Đề bài - bài 62 trang 219 sgk đại số 10 nâng cao

Đề bài

\({{\sin {\pi \over {15}}\cos {\pi \over 10} + \sin {\pi \over {10}}\cos {\pi \over 15}} \over {\cos {{2\pi } \over {15}}\cos {\pi \over {5}} - \sin {{2\pi } \over {15}}\sin {\pi \over {5}}}}\)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)