Bài 24 trang 218 Sách bài tập (SBT) Toán Đại số 10

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

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

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

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

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