Bài 21 trang 214 SBT đại số 10

LG a

$4{\cos ^4}a - 2\cos 2a - \dfrac{1}{2}\cos 4a$;

Lời giải chi tiết:

$4{\cos ^4}a - 2\cos 2a - \dfrac{1}{2}\cos 4a$

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

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

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

LG b

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

Lời giải chi tiết:

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

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

=${\sin ^2}a + {\cos ^2}a + 2{\sin ^2}a\dfrac{{\cos a}}{{\sin a}} - 1$ $= \sin 2a$

LG c

$\dfrac{{\cos 2a}}{{{{\cos }^4}a - {{\sin }^4}a}} - \dfrac{{{{\cos }^4}a + {{\sin }^4}a}}{{1 - \dfrac{1}{2}{{\sin }^2}2a}}$.

Lời giải chi tiết:

$\dfrac{{\cos 2a}}{{{{\cos }^4}a - {{\sin }^4}a}} - \dfrac{{{{\cos }^4}a + {{\sin }^4}a}}{{1 - \dfrac{1}{2}{{\sin }^2}2a}}$

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

$\begin{array}{l} = 1 - \dfrac{{{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right)}^2} - 2{{\sin }^2}a{{\cos }^2}a}}{{1 - 2{{\sin }^2}a{{\cos }^2}a}}\\ = 1 - \dfrac{{1 - 2{{\sin }^2}a{{\cos }^2}a}}{{1 - 2{{\sin }^2}a{{\cos }^2}a}}\\ = 1 - 1 = 0 \end{array}$

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