Bài 7 trang 161 SGK Đại số 10

LG a

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

Lời giải chi tiết:

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

LG b

$\displaystyle {{\sin a + \sin 3a + \sin 5a} \over {\cos a + \cos 3a + \cos 5a}} = \tan 3a$

Lời giải chi tiết:

$\begin{array}{l} \dfrac{{\sin a + \sin 3a + \sin 5a}}{{\cos a + \cos 3a + \cos 5a}}\\ = \dfrac{{\left( {\sin 5a + \sin a} \right) + \sin 3a}}{{\left( {\cos 5a + \cos a} \right) + \cos 3a}}\\ = \dfrac{{2\sin \dfrac{{5a + a}}{2}\cos \dfrac{{5a - a}}{2} + \sin 3a}}{{2\cos \dfrac{{5a + a}}{2}\cos \dfrac{{5a - a}}{2} + \cos 3a}}\\ = \dfrac{{2\sin 3a\cos 2a + \sin 3a}}{{2\cos 3a\cos 2a + \cos 3a}}\\ = \dfrac{{\sin 3a\left( {2\cos 2a + 1} \right)}}{{\cos 3a\left( {2\cos 2a + 1} \right)}}\\ = \dfrac{{\sin 3a}}{{\cos 3a}}\\ = \tan 3a \end{array}$

LG c

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

Lời giải chi tiết:

$\begin{array}{l} \dfrac{{{{\sin }^4}a - {{\cos }^4}a + {{\cos }^2}a}}{{2\left( {1 - \cos a} \right)}}\\ = \dfrac{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right)\left( {{{\sin }^2}a - {{\cos }^2}a} \right) + {{\cos }^2}a}}{{2\left( {1 - \cos a} \right)}}\\ = \dfrac{{{{\sin }^2}a - {{\cos }^2}a + {{\cos }^2}a}}{{2\left( {1 - \cos a} \right)}}\\ = \dfrac{{{{\sin }^2}a}}{{2\left[ {1 - \left( {1 - 2{{\sin }^2}\dfrac{a}{2}} \right)} \right]}}\\ = \dfrac{{{{\left( {2\sin \dfrac{a}{2}\cos \dfrac{a}{2}} \right)}^2}}}{{2.2{{\sin }^2}\dfrac{a}{2}}}\\ = \dfrac{{4{{\sin }^2}\dfrac{a}{2}{{\cos }^2}\dfrac{a}{2}}}{{4{{\sin }^2}\dfrac{a}{2}}}\\ = {\cos ^2}\dfrac{a}{2} \end{array}$

LG d

 $\displaystyle {{\tan 2x\tan x} \over {\tan 2x - \tan x}} = \sin 2x$

Lời giải chi tiết:

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

Cách khác:

$\begin{array}{l} \tan 2x = \tan \left( {x + x} \right)\\ = \dfrac{{\tan x + \tan x}}{{1 - \tan x.\tan x}}\\ = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\\ \Rightarrow \dfrac{{\tan 2x\tan x}}{{\tan 2x - \tan x}}\\ = \dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}.\tan x}}{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} - \tan x}}\\ = \dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}:\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} - \tan x} \right)\\ = \dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}:\dfrac{{2\tan x - \tan x + {{\tan }^3}x}}{{1 - {{\tan }^2}x}}\\ = \dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}:\dfrac{{\tan x + {{\tan }^3}a}}{{1 - {{\tan }^2}x}}\\ = \dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}.\dfrac{{1 - {{\tan }^2}x}}{{\tan x\left( {1 + {{\tan }^2}x} \right)}}\\ = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}\\ = 2\tan x.\dfrac{1}{{\dfrac{1}{{{{\cos }^2}x}}}}\\ = 2\tan x.{\cos ^2}x\\ = 2.\dfrac{{\sin x}}{{\cos x}}.{\cos ^2}x\\ = 2\sin x\cos x\\ = \sin 2x \end{array}$

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