Rút gọn các biểu thức (không dùng bảng số và máy tính)
LG a
\({\sin ^2}({180^0} - \alpha )\) \( + \tan ^2({180^0} - \alpha ){\tan ^2}({270^0} - \alpha ) \) \( + \sin ({90^0} + \alpha )\cos(\alpha - {360^0})\)
Lời giải chi tiết:
\({\sin ^2}({180^0} - \alpha ) \) \( + \tan ^2({180^0} - \alpha ){\tan ^2}({270^0} - \alpha ) \) \( + \sin ({90^0} + \alpha )\cos(\alpha - {360^0})\)
= \({\sin ^2}\alpha + {\tan ^2}\alpha {\cot ^2}\alpha + {\cos ^2}\alpha = 2\)
LG b
\({{\cos (\alpha - {{90}^0})} \over {\sin ({{180}^0} - \alpha )}} + {{\tan (\alpha - {{180}^0})c{\rm{os(18}}{{\rm{0}}^0} + \alpha )\sin ({{270}^0} + \alpha )} \over {\tan ({{270}^0} + \alpha )}}\)
Lời giải chi tiết:
\({{\cos (\alpha - {{90}^0})} \over {\sin ({{180}^0} - \alpha )}} + {{\tan (\alpha - {{180}^0})c{\rm{os(18}}{{\rm{0}}^0} + \alpha )\sin ({{270}^0} + \alpha )} \over {\tan ({{270}^0} + \alpha )}}\)
\(\begin{array}{l}
= \frac{{\cos \left( {{{90}^0} - \alpha } \right)}}{{\sin \alpha }} + \frac{{\tan \alpha .\left( { - \cos \alpha } \right)\sin \left( { - {{90}^0} + \alpha } \right)}}{{\tan \left( {{{90}^0} + \alpha } \right)}}\\
= \frac{{\sin \alpha }}{{\sin \alpha }} + \frac{{\tan \alpha .\left( { - \cos \alpha } \right)\left[ { - \sin \left( {{{90}^0} - \alpha } \right)} \right]}}{{\tan \left[ {{{90}^0} - \left( { - \alpha } \right)} \right]}}\\
= 1 + \frac{{\tan \alpha .\left( { - \cos \alpha } \right)\left( { - \cos \alpha } \right)}}{{\cot \left( { - \alpha } \right)}}\\
= 1 + \frac{{\frac{{\sin \alpha }}{{\cos \alpha }}.{{\cos }^2}\alpha }}{{ - \cot \alpha }}\\
= 1 - \frac{{\sin \alpha \cos \alpha }}{{\frac{{\cos \alpha }}{{\sin \alpha }}}}\\
= 1 - {\sin ^2}\alpha \\
= {\cos ^2}\alpha
\end{array}\)
LG c
\({{\cos ( - {{288}^0})cot{{72}^0}} \over {tan( - {{162}^0})\sin {{108}^0}}} + \tan {18^0}\)
Lời giải chi tiết:
\({{\cos ( - {{288}^0})cot{{72}^0}} \over {tan( - {{162}^0})\sin {{108}^0}}} + \tan {18^0}\)
\( = {{\cos ({{72}^0} - {{360}^0})\cot {{72}^0}} \over {\tan ({{18}^0} - {{180}^0})\sin ({{180}^0} - {{72}^0})}} - \tan {18^0}\)
= \({{{\rm{cos7}}{{\rm{2}}^0}\cot {{72}^0}} \over {\tan {{18}^0}\sin {{72}^0}}} - \tan {18^0}\)
= \({{{{\cot }^2}{{72}^0}} \over {\tan {{18}^0}}} - \tan {18^0} \) \(= {{{{\tan }^2}{{18}^0}} \over {\tan {{18}^0}}} - \tan {18^0} = 0\)
LG d
\({{\sin {{20}^0}\sin {\rm{3}}{{\rm{0}}^0}\sin {{40}^0}\sin {{50}^0}\sin {{60}^0}\sin {{70}^0}} \over {cos{{10}^0}{\rm{cos5}}{{\rm{0}}^0}}}\)
Lời giải chi tiết:
Ta có: \(\sin {70^0} = \cos {20^0},\sin {50^0} = \cos 4{{\rm{0}}^0};\) \(\sin {40^0} = cos{50^0}\).
Vì vậy
\({{\sin {{20}^0}\sin {\rm{3}}{{\rm{0}}^0}\sin {{40}^0}\sin {{50}^0}\sin {{60}^0}\sin {{70}^0}} \over {cos{{10}^0}{\rm{cos5}}{{\rm{0}}^0}}}\)
= \(\eqalign{
& {{{1 \over 2}.{{\sqrt 3 } \over 2}.\sin {{20}^0}\cos {\rm{2}}{{\rm{0}}^0}\cos {{50}^0}\cos {{40}^0}} \over {cos{{10}^0}{\rm{cos5}}{{\rm{0}}^0}}} \cr
& = {{{1 \over 2}.{{\sqrt 3 } \over 4}\sin {{40}^0}.cos{{40}^0}} \over {{\rm{cos1}}{{\rm{0}}^0}}} \cr} \)
= \({{{{\sqrt 3 } \over {16}}\sin {{80}^0}} \over {cos{{10}^0}}} = {{\sqrt 3 } \over {16}}\)
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