# 三角函数给角求值

## 相关变形

• 切化弦[整式变分式]，1的代换，分式通分约分，根式升幂；配方展开，提取公因式，公式的逆用，变用，
• 常用的互余、互补代换：$$sin70^{\circ}=cos20^{\circ}$$$$cos40^{\circ}=sin50^{\circ}$$$$sin140^{\circ}=sin40^{\circ}$$$$cos110^{\circ}=-sin70^{\circ}=-cos20^{\circ}$$
• 常见的角的拆分：

$$47^{\circ}=17^{\circ}+30^{\circ}$$$$8^{\circ}=15^{\circ}-7^{\circ}$$

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

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

• 常见的互余，倍角等

$$(\cfrac{\pi}{4}+\theta)+(\cfrac{\pi}{4}-\theta)=\cfrac{\pi}{2}$$$$(\cfrac{\pi}{3}+\theta)+(\cfrac{\pi}{6}-\theta)=\cfrac{\pi}{2}$$

$$2x\pm\cfrac{\pi}{2}=2(x\pm\cfrac{\pi}{4})$$$$2\alpha\pm\cfrac{\pi}{3}=2(\alpha\pm\cfrac{\pi}{6})$$

• 常见的配角技巧：

$$2\alpha=(\alpha+\beta)+(\alpha-\beta)$$$$2\beta=(\alpha+\beta)-(\alpha-\beta)$$

$$3\alpha-\beta=2(\alpha-\beta)+(\alpha-\beta)$$$$3\alpha+\beta=2(\alpha+\beta)+(\alpha-\beta)$$

$$\alpha=(\alpha+\beta)-\beta$$$$\beta=\alpha-(\alpha-\beta)$$

$$\alpha=\cfrac{\alpha+\beta}{2}+\cfrac{\alpha-\beta}{2}$$$$\beta=\cfrac{\alpha+\beta}{2}-\cfrac{\alpha-\beta}{2}$$

$$\alpha=(\alpha+\beta)-\beta$$$$(\cfrac{\pi}{6}-\alpha)+(\cfrac{\pi}{3}+\alpha)=\cfrac{\pi}{2}$$$$(\cfrac{\pi}{4}-\alpha)+(\cfrac{\pi}{4}+\alpha)=\cfrac{\pi}{2}$$

$$(\cfrac{\pi}{3}-\alpha)+(\cfrac{2\pi}{3}+\alpha)=\pi$$$$(\cfrac{\pi}{4}-\alpha)+(\cfrac{3\pi}{4}+\alpha)=\pi$$

## 难点变形

• $$\tan\theta-\sqrt{3}=\cfrac{\sin\theta}{\cos\theta}-\cfrac{\sqrt{3}\cos\theta}{\cos\theta}=\cfrac{2(\sin\theta\cdot \cfrac{1}{2}-\cos\theta\cdot\cfrac{\sqrt{3}}{2})}{\cos\theta}$$
• $$1+\sqrt{3}\tan\theta=\cfrac{\cos\theta}{\cos\theta}+\cfrac{\sqrt{3}\sin\theta}{\cos\theta}=\cfrac{\cos\theta+\sqrt{3}\sin\theta}{\cos\theta}=\cfrac{2(\cos\theta\cdot \cfrac{1}{2}+\sin\theta\cdot\cfrac{\sqrt{3}}{2})}{\cos\theta}$$

## 典例剖析

$$=\cfrac{sin(30^{\circ}-25^{\circ})+\cfrac{\sqrt{3}}{2}sin25^{\circ}}{cos25^{\circ}}$$$$=\cfrac{\cfrac{1}{2}cos25^{\circ}}{cos25^{\circ}}=\cfrac{1}{2}$$

$$=\cfrac{1+sin10^{\circ}}{2(1+sin10^{\circ})}=\cfrac{1}{2}$$

$$=\cfrac{4cos50^{\circ}cos40^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{4sin40^{\circ}cos40^{\circ}-sin40^{\circ}}{cos40^{\circ}}$$

$$=\cfrac{2sin80^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{2cos10^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{2cos(40^{\circ}-30^{\circ})-sin40^{\circ}}{cos40^{\circ}}$$

$$=\cfrac{2cos40^{\circ}\cdot \cfrac{\sqrt{3}}{2}+2sin40^{\circ}\cdot \cfrac{1}{2}-sin40^{\circ}}{cos40^{\circ}}=\sqrt{3}$$.

