Bài 6.18 trang 185 sbt đại số 10

LG a

\(\sin \alpha + \cos \alpha \)

Lời giải chi tiết:

Vì\(\pi < \alpha < {{3\pi } \over 2}\)

Nên\(\cos \alpha < 0,\sin \alpha < 0\) và\(\tan \alpha > 0\)

Ta có: \(\tan \alpha - 3\cot \alpha = 6 \Leftrightarrow \tan \alpha - {3 \over {\tan \alpha }} - 6 = 0\)

\( \Leftrightarrow {\tan ^2}\alpha - 6\tan \alpha - 3 = 0\)

Vì\(\tan \alpha > 0\) nên \(\tan \alpha = 3 + 2\sqrt 3\)

\({\rm{co}}{{\rm{s}}^2}\alpha = {1 \over {1 + {{\tan }^2}\alpha }} = {1 \over {22 + 12\sqrt 3 }}\)

Suy ra\({\rm{cos}}\alpha {\rm{ = - }}{1 \over {\sqrt {22 + 12\sqrt 3 } }},\sin \alpha = - {{3 + 2\sqrt 3 } \over {\sqrt {22 + 12\sqrt 3 } }}.\)

Vậy\(\sin \alpha + c{\rm{os}}\alpha {\rm{ = - }}{{4 + 2\sqrt 3 } \over {\sqrt {22 + 12\sqrt 3 } }}\)

LG b

\({{2\sin \alpha - \tan \alpha } \over {{\rm{cos}}\alpha {\rm{ + cot}}\alpha }}\)

Lời giải chi tiết:

\(\eqalign{
& {{2\sin \alpha - \tan \alpha } \over {{\rm{cos}}\alpha {\rm{ + cot}}\alpha }} = {{\sin \alpha (2 - {1 \over {{\rm{cos}}\alpha }})} \over {{\rm{cos \alpha (1 + }}{1 \over {\sin \alpha }})}} \cr
& = \tan \alpha .{{2\cos \alpha - 1} \over {{\rm{cos}}\alpha }}.{{\sin \alpha } \over {\sin \alpha + 1}} \cr &= {\tan ^2}\alpha .{{2\cos \alpha - 1} \over {\sin \alpha + 1}} \cr} \)

\(\eqalign{
& ={(3 + 2\sqrt 3 )^2}.{{ - {2 \over {\sqrt {22 + 12\sqrt 3 } }}-1} \over { - {{3 + 2\sqrt 3 } \over {\sqrt {22 + 12\sqrt 3 } }} + 1}} \cr
& = (21 + 12\sqrt 3 ).{{2 + \sqrt {22 + 12\sqrt 3 } } \over {3 + 2\sqrt 3 - \sqrt {22 + 12\sqrt 3 } }} \cr} \)