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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) \(4{\cos ^4}a – 2\cos 2a – {1 \over 2}\cos 4a\)
\( = 4{\cos ^4}a – 2(2{\cos ^2}a – 1) – {1 \over 2}(2{\cos ^2}2a – 1)\)
\( = 4{\cos ^4}a – 4{\cos ^2}a + 2 – {(2{\cos ^2}a – 1)^2} + {1 \over 2}\)
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\( = 4{\cos ^4}a – 4{\cos ^2}a + {5 \over 2} – 4{\cos ^4}a + 4{\cos ^2}a – 1 = {3 \over 2}\)
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) \({{\cos 2a} \over {{{\cos }^4}a – {{\sin }^4}a}} – {{{{\cos }^4}a + {{\sin }^4}a} \over {1 – {1 \over 2}{{\sin }^2}2a}}\)
\(= {{{{\cos }^2}a – {{\sin }^2}a} \over {({{\cos }^2}a + {{\sin }^2}a)({{\cos }^2}a – {{\sin }^2}a)}} – {{{{\cos }^4}a + {{\sin }^4}a} \over {1 – {1 \over 2}{{(2\sin a\cos a)}^2}}}\)
\( = 1 – {{{{\cos }^4}a + {{\sin }^4}a} \over {{{\sin }^2}a – si{n^2}aco{s^2}a + {{\cos }^2}a – {{\sin }^2}a{{\cos }^2}a}}\)
\( = 1 – {{{{\cos }^4}a + {{\sin }^4}a} \over {{{\sin }^2}a(1 – co{s^2}a) + {{\cos }^2}a(1 – {{\sin }^2}a)}} = 0\)
Mục lục môn Toán 10 (SBT)