Bài 8 trang 161 SGK Đại số 10

LG a

$\displaystyle {{1 + \sin 4a - \cos 4a} \over {1 + \cos 4a + \sin 4a}}$

Lời giải chi tiết:

$\begin{array}{l} \dfrac{{1 + \sin 4a - \cos 4a}}{{1 + \cos 4a + \sin 4a}}\\ = \dfrac{{1 + \sin 4a - \left( {1 - 2{{\sin }^2}2a} \right)}}{{1 + \sin 4a + \left( {2{{\cos }^2}2a - 1} \right)}}\\ = \dfrac{{\sin 4a + 2{{\sin }^2}2a}}{{\sin 4a + 2{{\cos }^2}2a}}\\ = \dfrac{{2\sin 2a\cos 2a + 2{{\sin }^2}2a}}{{2\sin 2a\cos 2a + 2{{\cos }^2}2a}}\\ = \dfrac{{2\sin 2a\left( {\cos 2a + \sin 2a} \right)}}{{2\cos 2a\left( {\cos 2a + \sin 2a} \right)}}\\ = \dfrac{{\sin 2a}}{{\cos 2a}}\\ = \tan 2a \end{array}$

LG b

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

Lời giải chi tiết:

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

LG c

$\displaystyle {{\cos 2x - \sin 4x - \cos 6x} \over {\cos 2x + \sin 4x - \cos 6x}}$

Lời giải chi tiết:

$\begin{array}{l} \dfrac{{\cos 2x - \sin 4x - \cos 6x}}{{\cos 2x + \sin 4x - \cos 6x}}\\ = \dfrac{{\left( {\cos 2x - \cos 6x} \right) - \sin 4x}}{{\left( {\cos 2x - \cos 6x} \right) + \sin 4x}}\\ = \dfrac{{ - 2\sin \dfrac{{2x + 6x}}{2}\sin \dfrac{{2x - 6x}}{2} - \sin 4x}}{{ - 2\sin \dfrac{{2x + 6x}}{2}\sin \dfrac{{2x - 6x}}{2} + \sin 4x}}\\ = \dfrac{{ - 2\sin 4x\sin \left( { - 2x} \right) - \sin 4x}}{{ - 2\sin 4x\sin \left( { - 2x} \right) + \sin 4x}}\\ = \dfrac{{ - 2\sin 4x.\left( { - \sin 2x} \right) - \sin 4x}}{{ - 2\sin 4x.\left( { - \sin 2x} \right) + \sin 4x}}\\ = \dfrac{{2\sin 4x\sin 2x - \sin 4x}}{{2\sin 4x\sin 2x + \sin 4x}}\\ = \dfrac{{\sin 4x\left( {2\sin 2x - 1} \right)}}{{\sin 4x\left( {2\sin 2x + 1} \right)}}\\ = \dfrac{{2\sin 2x - 1}}{{2\sin 2x + 1}} \end{array}$

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