《2025年高考帮数学》

1.两角和与差的正弦、余弦、正切公式
$S_{(\alpha\pm\beta)}:\sin(\alpha\pm\beta)=$①________________.
$C_{(\alpha\pm\beta)}:\cos(\alpha\pm\beta)=$②________________.
$T_{(\alpha\pm\beta)}:\tan(\alpha\pm\beta)=$③________________$(\alpha,\beta,\alpha\pm\beta\neq k\pi+\frac{\pi}{2},k\in\mathbf{Z})$.

答案： ①$\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$ ②$\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$ ③$\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}$

2.二倍角公式
$S_{2\alpha}:\sin2\alpha=$④__________.
$C_{2\alpha}:\cos2\alpha=$⑤__________ =⑥__________ =⑦__________.
$T_{2\alpha}:\tan2\alpha=$⑧__________$(\alpha\neq k\pi+\frac{\pi}{2}且\alpha\neq\frac{k\pi}{2}+\frac{\pi}{4},k\in\mathbf{Z})$.

答案： ④$2\sin\alpha\cos\alpha$ ⑤$\cos^{2}\alpha-\sin^{2}\alpha$ ⑥$2\cos^{2}\alpha - 1$ ⑦$1 - 2\sin^{2}\alpha$ ⑧$\frac{2\tan\alpha}{1 - \tan^{2}\alpha}$

3.辅助角公式
$a\sin\alpha + b\cos\alpha=\sqrt{a^{2}+b^{2}}\sin(\alpha+\varphi)$(其中$a\neq0,\sin\varphi=\frac{b}{\sqrt{a^{2}+b^{2}}},\cos\varphi=\frac{a}{\sqrt{a^{2}+b^{2}}},\tan\varphi=$⑨_____).

答案： ⑨$\frac{b}{a}$

1.[2023北京海淀区月考]若$\tan(\alpha-\frac{5\pi}{12})=\frac{1}{2}$,则$\tan(\alpha-\frac{\pi}{6})$的值为 ( )
A.3
B.$\frac{1}{3}$
C.-3
D.$-\frac{1}{3}$

答案： 1.A 因为$\tan(\alpha - \frac{5\pi}{12}) = \tan[(\alpha - \frac{\pi}{6}) - \frac{\pi}{4}] = \frac{\tan(\alpha - \frac{\pi}{6}) - 1}{1 + \tan(\alpha - \frac{\pi}{6})} = \frac{1}{2}$，所以$\tan(\alpha - \frac{\pi}{6}) = 3$.

2.已知$\sin\alpha=\frac{15}{17},\alpha\in(\frac{\pi}{2},\pi)$,则$\cos(\frac{\pi}{4}-\alpha)$的值为________.

答案： 2.$\frac{7\sqrt{2}}{34}$ $\because\sin\alpha = \frac{15}{17},\alpha\in(\frac{\pi}{2},\pi),\therefore\cos\alpha = -\sqrt{1 - \sin^{2}\alpha} = -\sqrt{1 - (\frac{15}{17})^{2}} = -\frac{8}{17},\therefore\cos(\frac{\pi}{4} - \alpha) = \cos\frac{\pi}{4}\cos\alpha + \sin\frac{\pi}{4}\sin\alpha = \frac{\sqrt{2}}{2}\times(-\frac{8}{17}) + \frac{\sqrt{2}}{2}\times\frac{15}{17} = \frac{7\sqrt{2}}{34}$.