Bài tập 7 trang 155 SGK Đại số 10

Câu a:

Ta có:

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

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

Câu b:

Ta có:

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

Câu c:

\(1 - 2c{\rm{os}}x = 2\left( {\frac{1}{2} + c{\rm{osx}}} \right) = 2\left( {c{\rm{os}}\frac{\pi }{3} - c{\rm{osx}}} \right)\)

\( = 4\left( {{\rm{cos}}\frac{{\frac{\pi }{3} + x}}{2}} \right).c{\rm{os}}\left( {{\rm{cos}}\frac{{\frac{\pi }{3} - x}}{2}} \right) = 4\cos \left( {\frac{\pi }{6} + \frac{x}{2}} \right).c{\rm{os}}\left( {\frac{\pi }{6} - \frac{x}{2}} \right)\)

Câu d:

 \(1 - 2{\mathop{\rm sinx}\nolimits}  = 2\left( {\frac{1}{2} - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) = 2\left( {\sin \frac{\pi }{6} - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)\)

\( = 2.2\cos \left( {\frac{{\frac{\pi }{6} + x}}{2}} \right).\sin \left( {\frac{{\frac{\pi }{6} - x}}{2}} \right) = 4\cos \left( {\frac{\pi }{{12}} + \frac{x}{2}} \right).\sin \left( {\frac{\pi }{{12}} - \frac{x}{2}} \right)\)

-- Mod Toán 10 HỌC247