Đề bài
\[{{\sin {\pi \over {15}}\cos {\pi \over 10} + \sin {\pi \over {10}}\cos {\pi \over 15}} \over {\cos {{2\pi } \over {15}}\cos {\pi \over {5}} - \sin {{2\pi } \over {15}}\sin {\pi \over {5}}}}\]bằng:
\[[A]\,\sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;[B]\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\]
\[[C]\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[D]\, {1 \over 2}\]
Lời giải chi tiết
Ta có:
\[\eqalign{
& {{\sin {\pi \over {15}}\cos {\pi \over 10} + \sin {\pi \over {10}}\cos {\pi \over 15}} \over {\cos {{2\pi } \over {15}}\cos {\pi \over {5}} - \sin {{2\pi } \over {15}}\sin {\pi \over {5}}}} \cr &= {{\sin [{\pi \over {15}} + {\pi \over {10}}]} \over {\cos [{{2\pi } \over {15}} + {\pi \over 5}]}} \cr
& = {{\sin {\pi \over 6}} \over {\cos {\pi \over 3}}} = 1 \cr} \]
Chọn [B]