《2025年创新设计高考总复习数学浙江专版》

考点三 角的变换问题
例 3 (1)(2020·全国 III 卷)已知 $ \sin \theta + \sin(\theta + \frac{\pi}{3}) = 1 $，则 $ \sin(\theta + \frac{\pi}{6}) = $(B)

A.$ \frac{1}{2} $
B.$ \frac{\sqrt{3}}{3} $
C.$ \frac{2}{3} $
D.$ \frac{\sqrt{2}}{2} $

答案： 例3
(1)B[
(1)因为$\sin \theta + \sin(\theta + \frac{\pi}{3})= \sin(\theta + \frac{\pi}{6} - \frac{\pi}{6}) + \sin(\theta + \frac{\pi}{6} + \frac{\pi}{6})= \sin(\theta + \frac{\pi}{6}) \cos \frac{\pi}{6} - \cos(\theta + \frac{\pi}{6}) \sin \frac{\pi}{6} + \sin(\theta + \frac{\pi}{6}) \cos \frac{\pi}{6} + \cos(\theta + \frac{\pi}{6}) \sin \frac{\pi}{6}= 2\sin(\theta + \frac{\pi}{6}) \cos \frac{\pi}{6} = \sqrt{3} \sin(\theta + \frac{\pi}{6}) = 1,$所以$\sin(\theta + \frac{\pi}{6}) = \frac{\sqrt{3}}{3}.]$

(2)(2025·浙江名校联考)已知 $ \alpha \in (\frac{\pi}{2}, \pi) $，$ \beta \in (0, \frac{\pi}{2}) $，若 $ \sin(\alpha + \beta) = \frac{1}{3} $，$ \cos \beta = \frac{\sqrt{3}}{3} $，则 $ \cos 2\alpha = $(D)

A.$ \frac{1}{3} $
B.$ -\frac{1}{3} $
C.$ \frac{23}{27} $
D.$ -\frac{23}{27}$$$

答案： 例3
(2)D[
(2)由于$\alpha \in (\frac{\pi}{2}, \pi), \beta \in (0, \frac{\pi}{2}),$则$\alpha + \beta \in (\frac{\pi}{2}, \frac{3\pi}{2}),$而$\sin(\alpha + \beta) = \frac{1}{3},$故$\alpha + \beta \in (\frac{\pi}{2}, \pi),$所以$\cos(\alpha + \beta) = - \sqrt{1 - \sin^{2}(\alpha + \beta)} = - \frac{2\sqrt{2}}{3}.$由$\cos \beta = \frac{\sqrt{3}}{3}, \beta \in (0, \frac{\pi}{2}),$可得$\sin \beta = \frac{\sqrt{6}}{3}$则$\cos \alpha = \cos[(\alpha + \beta) - \beta]= \cos(\alpha + \beta) \cos \beta + \sin(\alpha + \beta) \sin \beta= \frac{2\sqrt{2}}{3} × \frac{\sqrt{3}}{3} + \frac{1}{3} × \frac{\sqrt{6}}{3} = \frac{\sqrt{6}}{9}$故$\cos 2\alpha = 2\cos^{2} \alpha - 1= 2 × (\frac{\sqrt{6}}{9})^{2} - 1 = - \frac{23}{27}.]$

训练 3 (1)(2025·沈阳模拟)已知 $ \alpha \in (-\frac{\pi}{2}, 0) $ 且 $ \tan(\frac{\pi}{4} - \alpha) = 3\cos 2\alpha $，则 $ \sin 2\alpha = $- \frac{2}{3}.

答案： 训练$3(1)- \frac{2}{3}(2)6[(1)$由$\tan(\frac{\pi}{4} - \alpha) =3\cos 2\alpha,\frac{\sin(\frac{\pi}{4} - \alpha)}{\cos(\frac{\pi}{4} - \alpha)} = 3\cos 2\alpha = 3\sin(\frac{\pi}{2} - 2\alpha),$即$\frac{\sin(\frac{\pi}{4} - \alpha)}{\cos(\frac{\pi}{4} - \alpha)} = 6\sin(\frac{\pi}{4} - \alpha) \cos(\frac{\pi}{4} - \alpha),$由于$\alpha \in (-\frac{\pi}{2}, 0),$故$\frac{\pi}{4} - \alpha \in (\frac{\pi}{4}, \frac{3\pi}{4}),$则$\sin(\frac{\pi}{4} - \alpha) \neq 0,$故$\frac{1}{\cos(\frac{\pi}{4} - \alpha)} = 6\cos(\frac{\pi}{4} - \alpha),$即$2\cos^{2}(\frac{\pi}{4} - \alpha) = \frac{1}{3}$则$1 + \cos(\frac{\pi}{2} - 2\alpha) = \frac{1}{3}$即$1 + \sin 2\alpha = \frac{1}{3},$即$\sin 2\alpha = - \frac{2}{3}.(2)$因为$\tan \frac{\alpha}{2} = 1 - \tan^{2} \frac{\alpha}{2},$所以$\tan \alpha = \frac{2\tan \frac{\alpha}{2}}{1 - \tan^{2} \frac{\alpha}{2}} = 2.$又$\sin[(\alpha + \beta) + \alpha] = 2\sin[(\alpha + \beta) - \alpha],$所以$\sin(\alpha + \beta) \cos \alpha + \cos(\alpha + \beta) \sin \alpha= 2\sin(\alpha + \beta) \cos \alpha - 2\cos(\alpha + \beta) \sin \alpha,$即$\sin(\alpha + \beta) \cos \alpha - \cos(\alpha + \beta) \sin \alpha= \sin(\alpha + \beta - \alpha) = 0,$等号两边同时除以$\cos \alpha \cos(\alpha + \beta),$得$\tan(\alpha + \beta) = 3\tan \alpha = 6.]$

(2)(2025·河南名校联考)已知 $ \sin(2\alpha + \beta) = 2\sin \beta $，且 $ \tan \frac{\alpha}{2} = 1 - \tan^2 \frac{\alpha}{2} $，则 $ \tan(\alpha + \beta) = $____.

