Bài 64 trang 219 SGK Đại số 10 Nâng cao

Đề bài

\(\sin {{{{90}^0}} \over 4}\cos {{{{270}^0}} \over 4}\) bằng: 

\(\eqalign{
& (A)\,{1 \over 2}(1 - {{\sqrt 2 } \over 2})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(B)\,{1 \over 2}({{\sqrt 2 } \over 2} - 1)\,\,\, \cr 
& (C)\,{1 \over 2}(1 + {{\sqrt 2 } \over 2})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(D)\,\sqrt 2 - 1 \cr} \)

Lời giải chi tiết

Ta có:

 \(\sin {{{{90}^0}} \over 4}\cos {{{{270}^0}} \over 4} = {1 \over 2}\sin ({90^0} - \sin {45^0}) \)

\(= {1 \over 2}(1 - {{\sqrt 2 } \over 2})\)

Chọn (A)