$$=\cfrac{sin(15^{\circ}-7^{\circ})+sin7^{\circ}cos15^{\circ}}{cos(15^{\circ}-7^{\circ})-sin7^{\circ}sin15^{\circ}}$$

$$=\cfrac{sin15^{\circ}}{cos15^{\circ}}=tan15^{\circ}=2-\sqrt{3}$$

$$=\cfrac{2\sin\cfrac{\pi}{17}\cos\cfrac{\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2\sin\cfrac{\pi}{17}}$$

$$=\cfrac{\sin\cfrac{2\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2\sin\cfrac{\pi}{17}}$$

$$=\cfrac{2\sin\cfrac{2\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2^2\sin\cfrac{\pi}{17}}$$

$$=\cfrac{2\sin\cfrac{4\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2^3\sin\cfrac{\pi}{17}}$$

$$=\cfrac{2\sin\cfrac{8\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2^4\sin\cfrac{\pi}{17}}$$

$$=\cfrac{\sin\cfrac{16\pi}{17}}{2^4\sin\cfrac{\pi}{17}}$$

$$=\cfrac{sin\cfrac{\pi}{17}}{2^4\sin\cfrac{\pi}{17}}$$

$$=\cfrac{1}{16}$$

$$1+2sinxcosx=\cfrac{1}{2}$$，故$$2sinxcosx=-\cfrac{1}{2}$$

$$sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sinx^2cos^2x$$

$$=1-2sinx^2cos^2x=1-\cfrac{1}{2}(2sinxcosx)^2=1-\cfrac{1}{8}=\cfrac{7}{8}$$

$$=\cfrac{sin(30^{\circ}+17^{\circ})-sin17^{\circ}cos30^{\circ}}{cos17^{\circ}}$$

$$=\cfrac{sin30^{\circ}cos17^{\circ}}{cos17^{\circ}}$$

$$=\sin30^{\circ}=\cfrac{1}{2}$$

$$=\cfrac{cos10^{\circ}}{2sin10^{\circ}}-sin10^{\circ}(\cfrac{cos^25^{\circ}-sin^25^{\circ}}{sin5^{\circ}cos5^{\circ}})$$

$$=\cfrac{cos10^{\circ}}{2sin10^{\circ}}-sin10^{\circ}\cfrac{2cos10^{\circ}}{2sin5^{\circ}cos5^{\circ}})$$

$$=\cfrac{cos10^{\circ}}{2sin10^{\circ}}-2cos10^{\circ}$$

$$==\cfrac{cos10^{\circ}}{2sin10^{\circ}}-\cfrac{2cos10^{\circ}\cdot 2sin10^{\circ}}{2sin10^{\circ}}$$

$$=\cfrac{cos10^{\circ}-2sin20^{\circ}}{2sin10^{\circ}}$$

$$=\cfrac{cos10^{\circ}-2sin(30^{\circ}-10^{\circ})}{2sin10^{\circ}}$$

$$=\cfrac{cos10^{\circ}-cos10^{\circ}+2\cdot \cfrac{\sqrt{3}}{2}sin10^{\circ}}{2sin10^{\circ}}$$

$$=\cfrac{\sqrt{3}}{2}$$

$$=\cfrac{cos10^{\circ}+\sqrt{3}sin10^{\circ}}{\sqrt{1-sin10^{\circ}}}$$

$$=\cfrac{cos10^{\circ}+\sqrt{3}sin10^{\circ}}{\sqrt{(cos5^{\circ}-sin5^{\circ})^2}}$$