答案： 训练3
(2)6[
(2)因为$\tan \frac{\alpha}{2} = 1 - \tan^{2} \frac{\alpha}{2},$所以$\tan \alpha = \frac{2\tan \frac{\alpha}{2}}{1 - \tan^{2} \frac{\alpha}{2}} = 2.$又$\sin[(\alpha + \beta) + \alpha] = 2\sin[(\alpha + \beta) - \alpha],$所以$\sin(\alpha + \beta) \cos \alpha + \cos(\alpha + \beta) \sin \alpha= 2\sin(\alpha + \beta) \cos \alpha - 2\cos(\alpha + \beta) \sin \alpha,$即$\sin(\alpha + \beta) \cos \alpha - \cos(\alpha + \beta) \sin \alpha= \sin(\alpha + \beta - \alpha) = 0,$等号两边同时除以$\cos \alpha \cos(\alpha + \beta),$得$\tan(\alpha + \beta) = 3\tan \alpha = 6.]$

典例
已知$\alpha ,\beta \in (0,\pi )$，$\tan \dfrac{\alpha }{2}=\dfrac{1}{2}$，$\sin (\alpha -\beta )=\dfrac{5}{13}$，则$\cos \beta =$______.

答案： 典例 $\frac{56}{65}$ $\because \left[\because \tan \frac{\alpha}{2}=\frac{1}{2}\right.$ $\therefore \sin \alpha=\frac{2\tan\frac{\alpha}{2}}{1+\tan^{2}\frac{\alpha}{2}}=\frac{2 × \frac{1}{2}}{1+(\frac{1}{2})^{2}}=\frac{4}{5}$ $\cos \alpha=\frac{1-\tan^{2}\frac{\alpha}{2}}{1+\tan^{2}\frac{\alpha}{2}}=\frac{1-(\frac{1}{2})^{2}}{1+(\frac{1}{2})^{2}}=\frac{3}{5}$ $\because \alpha,\beta \in (0,\pi),\cos \alpha＞0,\therefore \alpha \in \left(0,\frac{\pi}{2}\right)$, $\therefore \alpha-\beta \in \left(-\pi,\frac{\pi}{2}\right)$, $\because \sin (\alpha-\beta)=\frac{5}{13}＞0,\therefore \alpha-\beta \in \left(0,\frac{\pi}{2}\right)$, $\therefore \cos (\alpha-\beta)=\frac{12}{13}$, $\therefore \cos \beta=\cos \left[-\beta\right]=\cos \left[(\alpha-\beta)-\alpha\right]$ $=\cos (\alpha-\beta)\cos \alpha+\sin (\alpha-\beta)\sin \alpha$ $=\frac{12}{13} × \frac{3}{5}+\frac{5}{13} × \frac{4}{5}=\frac{56}{65}$.

训练
(2024·济宁质检)已知$6\sin ^{2}\alpha +\sin \alpha \cos \alpha -2\cos ^{2}\alpha =0$，$\alpha \in \left(\dfrac{\pi }{2},\pi \right)$，则$\tan \alpha =$______，$\sin \left(2\alpha +\dfrac{\pi }{3}\right)=$______.

答案： 训练 $\frac{2}{3}-\frac{5\sqrt{3}-12}{26}$ $\left[\because 6\sin^{2}\alpha+\sin \alpha \cos \alpha-2\cos^{2}\alpha=\frac{6\sin^{2}\alpha+\sin \alpha \cos \alpha-2\cos^{2}\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}\right.$ $=\frac{6\tan^{2}\alpha+\tan \alpha-2}{\tan^{2}\alpha+1}=0$, 即 $6\tan^{2}\alpha+\tan \alpha-2=0$, 解得 $\tan \alpha=-\frac{2}{3}$ 或 $\tan \alpha=\frac{1}{2}$, $\because \alpha \in \left(\frac{\pi}{2},\pi\right)$, $\therefore \tan \alpha=-\frac{2}{3}$. $\because \sin 2\alpha=\frac{2\tan \alpha}{1+\tan^{2}\alpha}=\frac{12}{13}$, $\cos 2\alpha=\frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha}=\frac{5}{13}$, $\therefore \sin \left(2\alpha+\frac{\pi}{3}\right)=\sin 2\alpha \cos \frac{\pi}{3}+\cos 2\alpha \sin \frac{\pi}{3}$ $=-\frac{12}{13} × \frac{1}{2}+\frac{5}{13} × \frac{\sqrt{3}}{2}=\frac{5\sqrt{3}-12}{26}$.