[例2](1)(2020·全国卷Ⅲ)已知 sin theta +sin (theta +dfrac (pi )(3))=1,-|||-则 sin (theta +dfrac (pi )(6))= ()-|||-A. dfrac (1)(2) B. dfrac (sqrt {3)}(3)-|||-C. dfrac (2)(3) D. dfrac (sqrt {2)}(2)

步骤 1：展开 $\sin (\theta +\dfrac {\pi }{3})$
根据和角公式，$\sin (\theta +\dfrac {\pi }{3})$ 可以展开为 $\sin \theta \cos \dfrac {\pi }{3} + \cos \theta \sin \dfrac {\pi }{3}$。由于 $\cos \dfrac {\pi }{3} = \dfrac {1}{2}$ 和 $\sin \dfrac {\pi }{3} = \dfrac {\sqrt {3}}{2}$，所以 $\sin (\theta +\dfrac {\pi }{3}) = \dfrac {1}{2}\sin \theta + \dfrac {\sqrt {3}}{2}\cos \theta$。
步骤 2：代入原方程
将 $\sin (\theta +\dfrac {\pi }{3})$ 的表达式代入原方程 $\sin \theta +\sin (\theta +\dfrac {\pi }{3})=1$，得到 $\sin \theta + \dfrac {1}{2}\sin \theta + \dfrac {\sqrt {3}}{2}\cos \theta = 1$。
步骤 3：化简方程
化简方程得到 $\dfrac {3}{2}\sin \theta + \dfrac {\sqrt {3}}{2}\cos \theta = 1$。为了求 $\sin (\theta +\dfrac {\pi }{6})$，我们注意到 $\sin (\theta +\dfrac {\pi }{6}) = \sin \theta \cos \dfrac {\pi }{6} + \cos \theta \sin \dfrac {\pi }{6}$。由于 $\cos \dfrac {\pi }{6} = \dfrac {\sqrt {3}}{2}$ 和 $\sin \dfrac {\pi }{6} = \dfrac {1}{2}$，所以 $\sin (\theta +\dfrac {\pi }{6}) = \dfrac {\sqrt {3}}{2}\sin \theta + \dfrac {1}{2}\cos \theta$。
步骤 4：求解 $\sin (\theta +\dfrac {\pi }{6})$
将 $\dfrac {3}{2}\sin \theta + \dfrac {\sqrt {3}}{2}\cos \theta = 1$ 乘以 $\dfrac {2}{\sqrt {3}}$，得到 $\sqrt {3}\sin \theta + \cos \theta = \dfrac {2}{\sqrt {3}}$。由于 $\sin (\theta +\dfrac {\pi }{6}) = \dfrac {\sqrt {3}}{2}\sin \theta + \dfrac {1}{2}\cos \theta$，所以 $\sin (\theta +\dfrac {\pi }{6}) = \dfrac {1}{\sqrt {3}}(\sqrt {3}\sin \theta + \cos \theta) = \dfrac {1}{\sqrt {3}} \cdot \dfrac {2}{\sqrt {3}} = \dfrac {2}{3}$。