$$=\cfrac{cos10^{\circ}+\sqrt{3}sin10^{\circ}}{(cos5^{\circ}-sin5^{\circ})^2}$$

$$=\cfrac{2sin(10^{\circ}+30^{\circ})}{-\sqrt{2}sin(5^{\circ}-45^{\circ})}$$

$$=\cfrac{2sin40^{\circ}}{\sqrt{2}sin40^{\circ}}=\sqrt{2}$$

$$=\cfrac{cos40^{\circ}}{cos25^{\circ}\cdot |sin20^{\circ}-cos20^{\circ}|}$$

$$=\cfrac{cos^220^{\circ}-sin^220^{\circ}}{cos25^{\circ}(cos20^{\circ}-sin20^{\circ})}$$

$$=\cfrac{cos20^{\circ}+sin20^{\circ}}{cos25^{\circ}}$$

$$=\cfrac{\sqrt{2}sin(20^{\circ}+45^{\circ})}{cos25^{\circ}}$$

$$=\cfrac{\sqrt{2}sin65^{\circ}}{cos25^{\circ}}=\sqrt{2}$$.

$$=\cfrac{\sqrt{3}\cdot \cfrac{sin12^{\circ}-\sqrt{3}cos12^{\circ}}{cos12^{\circ}}}{2cos24^{\circ}sin12^{\circ}}$$

$$=\cfrac{\sqrt{3}\cdot 2sin(12^{\circ}-60^{\circ})}{2cos24^{\circ}sin12^{\circ}cos12^{\circ}}$$

$$=\cfrac{2\sqrt{3}sin(-48^{\circ})}{sin24^{\circ}cos24^{\circ}}=-4\sqrt{3}$$

## 题型变化

【2020宝鸡市质检三文科第11题】著名数学家华罗庚先生被誉为“中国现代数学之父”，他倡导的“$$0.618$$优选法"在生产和科研实践中得到了非常广泛的应用，黄金分割比$$t=\cfrac{\sqrt{5}-1}{2}\approx 0.618$$，还可以表示成$$2\sin18^{\circ}$$，则$$\cfrac{2\cos^{2}27^{\circ}-1}{t\sqrt{4-t^{2}}}=$$$$\qquad$$

$A.4$ $B.\sqrt{5}-1$ $C.2$ $D.\cfrac{1}{2}$

$$\cfrac{2\cos^{2}27^{\circ}-1}{t\sqrt{4-t^{2}}}=\cfrac{\cos54^{\circ}}{2\sin18^{\circ}\sqrt{4-4\sin^{2}18^{\circ}}}=\cfrac{\cos54^{\circ}}{2\sin18^{\circ}\sqrt{4(1-\sin^{2}18^{\circ})}}$$

$$=\cfrac{\sin36^{\circ}}{2\sin18^{\circ}\cdot 2\cos18^{\circ}}=\cfrac{\sin36^{\circ}}{4\sin18^{\circ}\cos18^{\circ}}=\cfrac{1}{2}$$，故选$$D$$.

【2021届高三文科三轮模拟题】公元前 $$6$$ 世纪，古希腊的毕达哥拉斯学派通过研究正五边形和正十边形的作图，发现了黄金分割值约为 $$0.618$$，这一数值也可以表示为$$S=2\sin18^{\circ}$$，若 $$S^2+t=4$$ ，则 $$\cfrac{S\sqrt{t}}{2\sin^2207^{\circ}-1}$$=_____________.

$$\cfrac{S\sqrt{t}}{2\sin^2207^{\circ}-1}=\cfrac{2\sin18^{\circ}\sqrt{4-4\sin^218^{\circ}}}{2\sin^2207^{\circ}-1}$$

$$=\cfrac{2\sin18^{\circ}\cdot 2\cos18^{\circ}}{-\cos414^{\circ}}$$ $$=\cfrac{2\sin36^{\circ}}{-\cos54^{\circ}}=-2$$

## 对应练习

posted @ 2018-01-31 17:54  静雅斋数学  阅读(1194)  评论(0编辑  收藏  举报

----静雅斋