Nội dung bài giảng
Nếu \(\sin \alpha + \cos \alpha = {1 \over 2}\) thì sin2α bằng:
\(\eqalign{
& (A)\,{3 \over 8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(B)\, - {3 \over 4} \cr
& (C)\,{1 \over {\sqrt 2 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(D)\,{3 \over 4} \cr} \)
Đáp án
Ta có:
\(\eqalign{
& \sin \alpha + \cos \alpha = {1 \over 2} \Rightarrow 1 + 2\sin \alpha \cos \alpha = {1 \over 4} \cr
& \Rightarrow \sin 2\alpha = - {3 \over 4} \cr} \)
Chọn (B)