Cho \(\cos\alpha=\dfrac{3}{5}\) với \(\alpha\in\left(\dfrac{3\pi}{2};2\pi\right)\). Tính \(\sin2a\).
\(\dfrac{24}{25}\).\(-\dfrac{24}{25}\).\(\dfrac{4}{5}\).\(-\dfrac{4}{5}\).Hướng dẫn giải:Áp dụng \(\sin^2\alpha+\cos^2\alpha=1\) \(\Rightarrow\sin\alpha=\pm\dfrac{4}{5}\)
Do \(\alpha\in\left(\dfrac{3\pi}{2};2\pi\right)\)nên \(\sin\alpha< 0\) suy ra \(\Rightarrow\sin\alpha=-\dfrac{4}{5}\)
Áp dụng \(\sin2\alpha=2\sin\alpha\cos\alpha=2.\dfrac{3}{5}.\left(-\dfrac{4}{5}\right)=-\dfrac{24}{25}\)