LG a - bài 18 trang 214 sbt đại số 10

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:

\(\dfrac{{\cot a + \tan a}}{{1 + \tan 2a\tan a}} \) \(= \dfrac{{\dfrac{{\cos a}}{{\sin a}} + \dfrac{{\sin a}}{{\cos a}}}}{{1 + \dfrac{{\sin 2a\sin a}}{{\cos 2a\cos a}}}}\)

=\(\dfrac{1}{{\sin a\cos a}}.\dfrac{{\cos acos2a}}{{\cos 2a\cos a + \sin 2a\sin a}}\)

=\(\dfrac{2}{{\sin 2a}}.\dfrac{{\cos acos2a}}{{\cos (2a - a)}} = 2\cot 2a\)