Cho \(\sin\alpha+\cos\alpha=m\) với \(\alpha\in\left(\dfrac{3\pi}{2};2\pi\right)\). Khi đó \(\sin\alpha.\cos\alpha\) bằng
\(\dfrac{1-2m^2}{2}\).\(\dfrac{m^2-1}{2}\).\(\dfrac{2m^2-1}{2}\).\(\dfrac{1-m^2}{2}\).Hướng dẫn giải:\(\sin\alpha+\cos\alpha=m\) \(\Rightarrow\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha=m^2\)
\(\Rightarrow1+2\sin\alpha.\cos\alpha=m^2\)
\(\Rightarrow\) \(\sin\alpha.\cos\alpha\) = \(\dfrac{m^2-1}{2}\).
