三角函数给角求值
前言
三角函数中的给角求值类问题,大多给定的是分式形式,或者可以化为分式形式的,比如含有弦和切,当切化弦后就变成了分式;并且这类题目往往需要将非特殊角拆分,然后最后一步约掉含有非特殊角的代数式,就得到了最终的值。 注意高频变形:分式约分,和加减抵消;
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
常涉及“切化弦”,“分式通分”,“辅助角公式”等高频变形;
- \(\tan\theta-\sqrt{3}=\cfrac{\sin\theta}{\cos\theta}-\cfrac{\sqrt{3}\cos\theta}{\cos\theta}=\cfrac{2(\sin\theta\cdot \cfrac{1}{2}-\cos\theta\cdot\cfrac{\sqrt{3}}{2})}{\cos\theta}\)
- \(1+\sqrt{3}\tan\theta=\cfrac{\cos\theta}{\cos\theta}+\cfrac{\sqrt{3}\sin\theta}{\cos\theta}=\cfrac{\cos\theta+\sqrt{3}\sin\theta}{\cos\theta}=\cfrac{2(\cos\theta\cdot \cfrac{1}{2}+\sin\theta\cdot\cfrac{\sqrt{3}}{2})}{\cos\theta}\)
注:在具体题目中,角\(\theta\)可以是具体的值,比如\(\tan12^{\circ}-\sqrt{3}\),或\(1+\sqrt{3}\tan21^{\circ}\)
典例剖析
分析:这类题目往往需要将非特殊角拆分,然后约掉含有非特殊角的代数式,就得到了最终的值。
原式=\(\cfrac{cos(90^{\circ}-5^{\circ})+\cfrac{\sqrt{3}}{2}sin25^{\circ}}{cos25^{\circ}}\)\(=\cfrac{sin5^{\circ}+\cfrac{\sqrt{3}}{2}sin25^{\circ}}{cos25^{\circ}}\)
\(=\cfrac{sin(30^{\circ}-25^{\circ})+\cfrac{\sqrt{3}}{2}sin25^{\circ}}{cos25^{\circ}}\)\(=\cfrac{\cfrac{1}{2}cos25^{\circ}}{cos25^{\circ}}=\cfrac{1}{2}\)
分析:原式=\(\cfrac{1-cos100^{\circ}}{2(1+sin10^{\circ})}\)\(=\cfrac{1-cos(90^{\circ}+10^{\circ})}{2(1+sin10^{\circ})}\)
\(=\cfrac{1+sin10^{\circ}}{2(1+sin10^{\circ})}=\cfrac{1}{2}\)
分析:\(\cfrac{3-sin70^{\circ}}{2-cos^210^{\circ}}=\cfrac{3-cos20^{\circ}}{2-cos^210^{\circ}}=\cfrac{3-(2cos^210^{\circ}-1)}{2-cos^210^{\circ}}=\cfrac{2(2-cos^210^{\circ})}{2-cos^210^{\circ}}=2\)
分析:\(4cos50^{\circ}-tan40^{\circ}=4cos50^{\circ}-\cfrac{sin40^{\circ}}{cos40^{\circ}}\)
\(=\cfrac{4cos50^{\circ}cos40^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{4sin40^{\circ}cos40^{\circ}-sin40^{\circ}}{cos40^{\circ}}\)
\(=\cfrac{2sin80^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{2cos10^{\circ}-sin40^{\circ}}{cos40^{\circ}}=\cfrac{2cos(40^{\circ}-30^{\circ})-sin40^{\circ}}{cos40^{\circ}}\)
\(=\cfrac{2cos40^{\circ}\cdot \cfrac{\sqrt{3}}{2}+2sin40^{\circ}\cdot \cfrac{1}{2}-sin40^{\circ}}{cos40^{\circ}}=\sqrt{3}\).
