Trong các hệ thức sau, hệ thức nào sai?
Hướng dẫn giải:
Đáp án đúng là: C
Xét đáp án A.
Ta có \(4\sin \left( {\frac{x}{2} + 15^\circ } \right)\sin \left( {\frac{x}{2} - 15^\circ } \right)\)
\( = 4 \cdot \frac{1}{2} \cdot \left\{ {co{\mathop{\rm s}\nolimits} \left[ {\left( {\frac{x}{2} + 15^\circ } \right) - \left( {\frac{x}{2} - 15^\circ } \right)} \right] - cos\left[ {\left( {\frac{x}{2} + 15^\circ } \right) + \left( {\frac{x}{2} - 15^\circ } \right)} \right]} \right\}\)
\( = 4 \cdot \frac{1}{2} \cdot \left( {cos30^\circ - c{\rm{os}}x} \right)\)
\( = \sqrt 3 - 2c{\rm{os}}x\).
Vậy đáp án A đúng
Xét đáp án B
\(\frac{{4\sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = \frac{{4 \cdot \frac{1}{2} \cdot \left\{ {co{\mathop{\rm s}\nolimits} \left[ {\left( {x + \frac{\pi }{3}} \right) - \left( {x - \frac{\pi }{3}} \right)} \right] - co{\mathop{\rm s}\nolimits} \left[ {\left( {x + \frac{\pi }{3}} \right) + \left( {x - \frac{\pi }{3}} \right)} \right]} \right\}}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = \frac{{4 \cdot \frac{1}{2} \cdot \left( {co{\mathop{\rm s}\nolimits} \frac{{2\pi }}{3} - co{\mathop{\rm s}\nolimits} 2x} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = \frac{{4 \cdot \frac{1}{2} \cdot \left( { - \frac{1}{2} - co{\mathop{\rm s}\nolimits} 2x} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = \frac{{ - 1 - 2co{\mathop{\rm s}\nolimits} 2x}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = \frac{{ - \left( {si{n^2}x + co{s^2}x} \right) - 2\left( {co{s^2}x - si{n^2}x} \right)}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = \frac{{si{n^2}x - 3co{s^2}x}}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\( = {\tan ^2}x - 3\).
Vậy đáp án B đúng.
Xét đáp án C
cos12xcos2x = \(\frac{1}{2}\left( {cos10x + cos14x} \right)\)
\( = \frac{1}{2}\left( {2{{\cos }^2}5x - 1 + 2co{s^2}7x - 1} \right)\)
\( = co{s^2}5x + co{s^2}7x - 1\).
Vậy đáp án C sai.
Xét đáp án D
\(2\sqrt 2 {\rm{cos}}\frac{x}{2}cos\left( {\frac{x}{2} - \frac{\pi }{4}} \right)\)
\( = 2\sqrt 2 \cdot \frac{1}{2}\left\{ {{\rm{cos}}\left[ {\frac{x}{2} - \left( {\frac{x}{2} - \frac{\pi }{4}} \right)} \right] + {\rm{cos}}\left[ {\frac{x}{2} + \left( {\frac{x}{2} - \frac{\pi }{4}} \right)} \right]} \right\}\)
\( = 2\sqrt 2 \cdot \frac{1}{2}\left[ {{\rm{cos}}\frac{\pi }{4} + {\rm{cos}}\left( {x - \frac{\pi }{4}} \right)} \right]\)
\( = 2\sqrt 2 \cdot \frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} + {\rm{cos}}\left( {x - \frac{\pi }{4}} \right)} \right]\)
\( = 1 + \sqrt 2 {\rm{cos}}\left( {x - \frac{\pi }{4}} \right)\)
\( = 1 + \sqrt 2 \left( {\cos x\cos \frac{\pi }{4} + \sin x\sin \frac{\pi }{4}} \right)\)
\( = 1 + \sqrt 2 \left( {\frac{{\sqrt 2 }}{2}\cos x + \frac{{\sqrt 2 }}{2}\sin x} \right)\)
\( = 1 + \cos x + \sin x\).
Vậy đáp án D đúng.