Tính giá trị biểu thức lượng giác dùng công thức cộng

Tính \(\sin\alpha\): \(\sin^2\alpha + \cos^2\alpha = 1\) \(\Leftrightarrow \sin^2\alpha + \left(-\dfrac{1}{\sqrt{3}}\right)^2 = 1\) \(\Leftrightarrow \sin^2\alpha = 1 - \dfrac{1}{3} = \dfrac{2}{3}\) Vì \(\dfrac{\pi}{2} < \alpha < \pi\) nên \(\sin\alpha > 0\), suy ra \(\sin\alpha = \sqrt{\dfrac{2}{3}} = \dfrac{\sqrt{6}}{3}\). a) \(\sin\!\left(\alpha + \dfrac{\pi}{6}\right) = \sin\alpha\cos\dfrac{\pi}{6} + \cos\alpha\sin\dfrac{\pi}{6}\) \(= \dfrac{\sqrt{6}}{3}\cdot\dfrac{\sqrt{3}}{2} + \left(-\dfrac{1}{\sqrt{3}}\right)\cdot\dfrac{1}{2}\) \(= \dfrac{\sqrt{18}}{6} - \dfrac{1}{2\sqrt{3}} = \dfrac{3\sqrt{2}}{6} - \dfrac{\sqrt{3}}{6} = \dfrac{3\sqrt{2} - \sqrt{3}}{6}\) b) \(\cos\!\left(\alpha + \dfrac{\pi}{6}\right) = \cos\alpha\cos\dfrac{\pi}{6} - \sin\alpha\sin\dfrac{\pi}{6}\) \(= \left(-\dfrac{1}{\sqrt{3}}\right)\cdot\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{6}}{3}\cdot\dfrac{1}{2}\) \(= -\dfrac{1}{2} - \dfrac{\sqrt{6}}{6} = -\dfrac{3}{6} - \dfrac{\sqrt{6}}{6} = -\dfrac{3 + \sqrt{6}}{6}\) c) \(\sin\!\left(\alpha - \dfrac{\pi}{3}\right) = \sin\alpha\cos\dfrac{\pi}{3} - \cos\alpha\sin\dfrac{\pi}{3}\) \(= \dfrac{\sqrt{6}}{3}\cdot\dfrac{1}{2} - \left(-\dfrac{1}{\sqrt{3}}\right)\cdot\dfrac{\sqrt{3}}{2}\) \(= \dfrac{\sqrt{6}}{6} + \dfrac{1}{2} = \dfrac{\sqrt{6}}{6} + \dfrac{3}{6} = \dfrac{3 + \sqrt{6}}{6}\) d) \(\cos\!\left(\alpha - \dfrac{\pi}{6}\right) = \cos\alpha\cos\dfrac{\pi}{6} + \sin\alpha\sin\dfrac{\pi}{6}\) \(= \left(-\dfrac{1}{\sqrt{3}}\right)\cdot\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{6}}{3}\cdot\dfrac{1}{2}\) \(= -\dfrac{1}{2} + \dfrac{\sqrt{6}}{6} = -\dfrac{3}{6} + \dfrac{\sqrt{6}}{6} = \dfrac{-3 + \sqrt{6}}{6}\)