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Câu hỏi: \(\sin {{{{90}^0}} \over 4}\cos {{{{270}^0}} \over 4}\) bằng:
\(\eqalign{
& (A) {1 \over 2}(1 - {{\sqrt 2 } \over 2}) (B) {1 \over 2}({{\sqrt 2 } \over 2} - 1) \cr
& (C) {1 \over 2}(1 + {{\sqrt 2 } \over 2}) (D) \sqrt 2 - 1 \cr} \)
\(\eqalign{
& (A) {1 \over 2}(1 - {{\sqrt 2 } \over 2}) (B) {1 \over 2}({{\sqrt 2 } \over 2} - 1) \cr
& (C) {1 \over 2}(1 + {{\sqrt 2 } \over 2}) (D) \sqrt 2 - 1 \cr} \)
Lời giải chi tiết
Ta có:
\(\sin {{{{90}^0}} \over 4}\cos {{{{270}^0}} \over 4}\) \(= \frac{1}{2}\left( {\sin \frac{{{{90}^0} + {{270}^0}}}{4} + \sin \frac{{{{90}^0} - {{270}^0}}}{4}} \right)\) \(= {1 \over 2}(\sin {90^0} - \sin {45^0}) \)
\(= {1 \over 2}(1 - {{\sqrt 2 } \over 2})\)
Chọn (A)
Ta có:
\(\sin {{{{90}^0}} \over 4}\cos {{{{270}^0}} \over 4}\) \(= \frac{1}{2}\left( {\sin \frac{{{{90}^0} + {{270}^0}}}{4} + \sin \frac{{{{90}^0} - {{270}^0}}}{4}} \right)\) \(= {1 \over 2}(\sin {90^0} - \sin {45^0}) \)
\(= {1 \over 2}(1 - {{\sqrt 2 } \over 2})\)
Chọn (A)