思路补充:\(\sin80^{\circ}=\sin(120^{\circ}-40^{\circ})\),\(\sin40^{\circ}=\sin(10^{\circ}+30^{\circ})\);
分析:\(\cfrac{sin8^{\circ}+sin7^{\circ}cos15^{\circ}}{cos8^{\circ}-sin7^{\circ}sin15^{\circ}}\)
\(=\cfrac{sin(15^{\circ}-7^{\circ})+sin7^{\circ}cos15^{\circ}}{cos(15^{\circ}-7^{\circ})-sin7^{\circ}sin15^{\circ}}\)
\(=\cfrac{sin15^{\circ}}{cos15^{\circ}}=tan15^{\circ}=2-\sqrt{3}\)。
分析:\(\cos\cfrac{\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}\)
\(=\cfrac{2\sin\cfrac{\pi}{17}\cos\cfrac{\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2\sin\cfrac{\pi}{17}}\)
\(=\cfrac{\sin\cfrac{2\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2\sin\cfrac{\pi}{17}}\)
\(=\cfrac{2\sin\cfrac{2\pi}{17}\cdot\cos\cfrac{2\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2^2\sin\cfrac{\pi}{17}}\)
\(=\cfrac{2\sin\cfrac{4\pi}{17}\cdot\cos\cfrac{4\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2^3\sin\cfrac{\pi}{17}}\)
\(=\cfrac{2\sin\cfrac{8\pi}{17}\cdot\cos\cfrac{8\pi}{17}}{2^4\sin\cfrac{\pi}{17}}\)
\(=\cfrac{\sin\cfrac{16\pi}{17}}{2^4\sin\cfrac{\pi}{17}}\)
\(=\cfrac{sin\cfrac{\pi}{17}}{2^4\sin\cfrac{\pi}{17}}\)
\(=\cfrac{1}{16}\)
分析:由题目可知,\((sinx+cosx)^2=(\cfrac{\sqrt{2}}{2})^2\),
即\(1+2sinxcosx=\cfrac{1}{2}\),故\(2sinxcosx=-\cfrac{1}{2}\)
\(sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sinx^2cos^2x\)
\(=1-2sinx^2cos^2x=1-\cfrac{1}{2}(2sinxcosx)^2=1-\cfrac{1}{8}=\cfrac{7}{8}\)
分析:\(\cfrac{sin47^{\circ}-sin17^{\circ}cos30^{\circ}}{cos17^{\circ}}\)
\(=\cfrac{sin(30^{\circ}+17^{\circ})-sin17^{\circ}cos30^{\circ}}{cos17^{\circ}}\)
\(=\cfrac{sin30^{\circ}cos17^{\circ}}{cos17^{\circ}}\)
\(=\sin30^{\circ}=\cfrac{1}{2}\)。
分析:原式\(=\cfrac{2cos^210^{\circ}}{2\cdot 2sin10^{\circ}cos10^{\circ}}-sin10^{\circ}(\cfrac{cos5^{\circ}}{sin5^{\circ}}-\cfrac{sin5^{\circ}}{cos5^{\circ}})\)
\(=\cfrac{cos10^{\circ}}{2sin10^{\circ}}-sin10^{\circ}(\cfrac{cos^25^{\circ}-sin^25^{\circ}}{sin5^{\circ}cos5^{\circ}})\)
\(=\cfrac{cos10^{\circ}}{2sin10^{\circ}}-sin10^{\circ}\cfrac{2cos10^{\circ}}{2sin5^{\circ}cos5^{\circ}})\)
\(=\cfrac{cos10^{\circ}}{2sin10^{\circ}}-2cos10^{\circ}\)
\(==\cfrac{cos10^{\circ}}{2sin10^{\circ}}-\cfrac{2cos10^{\circ}\cdot 2sin10^{\circ}}{2sin10^{\circ}}\)
