Xét tính đúng sai của mệnh đề sai.

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

a) Đúng. \(\cot x - \tan x - 2\tan 2x\)

\( = \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x}} - 2\tan 2x = \frac{{\cos 2x}}{{\frac{1}{2}\sin 2x}} - 2\tan 2x\)

\( = 2 \cdot \frac{{\cos 2x}}{{\sin 2x}} - 2\tan 2x = 2\cot 2x - 2\tan 2x = 2\left( {\cot 2x - \tan 2x} \right)\)

\( = 2 \cdot \left( {2\cot 4x} \right) = 4\cot 4x\).

b) Đúng. sin²(a + b) – sin²a – sin²b

= sin²(a + b) – (sin²a + sin²b) = sin²(a + b) − \(\frac{{1 - \cos 2a}}{2} - \frac{{1 - \cos 2b}}{2}\)

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

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

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

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

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

\( = \cos \left( {a + b} \right) \cdot \left( {2\sin a \cdot \sin b} \right) = 2\sin a \cdot \sin b \cdot \cos \left( {a + b} \right)\).

c) Sai. \(\sin \alpha \cdot \sin \left( {\frac{\pi }{3} - \alpha } \right) \cdot \sin \left( {\frac{\pi }{3} + \alpha } \right)\)

= \(\sin \alpha \cdot \left[ {\sin \left( {\frac{\pi }{3} - \alpha } \right) \cdot \sin \left( {\frac{\pi }{3} + \alpha } \right)} \right]\)

\( = \sin \alpha \cdot \frac{1}{2} \cdot \left[ {\cos \left( {\frac{\pi }{3} - \alpha - \frac{\pi }{3} - \alpha } \right) - \cos \left( {\frac{\pi }{3} - \alpha + \frac{\pi }{3} + \alpha } \right)} \right] = \sin \alpha \cdot \frac{1}{2} \cdot \left( {\cos 2\alpha - \cos \frac{{2\pi }}{3}} \right)\)

\(\)\( = \sin \alpha \cdot \frac{1}{2} \cdot \left[ {\cos 2\alpha - \left( { - \frac{1}{2}} \right)} \right] = \sin \alpha \cdot \left( {\frac{1}{2}\cos 2\alpha + \frac{1}{4}} \right) = \frac{1}{2}\sin \alpha \cdot \cos 2\alpha + \frac{1}{4}\sin \alpha \)

\( = \frac{1}{2} \cdot \frac{1}{2}\left( {\sin 3\alpha - \sin \alpha } \right) + \frac{1}{4}\sin \alpha = \frac{1}{4}\sin 3\alpha - \frac{1}{4}\sin \alpha + \frac{1}{4}\sin \alpha = \frac{1}{4}\sin 3\alpha \).

d) Sai. cos(a + b)∙sin(a – b) + cos(b + c)∙sin(b – c) + cos(a + c)∙sin(c – a)

\( = \frac{1}{2}\left( {\sin 2a - \sin 2b} \right) + \frac{1}{2}\left( {\sin 2b - \sin 2c} \right) + \frac{1}{2}\left( {\sin 2c - \sin 2a} \right)\)

\( = \frac{1}{2}\left( {\sin 2a - \sin 2b + \sin 2b - \sin 2c + \sin 2c - \sin 2a} \right) = \frac{1}{2} \cdot 0 = 0\).