[题目]-|||-化简(1) sin (420)^circ cos (750)^circ +sin (-(330)^circ )cos (-(660)^circ )-|||-(2) dfrac (sin (2pi -alpha )cos (pi +alpha )cos (dfrac {pi )(2)+alpha )cos (dfrac (11pi )(2)-alpha )}(cos (pi -alpha )sin (3pi -alpha )sin (-pi -alpha )sin (dfrac {9pi )(2)+alpha )}

考查要点：本题主要考查三角函数的诱导公式应用及角度化简能力，涉及正弦、余弦的周期性、奇偶性及和角公式的运用。

解题核心思路：

1. 角度化简：将超过$360^\circ$或负角度的三角函数转换为$0^\circ \sim 360^\circ$内的等效角，利用周期性简化计算。
2. 符号判断：根据“奇变偶不变，符号看象限”口诀，确定化简后的三角函数符号。
3. 公式应用：结合和角公式（如$\sin(A+B)$）或分式约分技巧，进一步化简表达式。

第（1）题

化简角度

• $\sin 420^\circ = \sin(360^\circ + 60^\circ) = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$
• $\cos 750^\circ = \cos(720^\circ + 30^\circ) = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$
• $\sin(-330^\circ) = -\sin 330^\circ = -\sin(360^\circ - 30^\circ) = -(-\sin 30^\circ) = \dfrac{1}{2}$
• $\cos(-660^\circ) = \cos 660^\circ = \cos(720^\circ - 60^\circ) = \cos 60^\circ = \dfrac{1}{2}$

代入并计算

原式化简为：
$\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$

第（2）题

分子化简

1. $\sin(2\pi - \alpha) = -\sin \alpha$
2. $\cos(\pi + \alpha) = -\cos \alpha$
3. $\cos\left(\dfrac{\pi}{2} + \alpha\right) = -\sin \alpha$
4. $\cos\left(\dfrac{11\pi}{2} - \alpha\right) = \cos\left(6\pi - \dfrac{\pi}{2} - \alpha\right) = \cos\left(\dfrac{\pi}{2} + \alpha\right) = -\sin \alpha$

分子整体为：
$(-\sin \alpha)(-\cos \alpha)(-\sin \alpha)(-\sin \alpha) = \sin^3 \alpha \cos \alpha$

分母化简

1. $\cos(\pi - \alpha) = -\cos \alpha$
2. $\sin(3\pi - \alpha) = \sin(\pi - \alpha) = \sin \alpha$
3. $\sin(-\pi - \alpha) = -\sin(\pi + \alpha) = \sin \alpha$
4. $\sin\left(\dfrac{9\pi}{2} + \alpha\right) = \sin\left(4\pi + \dfrac{\pi}{2} + \alpha\right) = \sin\left(\dfrac{\pi}{2} + \alpha\right) = \cos \alpha$

分母整体为：
$(-\cos \alpha)(\sin \alpha)(\sin \alpha)(\cos \alpha) = -\cos^2 \alpha \sin^2 \alpha$

约分结果

$\frac{\sin^3 \alpha \cos \alpha}{-\cos^2 \alpha \sin^2 \alpha} = -\frac{\sin \alpha}{\cos \alpha} = -\tan \alpha$