Giả sử \(M = {\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 2x + sin3x + \cos x + \cos 2x + \cos 3x\). Biết biểu thức M có dạng \(a\sqrt b \cos \left( {\frac{x}{2} + \frac{\pi }{6}} \right) \cdo
Đáp án: 6
\(M = {\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 2x + sin3x + \cos x + \cos 2x + \cos 3x\)
\( = \left( {\sin 3x + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) + \left( {\cos 3x + \cos x} \right) + \sin 2x + \cos 2x\)
\( = 2\sin 2x \cdot \cos x + 2\cos 2x \cdot \cos x + \sin 2x + \cos 2x\)
= \(\left( {2\sin 2x\cos x + \sin 2x} \right) + \left( {2\cos 2x\cos x + \cos 2x} \right)\)
\( = \sin 2x\left( {2\cos x + 1} \right) + \cos 2x\left( {2\cos x + 1} \right) = \left( {{\mathop{\rm s}\nolimits} {\rm{in2x}} + \cos 2x} \right) \cdot \left( {2\cos x + 1} \right)\)
\( = 2\left( {\cos x + \frac{1}{2}} \right) \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right) = 2\left( {\cos x + \cos \frac{\pi }{3}} \right) \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right)\)
\( = 2\left[ {2\cos \left( {\frac{{x + \frac{\pi }{3}}}{2}} \right) \cdot \cos \left( {\frac{{x - \frac{\pi }{3}}}{2}} \right)} \right] \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right)\)
\( = 4\cos \left( {\frac{x}{2} - \frac{\pi }{6}} \right) \cdot \cos \left( {\frac{x}{2} + \frac{\pi }{6}} \right) \cdot \sqrt 2 \cos \left( {2x - \frac{\pi }{4}} \right)\).
Vậy a + b = 6.