# 三角函数式的化简

## 升幂公式

$$\sin\alpha=2\sin\cfrac{\alpha}{2}\cos\cfrac{\alpha}{2}$$

$$\cos\alpha=\cos^2\cfrac{\alpha}{2}-\sin^2\cfrac{\alpha}{2}$$

$$\cos\alpha=2\cos^2\cfrac{\alpha}{2}-1$$

$$\cos\alpha=1-2\sin^2\cfrac{\alpha}{2}$$

## 降幂公式

$$\sin^2\alpha=\cfrac{1-\cos2\alpha}{2}$$

$$\cos^2\alpha=\cfrac{1+\cos2\alpha}{2}$$

$$\tan^2\alpha=\cfrac{1-\cos2\alpha}{1+\cos2\alpha}$$

## 化简要求

(1)使三角函数式的次数尽量低;

(2)使三角函数式中的项数尽量少;

(3)使三角函数的种类尽量少;

(4)使三角函数式中的分母尽量不含有三角函数;

(5)使三角函数式中尽量不含有根号和绝对值符号;

(6)能求值的，要求出具体的值,否则就用三角函数式来表示.

## 常用变形

$$1+\sin\theta+\cos\theta=(1+\cos\theta)+\sin\theta=(2cos^2\cfrac{\theta}{2}+2sin\cfrac{\theta}{2}cos\cfrac{\theta}{2})$$

$$\sqrt{1-\sin40^{\circ}}=\sqrt{sin^220^{\circ}-2\sin20^{\circ}\cdot \cos20^{\circ}+\cos^220^{\circ}}$$

$$=\sqrt{(\cos20^{\circ}-\sin20^{\circ})^2}=|\cos20^{\circ}-\sin20^{\circ}|$$

## 典例剖析：

$$=\sqrt{(sin2-cos2)^2}=|sin2-cos2|=sin2-cos2$$

$$=\sqrt{2}\cdot\sqrt{1+\cfrac{sin^2x}{cos^2x}}=\sqrt{2}\cdot\sqrt{\cfrac{cos^2x+sin^2x}{cos^2x}}$$

$$=\sqrt{2}\cdot\sqrt{\cfrac{1}{cos^2x}}=\sqrt{2}\cdot \cfrac{1}{|cosx|}=-\cfrac{\sqrt{2}}{cosx}$$

$$=\sqrt{(1-sin\alpha sin\beta-cos\alpha cos\beta)(1-sin\alpha sin\beta+cos\alpha cos\beta)}$$

$$=\sqrt{(1-cos(\alpha-\beta))(1+cos(\alpha+\beta)}$$

$$=\sqrt{2sin^2\cfrac{\alpha-\beta}{2}\cdot 2cos^2\cfrac{\alpha+\beta}{2}}$$

$$=|2sin\cfrac{\alpha-\beta}{2}cos\cfrac{\alpha+\beta}{2}|$$

$$-\cfrac{\pi}{2}<\cfrac{\alpha+\beta}{2}<\cfrac{\pi}{2}$$

$$cos\cfrac{\alpha+\beta}{2}>0$$

$$-\cfrac{\pi}{2}<\cfrac{\alpha-\beta}{2}<0$$，故$$sin\cfrac{\alpha-\beta}{2}<0$$

$$=\cfrac{(2cos^2\cfrac{\theta}{2}+2sin\cfrac{\theta}{2}cos\cfrac{\theta}{2})(sin\cfrac{\theta}{2}-cos\cfrac{\theta}{2})}{\sqrt{2\cdot 2cos^2\cfrac{\theta}{2}}}$$

$$=\cfrac{2cos\cfrac{\theta}{2}(cos\cfrac{\theta}{2}+sin\cfrac{\theta}{2})(sin\cfrac{\theta}{2}-cos\cfrac{\theta}{2})}{2cos\cfrac{\theta}{2}}$$

