Đáp án đúng: B
Phương pháp giải:
Sử dụng phương pháp thêm bớt và hằng đằng thức để tính giá trị của biểu thức.Giải chi tiết:\(\begin{array}{l}A = {\sin ^4}\dfrac{\pi }{{16}} + {\sin ^4}\dfrac{{3\pi }}{{16}} + {\sin ^4}\dfrac{{5\pi }}{{16}} + {\sin ^4}\dfrac{{7\pi }}{{16}}\\\,\,\,\,\, = {\sin ^4}\dfrac{\pi }{{16}} + {\sin ^4}\dfrac{{3\pi }}{{16}} + {\cos ^4}\dfrac{\pi }{{16}} + {\cos ^4}\dfrac{{3\pi }}{{16}}\\\,\,\,\,\, = \left( {{{\sin }^4}\dfrac{\pi }{{16}} + {{\cos }^4}\dfrac{\pi }{{16}}} \right) + \left( {{{\sin }^4}\dfrac{{3\pi }}{{16}} + {{\cos }^4}\dfrac{{3\pi }}{{16}}} \right)\\\,\,\,\,\, = {\left( {{{\sin }^2}\dfrac{\pi }{{16}} + {{\cos }^2}\dfrac{\pi }{{16}}} \right)^2} - 2{\sin ^2}\dfrac{\pi }{{16}}.{\cos ^2}\dfrac{\pi }{{16}} + {\left( {{{\sin }^2}\dfrac{{3\pi }}{{16}} + {{\cos }^2}\dfrac{{3\pi }}{{16}}} \right)^2} - 2{\sin ^2}\dfrac{{3\pi }}{{16}}.{\cos ^2}\dfrac{{3\pi }}{{16}}\end{array}\)
\(\begin{array}{l}\,\,\,\, = 1 - 2{\sin ^2}\dfrac{\pi }{{16}}.{\cos ^2}\dfrac{\pi }{{16}} + 1 - 2{\sin ^2}\dfrac{{3\pi }}{{16}}.{\cos ^2}\dfrac{{3\pi }}{{16}}\\\,\,\,\,\, = 2 - \dfrac{1}{2} \cdot \left( {4{{\sin }^2}\dfrac{\pi }{{16}}.{{\cos }^2}\dfrac{\pi }{{16}} + 4{{\sin }^2}\dfrac{{3\pi }}{{16}}.{{\cos }^2}\dfrac{{3\pi }}{{16}}} \right)\\\,\,\,\,\, = 2 - \dfrac{1}{2} \cdot \left( {{{\sin }^2}\dfrac{\pi }{8} + {{\cos }^2}\dfrac{\pi }{8}} \right)\\\,\,\,\,\, = 2 - \dfrac{1}{2}\\\,\,\,\,\, = \dfrac{3}{2}\end{array}\)
Vậy \(A = \dfrac{3}{2}\).
Chọn B.
