Chứng minh: \(\sin 5x\left( {1 - 4{{\sin }^2}x} \right) - \sin 3x - 2\sin x.\cos 6x = 0\)
Chứng minh: \(\sin 5x\left( {1 - 4{{\sin }^2}x} \right) - \sin 3x - 2\sin x.\cos 6x = 0\)
\(\sin 5x\left( {1 - 4{{\sin }^2}x} \right) - \sin 3x - 2\sin x.\cos 6x = 0\)
\( \Leftrightarrow VT = \sin 5x\left( {1 - 4.\dfrac{{1 - \cos 2x}}{2}} \right) - \sin 3x - 2\sin x.\cos 6x\)
\( \Leftrightarrow VT = \sin 5x\left( {2\cos 2x - 1} \right) - \sin 3x - 2\sin x.\cos 6x\)
\( \Leftrightarrow VT = 2\sin 5x.\cos 2x - \sin 5x - \sin 3x - 2\sin x.\cos 6x\)
\( \Leftrightarrow VT = 2.\dfrac{1}{2}\left[ {\sin \left( {5x - 2x} \right) + \sin \left( {5x + 2x} \right)} \right] - \sin 5x - \sin 3x - 2.\dfrac{1}{2}\left[ {\sin \left( {x - 6x} \right) + \sin \left( {x + 6x} \right)} \right]\)
\( \Leftrightarrow VT = \sin 3x + \sin 7x - \sin 5x - \sin 3x - \sin \left( { - 5x} \right) - \sin 7x\)
\( \Leftrightarrow VT = - \sin 5x + \sin 5x = 0 = VP\left( {dpcm} \right)\)
Hỗ trợ - Hướng dẫn
![](/themes/images/call.png)
-
024.7300.7989
-
1800.6947
(Thời gian hỗ trợ từ 7h đến 22h)
Email: lienhe@tuyensinh247.com