Chứng minh các đồng nhất thức
\(\begin{array}{l}
a)\frac{{1 - \cos x + \cos 2x}}{{\sin 2x - \sin x}} = \cot x\\
b)\frac{{\sin x + \sin \frac{x}{2}}}{{1 + \cos x + \cos \frac{x}{2}}} = \tan \frac{x}{2}\\
c)\frac{{2\cos 2x - \sin 4x}}{{2\cos 2x + \sin 4x}} = {\tan ^2}\left( {\frac{\pi }{4} - x} \right)\\
d)\tan x - \tan y = \frac{{\sin \left( {x - y} \right)}}{{\cos x.\cos y}}
\end{array}\)
\(\begin{array}{l}
{\rm{a)}}1 - \cos x + \cos 2x = 1 + \cos 2x - \cos x = 2{\cos ^2}x - \cos x = \cos x\left( {2\cos x - 1} \right)\\
\sin 2x - \sin x = 2\sin x\cos x - \sin x = \sin x\left( {2\cos x - 1} \right)\\
\Rightarrow \frac{{1 - \cos x + \cos 2x}}{{\sin 2x - \sin x}} = \frac{{\cos x}}{{\sin x}} = \cot x
\end{array}\)
\( \begin{array}{l}
b) \sin x + \sin \frac{x}{2} = 2\sin \frac{x}{2}\cos \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2}\left( {2\cos \frac{x}{2} + 1} \right)\\
1 + \cos x + \cos \frac{x}{2} = 2{\cos ^2}\frac{x}{2} + \cos \frac{x}{2} = \cos \frac{x}{2}\left( {2\cos \frac{x}{2} + 1} \right)\\
\Rightarrow \frac{{\sin x + \sin \frac{x}{2}}}{{1 + \cos x + \cos \frac{x}{2}}} = \tan \frac{x}{2}
\end{array}\)
\(\begin{array}{l}
c) 2\cos 2x - \sin 4x = 2\cos 2x - 2\sin 2x.\cos 2x = 2\cos 2x\left( {1 - \sin 2x} \right)\\
2\cos 2x - \sin 4x = 2\cos 2x\left( {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)\\
\Rightarrow \frac{{2\cos 2x - \sin 4x}}{{2\cos 2x + \sin 4x}} = \frac{{1 - \sin 2x}}{{1 + \sin 2x}} = \frac{{\sin \frac{\pi }{2} - \sin 2x}}{{\sin \frac{\pi }{2} + \sin 2x}} = \frac{{2\cos \left( {\frac{\pi }{4} + x} \right)\sin \left( {\frac{\pi }{4} - x} \right)}}{{2\sin \left( {\frac{\pi }{4} + x} \right)\sin \left( {\frac{\pi }{4} - x} \right)}} = \cot \left( {\frac{\pi }{4} + x} \right)\tan \left( {\frac{\pi }{4} - x} \right)
\end{array}\)
Mà \(\left( {\frac{\pi }{4} + x} \right) + \left( {\frac{\pi }{4} - x} \right) = \frac{\pi }{2} \Rightarrow \cot \left( {\frac{\pi }{4} + x} \right) = \tan \left( {\frac{\pi }{4} - x} \right)\)
\( \Rightarrow \frac{{2\cos 2x - \sin 4x}}{{2\cos 2x + \sin 4x}}{\tan ^2}\left( {\frac{\pi }{4} - x} \right)\)
\( \begin{array}{l}
d) \tan x - \tan y = \frac{{\sin x}}{{\cos x}} - \frac{{\sin y}}{{\cos y}} = \frac{{\sin x\cos y - \sin y\cos x}}{{\cos x.\cos y}} = \frac{{\sin \left( {x - y} \right)}}{{\cos x\cos y}}\\
\Rightarrow \tan x - \tan y = \frac{{\sin \left( {x - y} \right)}}{{\cos x\cos y}}
\end{array}\)
-- Mod Toán 10