《2025年金版新学案高三总复习数学北师大版》

1. 二倍角的正弦、余弦、正切公式
(1) 公式 $ S_{2\alpha} $：$\sin 2\alpha = $2sinαcosα.
(2) 公式 $ C_{2\alpha} $：$\cos 2\alpha = $cos²α - sin²α $ = $2cos²α - 1 $ = $1 - 2sin²α.
(3) 公式 $ T_{2\alpha} $：$\tan 2\alpha = $$\frac{2tanα}{1 - tan²α}$.

答案：
(1)$2\sin\alpha\cos\alpha$
(2)$\cos^{2}\alpha-\sin^{2}\alpha$ $2\cos^{2}\alpha - 1$ $1 - 2\sin^{2}\alpha$
(3)$\frac{2\tan\alpha}{1 - \tan^{2}\alpha}$

2. 半角公式(不要求记忆)
$\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}$；$\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}$；$\tan \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}$. 符号由 $\frac{\alpha}{2}$ 所在象限决定.

答案： 答案略

1. (多选题)下列说法中正确的是(ABD)

A.$\sin 80^{\circ} = 2\sin 40^{\circ} \cos 40^{\circ}$
B.存在实数 $\alpha$，使 $\tan 2\alpha = 2\tan \alpha$
C.若 $\alpha$ 为第二象限角，$\sin \alpha = \frac{5}{13}$，则 $\sin 2\alpha = \frac{120}{169}$
D.$\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha} = \frac{1 - \cos \alpha}{\sin \alpha}$

答案： 1.ABD

2. (2023·新课标Ⅱ卷)已知 $\alpha$ 为锐角，$\cos \alpha = \frac{1 + \sqrt{5}}{4}$，则 $\sin \frac{\alpha}{2} =$(D)

A.$\frac{3 - \sqrt{5}}{8}$
B.$\frac{-1 + \sqrt{5}}{8}$
C.$\frac{3 - \sqrt{5}}{4}$
D.$\frac{-1 + \sqrt{5}}{4}$

答案： 2.D 因为$\cos\alpha = 1 - 2\sin^{2}\frac{\alpha}{2} = \frac{1 + \sqrt{5}}{4}$,而$\alpha$为锐角,所以$\sin\frac{\alpha}{2} = \sqrt{\frac{3 - \sqrt{5}}{8}} = \sqrt{\frac{(\sqrt{5} - 1)^{2}}{16}} = \frac{\sqrt{5} - 1}{4} = \frac{- 1 + \sqrt{5}}{4}$. 故选D.

3. (链接北师必修二 P165T1，改编)化简 $\sin^4 15^{\circ} - \sin^4 75^{\circ} =$(B)

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

答案： 3.B $\sin^{4}15^{\circ} - \sin^{4}75^{\circ} = \sin^{4}15^{\circ} - \sin^{4}(90^{\circ} - 15^{\circ}) = \sin^{4}15^{\circ} - \cos^{4}15^{\circ} = (\sin^{2}15^{\circ} - \cos^{2}15^{\circ})(\sin^{2}15^{\circ} + \cos^{2}15^{\circ}) = \sin^{2}15^{\circ} - \cos^{2}15^{\circ} = - \cos30^{\circ} = - \frac{\sqrt{3}}{2}$. 故选B.

4. (链接北师必修二 P164 例 1，改编)若 $\cos (\theta + \frac{\pi}{2}) = -\frac{\sqrt{7}}{4}$，则 $\cos 2\theta$ 的值为$\frac{1}{8}$.

答案： 4.$\frac{1}{8}$ 因为$\cos(\theta + \frac{\pi}{2}) = - \frac{\sqrt{7}}{4}$,所以$\sin\theta = \frac{\sqrt{7}}{4}$,所以$\cos2\theta = 1 - 2\sin^{2}\theta = \frac{1}{8}$.

1. $2\sqrt{1 + \sin 4} + \sqrt{2 + 2\cos 4}$ 等于(B)

A.$2\cos 2$
B.$2\sin 2$
C.$4\sin 2 + 2\cos 2$
D.$2\sin 2 + 4\cos 2$

答案： 1.B 原式$= 2\sqrt{(\sin2 + \cos2)^{2} + \sqrt{2}·\cos^{2}2} = 2|\sin2 + \cos2| + 2|\cos2|$. 由于$\frac{\pi}{2} ＜ 2 ＜ \frac{3}{4}\pi$,所以$\cos2 ＜ 0,\sin2 + \cos2 ＞ 0$. 故原式$= 2(\sin2 + \cos2) - 2\cos2 = 2\sin2$. 故选B.

2. (2025·湖南岳阳模拟)若 $\alpha$ 为第三象限角，且 $\sin \alpha = -\frac{3}{5}$，则 $\tan \frac{\alpha}{2} =$(A)

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

答案： 2.A $\alpha$为第三象限角,且$\sin\alpha = - \frac{3}{5}$,则$\cos\alpha = - \frac{4}{5}$,得$\tan\frac{\alpha}{2} = \frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}} = \frac{\frac{\sin^{2}\frac{\alpha}{2}}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}}{\frac{\cos^{2}\frac{\alpha}{2}}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}} = \frac{1 - \cos\alpha}{\sin\alpha} = \frac{1 - ( - \frac{4}{5})}{- \frac{3}{5}} = - 3$. 故选A.

3. 化简：已知 $0 ＜ \theta ＜ \pi$，则 $\frac{(1 + \sin \theta + \cos \theta)(\sin \frac{\theta}{2} - \cos \frac{\theta}{2})}{\sqrt{2 + 2\cos \theta}} =$-cosθ.

答案： 3.$-\cos\theta$ 原式$= \frac{\cos\theta\frac{\theta}{2}(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2} + 2\cos^{2}\frac{\theta}{2})(\sin\frac{\theta}{2} - \cos\frac{\theta}{2})}{\cos\frac{\theta}{2}\sqrt{4\cos^{2}\frac{\theta}{2}}}$
$= \frac{\cos\theta\frac{\sin^{2}\frac{\theta}{2} - \cos^{2}\frac{\theta}{2}}{\cos\frac{\theta}{2}} - \cos\frac{\theta}{2}·\cos\theta}{\cos\frac{\theta}{2}}$. 因为$0 ＜ \theta ＜ \pi$,所以$0 ＜ \frac{\theta}{2} ＜ \frac{\pi}{2}$,所以$\cos\frac{\theta}{2} ＞ 0$. 所以原式$= - \cos\theta$.