\(=\cfrac{cos10^{\circ}-2sin20^{\circ}}{2sin10^{\circ}}\)
\(=\cfrac{cos10^{\circ}-2sin(30^{\circ}-10^{\circ})}{2sin10^{\circ}}\)
\(=\cfrac{cos10^{\circ}-cos10^{\circ}+2\cdot \cfrac{\sqrt{3}}{2}sin10^{\circ}}{2sin10^{\circ}}\)
\(=\cfrac{\sqrt{3}}{2}\)。
分析:原式\(=\cfrac{cos10^{\circ}-\sqrt{3}cos(100^{\circ})}{\sqrt{1-sin10^{\circ}}}\)
\(=\cfrac{cos10^{\circ}+\sqrt{3}sin10^{\circ}}{\sqrt{1-sin10^{\circ}}}\)
\(=\cfrac{cos10^{\circ}+\sqrt{3}sin10^{\circ}}{\sqrt{(cos5^{\circ}-sin5^{\circ})^2}}\)
\(=\cfrac{cos10^{\circ}+\sqrt{3}sin10^{\circ}}{(cos5^{\circ}-sin5^{\circ})^2}\)
\(=\cfrac{2sin(10^{\circ}+30^{\circ})}{-\sqrt{2}sin(5^{\circ}-45^{\circ})}\)
\(=\cfrac{2sin40^{\circ}}{\sqrt{2}sin40^{\circ}}=\sqrt{2}\)。
分析:原式\(=\cfrac{cos40^{\circ}}{cos25^{\circ}\cdot \sqrt{(sin20^{\circ}-cos20^{\circ})^2}}\)
\(=\cfrac{cos40^{\circ}}{cos25^{\circ}\cdot |sin20^{\circ}-cos20^{\circ}|}\)
\(=\cfrac{cos^220^{\circ}-sin^220^{\circ}}{cos25^{\circ}(cos20^{\circ}-sin20^{\circ})}\)
\(=\cfrac{cos20^{\circ}+sin20^{\circ}}{cos25^{\circ}}\)
\(=\cfrac{\sqrt{2}sin(20^{\circ}+45^{\circ})}{cos25^{\circ}}\)
\(=\cfrac{\sqrt{2}sin65^{\circ}}{cos25^{\circ}}=\sqrt{2}\).
分析:原式=\(\cfrac{\sqrt{3}\cfrac{sin12^{\circ}}{cos12^{\circ}}-3\cfrac{cos12^{\circ}}{cos12^{\circ}}}{2(2cos^212^{\circ}-1)sin12^{\circ}}\)
\(=\cfrac{\sqrt{3}\cdot \cfrac{sin12^{\circ}-\sqrt{3}cos12^{\circ}}{cos12^{\circ}}}{2cos24^{\circ}sin12^{\circ}}\)
\(=\cfrac{\sqrt{3}\cdot 2sin(12^{\circ}-60^{\circ})}{2cos24^{\circ}sin12^{\circ}cos12^{\circ}}\)
\(=\cfrac{2\sqrt{3}sin(-48^{\circ})}{sin24^{\circ}cos24^{\circ}}=-4\sqrt{3}\)。
题型变化
分析:由于\(t=2\sin18^{\circ}\),故有
\(\cfrac{2\cos^{2}27^{\circ}-1}{t\sqrt{4-t^{2}}}=\cfrac{\cos54^{\circ}}{2\sin18^{\circ}\sqrt{4-4\sin^{2}18^{\circ}}}=\cfrac{\cos54^{\circ}}{2\sin18^{\circ}\sqrt{4(1-\sin^{2}18^{\circ})}}\)
\(=\cfrac{\sin36^{\circ}}{2\sin18^{\circ}\cdot 2\cos18^{\circ}}=\cfrac{\sin36^{\circ}}{4\sin18^{\circ}\cos18^{\circ}}=\cfrac{1}{2}\),故选\(D\).
解析:由于 \(S^2+t=4\) ,则 \(t=4-S^2=4-4\sin^218^{\circ}\),
则 \(\cfrac{S\sqrt{t}}{2\sin^2207^{\circ}-1}=\cfrac{2\sin18^{\circ}\sqrt{4-4\sin^218^{\circ}}}{2\sin^2207^{\circ}-1}\)
\(=\cfrac{2\sin18^{\circ}\cdot 2\cos18^{\circ}}{-\cos414^{\circ}}\) \(=\cfrac{2\sin36^{\circ}}{-\cos54^{\circ}}=-2\)