Gợi ý làm bài
a) \({{\sin 2\alpha + \sin \alpha } \over {1 + c{\rm{os2}}\alpha {\rm{ + cos}}\alpha }} = {{\sin \alpha (2\cos \alpha + 1)} \over {2c{\rm{o}}{{\rm{s}}^2}\alpha {\rm{ + cos}}\alpha }}\)
= \({{\sin \alpha (2\cos \alpha + 1)} \over {c{\rm{os}}\alpha (2{\rm{cos}}\alpha + 1)}} = \tan \alpha \)
b) \({{4{{\sin }^2}\alpha } \over {1 - c{\rm{o}}{{\rm{s}}^2}{\alpha \over 2}}} = {{16{{\sin }^2}{\alpha \over 2}{{\cos }^2}{\alpha \over 2}} \over {{{\sin }^2}{\alpha \over 2}}} = 16{\cos ^2}{\alpha \over 2}\)
c) \({{1 + c{\rm{os}}\alpha - \sin \alpha } \over {1 - c{\rm{os}}\alpha - {\rm{sin}}\alpha }} = {{2{{\cos }^2}{\alpha \over 2} - 2\sin {\alpha \over 2}\cos {\alpha \over 2}} \over {2si{n^2}{\alpha \over 2} - 2\sin {\alpha \over 2}\cos {\alpha \over 2}}}\)
= \({{2\cos {\alpha \over 2}(\cos {\alpha \over 2} - \sin {\alpha \over 2})} \over {2\sin {\alpha \over 2}(sin{\alpha \over 2} - \cos {\alpha \over 2})}} = - \cot {\alpha \over 2}\)
d) \({{1 + \sin \alpha - 2{{\sin }^2}({{45}^0} - {\alpha \over 2})} \over {4c{\rm{os}}{\alpha \over 2}}} = {{\sin \alpha + \cos ({{90}^0} - \alpha )} \over {4\cos {\alpha \over 2}}}\)
=\({{\sin \alpha + \sin \alpha } \over {4\cos {\alpha \over 2}}} = {{4\sin {\alpha \over 2}\cos {\alpha \over 2}} \over {4\cos {\alpha \over 2}}} = \sin {\alpha \over 2}\)