$$=sin^2\cfrac{\theta}{2}-cos^2\cfrac{\theta}{2}=-cos\theta$$

$$=\cfrac{2cos\theta}{sin\theta}\cdot (1+tan\theta\cdot tan\cfrac{\theta}{2})$$

$$=\cfrac{2cos\theta}{sin\theta}+2\cdot tan\cfrac{\theta}{2}$$

$$=\cfrac{2cos\theta}{sin\theta}+\cfrac{2\cdot sin\frac{\theta}{2}\cdot\sin\frac{\theta}{2}\cdot 2}{ cos\frac{\theta}{2}\cdot sin\frac{\theta}{2}\cdot 2}$$

$$=\cfrac{2cos\theta}{sin\theta}+\cfrac{2(1-cos\theta)}{sin\theta}$$

$$=\cfrac{2}{sin\theta}$$

$$=\cfrac{1}{sin\alpha cos\alpha}\cdot sin\alpha cos\alpha-2cos^2\alpha$$

$$=1-2cos^2\alpha$$

$$=-cos2\alpha$$

$$=\sqrt{2}\sqrt{2cos^24}+2\sqrt{sin^24+cos^24-2sin4\cdot cos4}$$

$$=2|cos4|+2\sqrt{(sin4-cos4)^2}$$

$$=2|cos4|+2|sin4-cos4|$$

$$=-2cos4-2(sin4-cos4)=-2sin4$$

$$k=2n(n\in N)$$时，原式=$$\cfrac{sin\alpha\cdot cos\alpha}{sin\alpha\cdot cos\alpha}=1$$

$$k=2n+1(n\in N)$$时，原式=$$\cfrac{sin(\pi+\alpha)\cdot cos\alpha}{sin\alpha\cdot cos(\pi-\alpha)}=\cfrac{-sin\alpha\cdot cos\alpha}{sin\alpha\cdot(- cos\alpha)}=1$$

$$\cfrac{\pi}{4}<\theta<\cfrac{\pi}{2}$$，则$$\sqrt{1+sin2\theta}+\sqrt{1-sin2\theta}$$=$$\qquad$$

$A.2sin\theta$ $B.2cos\theta$ $C.-2sin\theta$ $D.-2cos\theta$

$$\sqrt{1+sin2\theta}+\sqrt{1-sin2\theta}$$

$$=\sqrt{sin^2\theta+2sin\theta\cdot cos\theta+cos^2\theta}+\sqrt{sin^2\theta-2sin\theta\cdot cos\theta+cos^2\theta}$$

$$=\sqrt{(sin\theta+cos\theta)^2}+\sqrt{(sin\theta-cos\theta)^2}$$

$$=|sin\theta+cos\theta|+|sin\theta-cos\theta|$$

$$=(sin\theta+cos\theta)+(sin\theta-cos\theta)=2sin\theta$$，故选$$A$$

$$\sqrt{1+sin2\theta}+\sqrt{1-sin2\theta}$$

$$=\sqrt{(\sqrt{1+sin2\theta}+\sqrt{1-sin2\theta})^2}$$

$$=\sqrt{2+2\sqrt{1+sin2\theta}\cdot \sqrt{1-sin2\theta}}$$

$$=\sqrt{2+2\sqrt{1^2-sin^22\theta}}$$

$$=\sqrt{2+2\sqrt{cos^22\theta}}=\sqrt{2+2|cos2\theta|}$$

$$=\sqrt{2-2cos2\theta}=\sqrt{2}\cdot \sqrt{1-cos2\theta}$$

$$=\sqrt{2}\cdot \sqrt{2sin^2\theta}=\sqrt{2}\cdot \sqrt{2}|sin\theta|$$

$$=2sin\theta$$，故选$$A$$ .

posted @ 2018-01-31 09:18  静雅斋数学  阅读(2273)  评论(0编辑  收藏  举报

----静雅斋