a - \dfrac{1}{2}\cos 4a\]
=\[4{\cos ^4}a - 2[2{\cos ^2}a - 1]\] \[ - \dfrac{1}{2}[2{\cos ^2}2a - 1]\]
=\[4{\cos ^4}a - 4{\cos ^2}a + 2 \] \[- {[2{\cos ^2}a - 1]^2} + \dfrac{1}{2}\]
=\[4{\cos ^4}a - 4{\cos ^2}a + \dfrac{5}{2} \] \[- 4{\cos ^4}a + 4{\cos ^2}a - 1 \] \[= \dfrac{3}{2}\]
LG b
\[{\sin ^2}a\left[ {1 + \dfrac{1}{{\sin a}} + \cot a} \right]\left[ {1 - \dfrac{1}{{\sin a}} + \cot a} \right]\];
Lời giải chi tiết:
\[{\sin ^2}a[1 + \dfrac{1}{{\sin a}} + \cot a][1 - \dfrac{1}{{\sin a}} + \cot a]\]
=\[{\sin ^2}a\left[ {{{[1 + cota]}^2} - \dfrac{1}{{{{\sin }^2}a}}} \right] \] \[= {\sin ^2}a[1 + {\cot ^2}a + 2\cot a] - 1\]
=\[{\sin ^2}a + {\cos ^2}a + 2{\sin ^2}a\dfrac{{\cos a}}{{\sin a}} - 1\] \[ = \sin 2a\]
LG c
\[\dfrac{{\cos 2a}}{{{{\cos }^4}a - {{\sin }^4}a}} - \dfrac{{{{\cos }^4}a + {{\sin }^4}a}}{{1 - \dfrac{1}{2}{{\sin }^2}2a}}\].
Lời giải chi tiết:
\[\dfrac{{\cos 2a}}{{{{\cos }^4}a - {{\sin }^4}a}} - \dfrac{{{{\cos }^4}a + {{\sin }^4}a}}{{1 - \dfrac{1}{2}{{\sin }^2}2a}}\]
=\[\dfrac{{{{\cos }^2}a - {{\sin }^2}a}}{{[{{\cos }^2}a + {{\sin }^2}a][{{\cos }^2}a - {{\sin }^2}a]}} \] \[- \dfrac{{{{\cos }^4}a + {{\sin }^4}a}}{{1 - \dfrac{1}{2}{{[2\sin a\cos a]}^2}}}\]
\[\begin{array}{l}
= 1 - \dfrac{{{{\left[ {{{\sin }^2}a + {{\cos }^2}a} \right]}^2} - 2{{\sin }^2}a{{\cos }^2}a}}{{1 - 2{{\sin }^2}a{{\cos }^2}a}}\\
= 1 - \dfrac{{1 - 2{{\sin }^2}a{{\cos }^2}a}}{{1 - 2{{\sin }^2}a{{\cos }^2}a}}\\
= 1 - 1 = 0
\end{array}\]