Video hướng dẫn giải
- LG a
- LG b
- LG c
- LG d
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:
\[\begin{array}{l}
\tan 2x = \tan \left[ {x + x} \right]\\
=\dfrac{{\tan x + \tan x}}{{1 - \tan x.\tan x}}\\
=\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\\
\Rightarrow\dfrac{{\tan 2x\tan x}}{{\tan 2x - \tan x}}\\
=\dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}.\tan x}}{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} - \tan x}}\\
=\dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}:\left[ {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} - \tan x} \right]\\
=\dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}:\dfrac{{2\tan x - \tan x + {{\tan }^3}x}}{{1 - {{\tan }^2}x}}\\
=\dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}:\dfrac{{\tan x + {{\tan }^3}a}}{{1 - {{\tan }^2}x}}\\
=\dfrac{{2{{\tan }^2}x}}{{1 - {{\tan }^2}x}}.\dfrac{{1 - {{\tan }^2}x}}{{\tan x\left[ {1 + {{\tan }^2}x} \right]}}\\
=\dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}\\
= 2\tan x.\dfrac{1}{{\dfrac{1}{{{{\cos }^2}x}}}}\\
= 2\tan x.{\cos ^2}x\\
= 2.\dfrac{{\sin x}}{{\cos x}}.{\cos ^2}x\\
= 2\sin x\cos x\\
= \sin 2x
\end{array}\]