《2025年创新设计高考总复习高三数学人教版A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}$.故$\cos2\alpha=2\cos^{2}\alpha-1$$=2×(-\frac{\sqrt{6}}{9})^{2}-1=-\frac{23}{27}$.]

(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)$-\frac{2}{3}$[
(1)由$\tan(\frac{\pi}{4}-\alpha)=\frac{\sin(\frac{\pi}{4}-\alpha)}{\cos(\frac{\pi}{4}-\alpha)}=3\cos2\alpha=3\sin(\frac{\pi}{2}-2\alpha)$,得$\frac{\sin(\frac{\pi}{4}-\alpha)}{\cos(\frac{\pi}{4}-\alpha)}=3\sin(\frac{\pi}{2}-2\alpha)$,即$\frac{1}{\cos(\frac{\pi}{4}-\alpha)}=6\sin(\frac{\pi}{4}-\alpha)\cos(\frac{\pi}{4}-\alpha)$,由于$\alpha\in(-\frac{\pi}{2},0)$,故$\frac{\pi}{4}-\alpha\in(\frac{\pi}{4},\frac{3\pi}{4})$,则$\sin(\frac{\pi}{4}-\alpha)\neq0$,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$1+\sin2\alpha=\frac{1}{3}$,即$\sin2\alpha=-\frac{2}{3}$.]

(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$\tan\frac{\alpha}{2}=\frac{1-\tan^{2}\frac{\alpha}{2}}{2\tan\frac{\alpha}{2}}$,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=3\cos(\alpha+\beta)\sin\alpha$,等号两边同时除以$\cos\alpha\cos(\alpha+\beta)$,得$\tan(\alpha+\beta)=3\tan\alpha=6$.]

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

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

答案： 训练$-\frac{2}{3}$$\frac{5\sqrt{3}-12}{26}$[
∵$6\sin^{2}\alpha+\sin\alpha\cos\alpha-2\cos^{2}\alpha$$=6\sin^{2}\alpha+\sin\alpha\cos\alpha-2\cos^{2}\alpha$$=\frac{6\tan^{2}\alpha+\tan\alpha-2}{\tan^{2}\alpha+1}=0$,即$6\tan^{2}\alpha+\tan\alpha-2=0$,解得$\tan\alpha=-\frac{2}{3}$或$\tan\alpha=\frac{1}{2}$,
∵$\alpha\in(\frac{\pi}{2},\pi)$,
∴$\tan\alpha=-\frac{2}{3}$.
∵$\sin2\alpha=\frac{2\tan\alpha}{1+\tan^{2}\alpha}=\frac{12}{13}$,$\cos2\alpha=\frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha}=\frac{5}{13}$,
∴$\sin(2\alpha+\frac{\pi}{3})=\sin2\alpha\cos\frac{\pi}{3}+\cos2\alpha\sin\frac{\pi}{3}$$=\frac{12}{13}×\frac{1}{2}+\frac{5}{13}×\frac{\sqrt{3}}{2}=\frac{5\sqrt{3}-12}{26}$.]