Bài 62 trang 219 SGK Đại số 10 Nâng cao

Đề bài

\({{\sin {\pi  \over {15}}\cos {\pi  \over 10} + \sin {\pi  \over {10}}\cos {\pi  \over 15}} \over {\cos {{2\pi } \over {15}}\cos {\pi  \over {15}} - \sin {{2\pi } \over {15}}\sin {\pi  \over {15}}}}\) bằng:

\((A)\,\sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;(B)\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\)

\((C)\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(D)\,  {1 \over 2}\)

Hướng dẫn giải

Ta có:

\(\eqalign{
& {{\sin {\pi \over {15}}\cos {\pi \over 10} + \sin {\pi \over {10}}\cos {\pi \over 15}} \over {\cos {{2\pi } \over {15}}\cos {\pi \over {15}} - \sin {{2\pi } \over {15}}\sin {\pi \over {15}}}} = {{\sin ({\pi \over {15}} + {\pi \over {10}})} \over {\cos ({{2\pi } \over {15}} + {\pi \over 5})}} \cr
& = {{\sin {\pi \over 6}} \over {\cos {\pi \over 3}}} = 1 \cr} \)

Chọn (B)