Bài 7 trang 161 SGK Đại số 10

Chứng minh các hệ thức sau:

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:

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