Câu hỏi 4 - Mục Bài tập trang 19

1. Nội dung câu hỏi

Chứng minh các đẳng thức lượng giác sau:

a) \(4\cos x\cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right) = \cos 3x\);

b) \(\frac{{\sin 2x\cos x}}{{\left( {1 + \cos x} \right)\left( {1 + \cos 2x} \right)}} = \tan \frac{x}{2}\);

c) \(\sin x\left( {1 + 2\cos 2x + 2\cos 4x + 2\cos 6x} \right) = \sin 7x\);

d) \(\frac{{{{\sin }^2}3x}}{{{{\sin }^2}x}} - \frac{{{{\cos }^2}3x}}{{{{\cos }^2}x}} = 8\cos 2x\).

3. Lời giải chi tiết 

a) \(4\cos x\cos \left( {\frac{\pi }{3} - x} \right)\cos \left( {\frac{\pi }{3} + x} \right) \) \( = 2\cos x\left( {\cos \frac{{2\pi }}{3} + \cos 2x} \right)\)

\( \) \( = 2\cos x.\cos 2x + 2.\frac{{ - 1}}{2}\cos x \) \( = \cos 3x + \cos x - \cos x \) \( = \cos 3x\)

b) \(\frac{{\sin 2x\cos x}}{{\left( {1 + \cos x} \right)\left( {1 + \cos 2x} \right)}} \) \( = \frac{{2\sin x{{\cos }^2}x}}{{\left( {1 + \cos x} \right)\left( {1 + 2{{\cos }^2}x - 1} \right)}} \) \( = \frac{{2\sin x{{\cos }^2}x}}{{\left( {1 + \cos x} \right)2{{\cos }^2}x}}\)

\( \) \( = \frac{{\sin x}}{{1 + \cos x}} \) \( = \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{1 + 2{{\cos }^2}\frac{x}{2} - 1}} \) \( = \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}} \) \( = \tan \frac{x}{2}\)

c) \(\sin x\left( {1 + 2\cos 2x + 2\cos 4x + 2\cos 6x} \right)\)

\( \) \( = \sin x + 2\sin x\cos 2x + 2\sin x\cos 4x + 2\sin x\cos 6x\)

\( \) \( = \sin x + \sin 3x - \sin x + \sin 5x - \sin 3x + \sin 7x - \sin 5x\)\( \) \( = \sin 7x\)

d) \(\frac{{{{\sin }^2}3x}}{{{{\sin }^2}x}} - \frac{{{{\cos }^2}3x}}{{{{\cos }^2}x}} \) \( = \frac{{{{\sin }^2}3x{{\cos }^2}x - {{\cos }^2}3x{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} \) \( = \frac{{{{\left( {\sin 3x\cos x} \right)}^2} - {{\left( {\cos 3x\sin x} \right)}^2}}}{{{{\sin }^2}x{{\cos }^2}x}}\)

\( \) \( = \frac{{\left( {\sin 3x\cos x + \cos 3x\sin x} \right)\left( {\sin 3x\cos x - \cos 3x\sin x} \right)}}{{{{\sin }^2}x{{\cos }^2}x}} \) \( = \frac{{\sin 4x\sin 2x}}{{\frac{1}{4}{{\sin }^2}2x}}\)

\( \) \( = \frac{{4\sin 4x}}{{\sin 2x}} \) \( = \frac{{8\sin 2x\cos 2x}}{{\sin 2x}} \) \( = 8\cos 2x\).