Bài 24 trang 218 SBT Toán Đại số 10: Rút gọn...

Rút gọn

a) ${{1 + \cos a} \over {1 - \cos a}}{\tan ^2}{a \over 2} - {\cos ^2}a$

b) $4{\cos ^4}a - 2\cos 2a - {1 \over 2}\cos 4a$

c) ${\sin ^2}a\left( {1 + {1 \over {\sin a}} + \cot a} \right)\left( {1 - {1 \over {\sin a}} + \cot a} \right)$

d) ${{\cos 2a} \over {{{\cos }^4}a - {{\sin }^4}a}} - {{{{\cos }^4}a + {{\sin }^4}a} \over {1 - {1 \over 2}{{\sin }^2}2a}}$

Gợi ý làm bài

a) 

$\eqalign{ & {{1 + \cos a} \over {1 - \cos a}}{\tan ^2}{a \over 2} - {\cos ^2}a \cr & = {{2{{\cos }^2}{a \over 2}} \over {2{{\sin }^2}{a \over 2}}}{\tan ^2}{a \over 2} - {\cos ^2}a = {\sin ^2}a \cr}$

b) $4{\cos ^4}a - 2\cos 2a - {1 \over 2}\cos 4a$

$= 4{\cos ^4}a - 2(2{\cos ^2}a - 1) - {1 \over 2}(2{\cos ^2}2a - 1)$

$= 4{\cos ^4}a - 4{\cos ^2}a + 2 - {(2{\cos ^2}a - 1)^2} + {1 \over 2}$

$= 4{\cos ^4}a - 4{\cos ^2}a + {5 \over 2} - 4{\cos ^4}a + 4{\cos ^2}a - 1 = {3 \over 2}$

c) ${\sin ^2}a(1 + {1 \over {\sin a}} + \cot a)(1 - {1 \over {\sin a}} + \cot a)$

$\eqalign{ & = {\sin ^2}a\left[ {{{(1 + cota)}^2} - {1 \over {{{\sin }^2}a}}} \right] \cr & = {\sin ^2}a(1 + {\cot ^2}a + 2\cot a) - 1 \cr}$

$\eqalign{ & = {\sin ^2}a + {\cos ^2}a + 2{\sin ^2}a{{\cos a} \over {\sin a}} - 1 \cr & = \sin 2a \cr}$

d) ${{\cos 2a} \over {{{\cos }^4}a - {{\sin }^4}a}} - {{{{\cos }^4}a + {{\sin }^4}a} \over {1 - {1 \over 2}{{\sin }^2}2a}}$

$= {{{{\cos }^2}a - {{\sin }^2}a} \over {({{\cos }^2}a + {{\sin }^2}a)({{\cos }^2}a - {{\sin }^2}a)}} - {{{{\cos }^4}a + {{\sin }^4}a} \over {1 - {1 \over 2}{{(2\sin a\cos a)}^2}}}$

$= 1 - {{{{\cos }^4}a + {{\sin }^4}a} \over {{{\sin }^2}a - si{n^2}aco{s^2}a + {{\cos }^2}a - {{\sin }^2}a{{\cos }^2}a}}$

$= 1 - {{{{\cos }^4}a + {{\sin }^4}a} \over {{{\sin }^2}a(1 - co{s^2}a) + {{\cos }^2}a(1 - {{\sin }^2}a)}} = 0$