Xét tính đúng sai của các mệnh đề sau:

Đáp án: a) Sai. b) Đúng. c) Sai. d) Đúng.

a) Sai. \(\frac{{\sin 2x + \sin 4x + \sin 6x}}{{2\left( {1 - \cos x} \right)}}\)

\( = \frac{{\left( {\sin 2x + \sin 6x} \right) + \sin 4x}}{{2\left( {1 - \cos x} \right)}} = \frac{{2\sin \frac{{2x + 6x}}{2} \cdot \cos \frac{{2x - 6x}}{2} + \sin 4x}}{{2\left( {1 - \cos x} \right)}}\)

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

b) Đúng. \(\frac{{1 + \cos x + \cos 2x + \cos 3x}}{{2{{\cos }^2}x + \cos x - 1}}\)

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

\( = \frac{{2\cos x \cdot \left( {\cos x + \cos 2x} \right)}}{{\cos x + \cos 2x}} = 2\cos x\).

c) Sai. \(\frac{{\cos x - \cos 2x + \cos 3x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} - \sin 2x + \sin 3x}}\)

\( = \frac{{\left( {\cos x + \cos 3x} \right) - \cos 2x}}{{\left( {{\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 3x} \right) - \sin 2x}} = \frac{{2\cos 2x\cos x - \cos 2x}}{{2\sin 2x\cos x - \sin 2x}} = \frac{{\cos 2x\left( {2\cos x - 1} \right)}}{{\sin 2x\left( {2\sin x - 1} \right)}} = \cot 2x\)

d) Đúng. \(D = \sin a + \sin b + \sin c - \sin \left( {a + b + c} \right)\)

\( = \left( {\sin a + \sin b} \right) - \left[ {\sin \left( {a + b + c} \right) - \sin c} \right]\)

\( = 2\sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} - \left[ {2\cos \frac{{\left( {c + a + b + c} \right)}}{2} \cdot \sin \frac{{\left( {a + b + c - c} \right)}}{2}} \right]\)

\( = 2\sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} - 2\cos \frac{{a + b + 2c}}{2} \cdot \sin \frac{{a + b}}{2}\)

\( = 2\sin \frac{{a + b}}{2} \cdot \left( {\cos \frac{{a - b}}{2} - \cos \frac{{a + b + 2c}}{2}} \right)\)

\( = 2\sin \frac{{a + b}}{2}\left[ { - 2 \cdot \sin \frac{{\frac{{a - b + a + b + 2c}}{2}}}{2} \cdot \sin \frac{{\frac{{a - b - 2c - a - b}}{2}}}{2}} \right]\)

\( = 2\sin \frac{{a + b}}{2} \cdot \left( { - 2} \right) \cdot \sin \frac{{a + c}}{2} \cdot \sin \frac{{ - \left( {b + c} \right)}}{2}\)

\( = 4\sin \frac{{a + b}}{2} \cdot \sin \frac{{b + c}}{2} \cdot \sin \frac{{c + a}}{2}\).