Gợi ý làm bài
a)
\(\eqalign{
& {{\tan 2\alpha } \over {\tan 4\alpha - \tan 2\alpha }} = {{\tan 2\alpha } \over {{{2\tan 2\alpha } \over {1 - {{\tan }^2}\alpha }} - \tan 2\alpha }} \cr
& = {{1 - {{\tan }^2}2\alpha } \over {1 + {{\tan }^2}2\alpha }} = \cos 4\alpha \cr} \)
b)
\(\eqalign{
& \sqrt {1 + \sin \alpha } - \sqrt {1 - \sin \alpha } \cr
& = \sqrt {{{\left( {cos{\alpha \over 2} + sin{\alpha \over 2}} \right)}^2}} - \sqrt {{{\left( {cos{\alpha \over 2} - sin{\alpha \over 2}} \right)}^2}} \cr} \)
Vì \(0 < \alpha < {\pi \over 2}\) nên \(0 < {\alpha \over 2} < {\pi \over 4}\)
Suy ra \(0 < \sin {\alpha \over 2} < \cos {\alpha \over 2}\)
Vậy \(\sqrt {1 + \sin \alpha } - \sqrt {1 - \sin \alpha } = cos{\alpha \over 2} + sin{\alpha \over 2} - (cos{\alpha \over 2} - sin{\alpha \over 2})\)
\( = 2sin{\alpha \over 2}\)
c) \({{3 - 4\cos 2\alpha + c{\rm{os4}}\alpha } \over {3 + 4\cos 2\alpha + \cos 4\alpha }} = {{3 - 4\cos 2\alpha + 2c{\rm{o}}{{\rm{s}}^2}{\rm{2}}\alpha - 1} \over {3 + 4\cos 2\alpha + 2c{\rm{o}}{{\rm{s}}^2}{\rm{2}}\alpha - 1}}\)
\( = {{2(c{\rm{o}}{{\rm{s}}^2}{\rm{2}}\alpha - 2\cos 2\alpha + 1)} \over {2(c{\rm{o}}{{\rm{s}}^2}{\rm{2}}\alpha + 2\cos 2\alpha + 1)}}\)
\( = {{{{(\cos 2\alpha - 1)}^2}} \over {{{(\cos 2\alpha + 1)}^2}}} = {{{{( - 2{{\sin }^2}\alpha )}^2}} \over {{{(2{{\cos }^2}\alpha )}^2}}} = {\tan ^4}\alpha \)
d)
\(\eqalign{
& {{\sin \alpha + \sin 3\alpha + \sin 5\alpha } \over {\cos \alpha + \cos 3\alpha + c{\rm{os5}}\alpha }} \cr
& = {{(\sin 5\alpha + \sin \alpha ) + \sin 3\alpha } \over {(\cos 5\alpha + \cos \alpha ) + c{\rm{os3}}\alpha }} \cr} \)
\( = {{\sin 3\alpha (2\cos 2\alpha + 1)} \over {c{\rm{os3}}\alpha (2\cos 2\alpha + 1)}} = \tan 3\alpha \)
