Nội dung bài giảng
Không sử dụng bảng số và máy tính, hãy tính
a) \({\sin ^4}{\pi \over {16}} + {\sin ^4}{{3\pi } \over {16}} + {\sin ^4}{{5\pi } \over {16}} + {\sin ^4}{{7\pi } \over {16}}\)
b) \(\cot 7,{5^0} + \tan 67,{5^0} - \tan 7,{5^0} - \cot 67,{5^0}\)
Gợi ý làm bài
a) \({\sin ^4}{\pi \over {16}} + {\sin ^4}{{3\pi } \over {16}} + {\sin ^4}{{5\pi } \over {16}} + {\sin ^4}{{7\pi } \over {16}}\)
\( = {\left( {{{1 - \cos {\pi \over 8}} \over 2}} \right)^2} + {\left( {{{1 - \cos {{3\pi } \over 8}} \over 2}} \right)^2} + {\left( {{{1 - \cos {{5\pi } \over 8}} \over 2}} \right)^2} + {\left( {{{1 - \cos {{7\pi } \over 8}} \over 2}} \right)^2}\)
\( = {1 \over 4}\left( {1 - 2\cos {\pi \over 8} + {{\cos }^2}{\pi \over 8} + 1 - 2\cos {{3\pi } \over 8} + {{\cos }^2}{{3\pi } \over 8} + 1 - 2\cos {{5\pi } \over 8} + {{\cos }^2}{{5\pi } \over 8} + 1 - 2\cos {{7\pi } \over 8} + {{\cos }^2}{{7\pi } \over 8}} \right)\)
\( = 1 - {1 \over 2}\left( {\cos {\pi \over 8} + \cos {{3\pi } \over 8} + \cos {{5\pi } \over 8} + \cos {{7\pi } \over 8}} \right) + {1 \over 4}\left( {{{1 + \cos {\pi \over 4}} \over 2} + {{1 + \cos {{3\pi } \over 4}} \over 2} + {{1 + \cos {{5\pi } \over 4}} \over 2} + {{1 + \cos {{7\pi } \over 4}} \over 2}} \right)$\)
=\(1 - {1 \over 2}\left( {\cos {\pi \over 8} + \cos {{3\pi } \over 8} - \cos {{3\pi } \over 8} - \cos {\pi \over 8}} \right) + {1 \over 8}\left( {4 + {{\sqrt 2 } \over 2} - {{\sqrt 2 } \over 2} - {{\sqrt 2 } \over 2} + {{\sqrt 2 } \over 2}} \right)\)
= \({3 \over 2}\)
b) \(\cot 7,{5^0} + \tan 67,{5^0} - \tan 7,{5^0} - \cot 67,{5^0}\)
= \({{\cos 7,{5^0}} \over {\sin 7,{5^0}}} - {{\sin 7,{5^0}} \over {\cos 7,{5^0}}} + {{\sin 67,{5^0}} \over {\cos 67,{5^0}}} - {{\cos 67,{5^0}} \over {\sin 67,{5^0}}}\)
= \({{{{\cos }^2}7,{5^0} - {{\sin }^2}7,{5^0}} \over {\sin 7,{5^0}\cos 7,{5^0}}} + {{{{\sin }^2}67,{5^0} - {{\cos }^2}67,{5^0}} \over {sin67,{5^0}\cos 67,{5^0}}}\)
= \(\eqalign{
& {{\cos {{15}^0}} \over {{1 \over 2}\sin {{15}^0}}} - {{\cos {{135}^0}} \over {{1 \over 2}\sin {{135}^0}}} \cr
& = {{2(\sin {{135}^0}\cos {{15}^0} - \cos {{135}^0}\sin {{15}^0})} \over {\sin {{15}^0}\sin {{135}^0}}} \cr} \)
= \({{\sin ({{135}^0} - {{15}^0})} \over {\sin ({{45}^0} - {{30}^0})\sin ({{180}^0} - {{45}^0})}}\)
= \({{2\sin {{120}^0}} \over {(\sin {{45}^0}\cos {{30}^0} - \cos {{45}^0}\sin {{30}^0})sin{{45}^0}}}\)
\(\eqalign{
& = {{\sqrt 3 } \over {{{\sqrt 2 } \over 2}({{\sqrt 3 } \over 2} - {1 \over 2}).{{\sqrt 2 } \over 2}}} \cr
& = {{4\sqrt 3 } \over {\sqrt 3 - 1}} = 6 + 2\sqrt 3 \cr} $\)