Bài 18 trang 213 SBT đại số 10

Câu hỏi: Rút gọn

Câu 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:
\(\dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - 4{{\sin }^2}\alpha {{\cos }^2}\alpha }}{{4 - {{\sin }^2}2\alpha - 4{{\sin }^2}\alpha }} \) \( = \dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - {{\left( {2\sin \alpha \cos \alpha } \right)}^2}}}{{4\left({1 - {{\sin }^2}\alpha } \right) -{{{\left({2\sin \alpha \cos \alpha } \right)}^2}} }}\)
\(= \dfrac{{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha - {{\sin }^2}2\alpha }}{{4{{\cos }^2}a - 4{{\sin }^2}\alpha {{\cos }^2}\alpha }}\)
=\(\dfrac{{4{{\sin }^4}\alpha }}{{4\cos ^2\alpha (1 - {{\sin }^2}\alpha)}}\) \( = \dfrac{{{{\sin }^4}\alpha }}{{{{\cos }^4}\alpha }}\) \(= {\tan ^4}\alpha \).

Câu b​

\(3 - 4\cos 2a + \cos 4a\);
Lời giải chi tiết:
\(3 - 4\cos 2a + \cos 4a \) \(= 3 - 4(1 - 2{\sin ^2}a) + (1 - 2{\sin ^2}2a)\)
\( = 3 - 4 + 8{\sin ^2}\alpha + 1 - 2{\left( {2\sin \alpha \cos \alpha } \right)^2}\)
\(=8{\sin ^2}a - 8{\sin ^2}a{\cos ^2}a \)
\(= 8{\sin ^2}a(1 - {\cos ^2}a)\)
\(=8{\sin ^4}a\)

Câu c​

$\cos 4 a-\sin 4 a \cot 2 a$
Lời giải chi tiết:
$\cos 4 a-\sin 4 a \cot 2 a=2 \cos ^{2} 2 a-1-2 \sin 2 a \cos 2 a \frac{\cos 2 a}{\sin 2 a}=-1$

Câu d​

\(\dfrac{{{\mathop{\rm cota}\nolimits} + \tan a}}{{1 + \tan 2a\tan a}}\)
Lời giải chi tiết:
\(\dfrac{{\cot a + \tan a}}{{1 + \tan 2a\tan a}} \) \(= \dfrac{{\dfrac{{\cos a}}{{\sin a}} + \dfrac{{\sin a}}{{\cos a}}}}{{1 + \dfrac{{\sin 2a\sin a}}{{\cos 2a\cos a}}}}\)
=\(\dfrac{1}{{\sin a\cos a}}.\dfrac{{\cos acos2a}}{{\cos 2a\cos a + \sin 2a\sin a}}\)
=\(\dfrac{2}{{\sin 2a}}.\dfrac{{\cos acos2a}}{{\cos (2a - a)}} = 2\cot 2a\)