1. Không dùng bảng hay máy tính cầm tay, chứng minh rằng
\(\sin 15^\circ + \sin 75^\circ > 1\) .
2. Cho \(\sin \alpha + \cos \alpha = \dfrac{1}{2}\) . Tính \({\sin ^3}\alpha + {\cos ^3}\alpha \)
1. Ta có:
\({\left( {\sin 15^\circ + sin75^\circ } \right)^2}\)
\(= {\left( {\sin 15^\circ + \cos 15^\circ } \right)^2}\)
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\(\begin{array}{l} = {\sin ^2}15^\circ + {\cos ^2}15^\circ + 2\sin 15^\circ \cos 15^\circ \\ = 1 + \sin 30^\circ = \dfrac{3}{2} > 1\end{array}\)
Mà \(\sin 15^\circ + \sin 75^\circ > 0\) nên suy ra \(\sin 15^\circ + \sin 75^\circ > 1\).
2. Ta có: \(\sin \alpha + \cos \alpha = \dfrac{1}{2} \)
\(\Rightarrow {\left( {\sin \alpha + \cos \alpha } \right)^2} = \dfrac{1}{4}\)
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\( \Rightarrow 1 + 2\sin \alpha \cos \alpha = \dfrac{1}{4}\)
\( \Rightarrow \sin \alpha \cos \alpha = – \dfrac{3}{8}\)
Do đó
\(\begin{array}{l}{\sin ^3}\alpha + {\cos ^3}\alpha \\= {\left( {\sin \alpha + \cos \alpha } \right)^3} – 3\sin \alpha \cos \alpha \left( {\sin \alpha + \cos \alpha } \right)\\ = \dfrac{1}{8} + \dfrac{9}{{16}} = \dfrac{{11}}{6}.\end{array}\)
