Xét tính đúng sai của các mệnh đề sau:

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

a) Đúng. \({\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) = 2 - {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) - {\sin ^2}\left( {x + \pi } \right)\)

\( \Leftrightarrow {\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) + {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) + {\sin ^2}\left( {x + \pi } \right) = 2\)

Xét biểu thức: \({\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) + {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) + {\sin ^2}\left( {x + \pi } \right)\)

\( = {\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\cos ^2}x + {\sin ^2}\left[ {\left( {x - \frac{\pi }{4}} \right) + \pi } \right] + {\sin ^2}x\)

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

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

b) Sai. \({\sin ^4}x\left( {1 + {{\sin }^2}x} \right) + {\cos ^4}x\left( {1 + {{\cos }^2}x} \right) + 5\left( {404 + {{\sin }^2}x \cdot {{\cos }^2}x} \right)\)

= \({\sin ^4}x + {\sin ^6}x + {\cos ^4}x + {\cos ^6}x + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x\)

\( = \left( {{{\sin }^4}x + {{\cos }^4}x} \right) + \left( {{{\sin }^6}x + {{\cos }^6}x} \right) + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x\)

\( = \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x \cdot {{\cos }^2}x} \right] + \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x \cdot {{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]\)

\( + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x\)

\( = 1 - 2{\sin ^2}x \cdot {\cos ^2}x + 1 - 3{\sin ^2}x \cdot {\cos ^2}x + 2020 + 5{\sin ^2}x \cdot {\cos ^2}x = 2022\).

c) Sai. \({\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x \cdot \cos 2x \cdot \cos 4x\)

\( = \left( {{\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x} \right) \cdot \cos 2x \cdot \cos 4x = \frac{1}{2}\sin 2x \cdot \cos 2x \cdot \cos 4x = \frac{1}{4}\sin 4x \cdot \cos 4x = \frac{1}{8}\sin 8x\)d) Đúng. \(\cos \left( {x - \frac{\pi }{3}} \right) \cdot \cos \left( {x + \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \cdot \cos \left( {x + \frac{{3\pi }}{4}} \right)\)

\( = \frac{1}{2}\left[ {\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right)} \right] + \frac{1}{2}\left[ {\cos \frac{{7\pi }}{{12}} + \cos \left( {2x + \frac{{11\pi }}{{12}}} \right)} \right]\)

\( = \frac{1}{2}\left[ {2\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right) + \cos \left( {2x + \frac{{11\pi }}{{12}}} \right)} \right]\)

\( = \frac{1}{2}\left[ {2\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right) + \cos \left( {2x - \frac{\pi }{{12}} + \pi } \right)} \right]\)

\( = \frac{1}{2}\left[ {2\cos \frac{{7\pi }}{{12}} + \cos \left( {2x - \frac{\pi }{{12}}} \right) - \cos \left( {2x - \frac{\pi }{{12}}} \right)} \right]\)

\( = \cos \frac{{7\pi }}{{12}} = \cos \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3} \cdot \cos \frac{\pi }{4} - \sin \frac{\pi }{3} \cdot \sin \frac{\pi }{4}\)

\( = \frac{1}{2} \cdot \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 3 }}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2} \cdot \left( {\frac{{1 - \sqrt 3 }}{2}} \right)\).