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Câu hỏi: Nếu \(\sin \alpha + \cos \alpha = {1 \over 2}\) thì sin2α bằng:
\(\eqalign{
& (A) {3 \over 8} (B) - {3 \over 4} \cr
& (C) {1 \over {\sqrt 2 }} (D) {3 \over 4} \cr} \)
\(\eqalign{
& (A) {3 \over 8} (B) - {3 \over 4} \cr
& (C) {1 \over {\sqrt 2 }} (D) {3 \over 4} \cr} \)
Lời giải chi tiết
Ta có:
\(\eqalign{
& \sin \alpha + \cos \alpha = {1 \over 2}\cr & \Rightarrow {\left({\sin \alpha + \cos \alpha } \right)^2} = \frac{1}{4}\cr & \Rightarrow {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = \frac{1}{4}\cr & \Rightarrow 1 + \sin 2\alpha = {1 \over 4} \cr
& \Rightarrow \sin 2\alpha = - {3 \over 4} \cr} \)
Chọn (B)
Ta có:
\(\eqalign{
& \sin \alpha + \cos \alpha = {1 \over 2}\cr & \Rightarrow {\left({\sin \alpha + \cos \alpha } \right)^2} = \frac{1}{4}\cr & \Rightarrow {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = \frac{1}{4}\cr & \Rightarrow 1 + \sin 2\alpha = {1 \over 4} \cr
& \Rightarrow \sin 2\alpha = - {3 \over 4} \cr} \)
Chọn (B)