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。

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 heta+cos heta=(1+cos heta)+sin heta=2cos^2cfrac{ heta}{2}+2sincfrac{ heta}{2}coscfrac{ heta}{2})

(1+sin heta-cos heta=(1-cos heta)+sin heta=2sin^2cfrac{ heta}{2}+2sincfrac{ heta}{2}coscfrac{ heta}{2})

• 常见的互余，倍角等

((cfrac{pi}{4}+ heta)+(cfrac{pi}{4}- heta)=cfrac{pi}{2})；((cfrac{pi}{3}+ heta)+(cfrac{pi}{6}- heta)=cfrac{pi}{2})；

(2xpmcfrac{pi}{2}=2(xpmcfrac{pi}{4}))；(2alphapmcfrac{pi}{3}=2(alphapmcfrac{pi}{6}))；

• 常见的配角技巧：

(2alpha=(alpha+eta)+(alpha-eta))；(2eta=(alpha+eta)-(alpha-eta))；

(alpha=(alpha+eta)-eta)；(eta=alpha-(alpha-eta))；

(alpha=cfrac{alpha+eta}{2}+cfrac{alpha-eta}{2})；(eta=cfrac{alpha+eta}{2}-cfrac{alpha-eta}{2})；

典例剖析

例1化简求值：(cfrac{3-sin70^{circ}}{2-cos^210^{circ}})

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)

例2化简求值：(4cos50^{circ}-tan40^{circ})

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}).

例3化简求值：(cfrac{sin8^{circ}+sin7^{circ}cos15^{circ}}{cos8^{circ}-sin7^{circ}sin15^{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})。

例4化简求值：(coscfrac{pi}{17}cdot coscfrac{2pi}{17}cdot coscfrac{4pi}{17}cdot coscfrac{8pi}{17})

分析：(coscfrac{pi}{17}cdot coscfrac{2pi}{17}cdot coscfrac{4pi}{17}cdot coscfrac{8pi}{17}=cfrac{2sincfrac{pi}{17}coscfrac{pi}{17}cdot coscfrac{2pi}{17}cdot coscfrac{4pi}{17}cdot coscfrac{8pi}{17}}{2sincfrac{pi}{17}})

(=cfrac{sincfrac{2pi}{17}cdot coscfrac{2pi}{17}cdot coscfrac{4pi}{17}cdot coscfrac{8pi}{17}}{2sincfrac{pi}{17}}=cfrac{2sincfrac{2pi}{17}cdot coscfrac{2pi}{17}cdot coscfrac{4pi}{17}cdot coscfrac{8pi}{17}}{2^2sincfrac{pi}{17}})

(=cfrac{2sincfrac{4pi}{17}cdot coscfrac{4pi}{17}cdot coscfrac{8pi}{17}}{2^3sincfrac{pi}{17}}=cfrac{2sincfrac{8pi}{17}cdot coscfrac{8pi}{17}}{2^4sincfrac{pi}{17}})

(=cfrac{sincfrac{16pi}{17}}{2^4sincfrac{pi}{17}}=cfrac{sincfrac{pi}{17}}{2^4sincfrac{pi}{17}}=cfrac{1}{16})

例5已知(sinx+cosx=cfrac{sqrt{2}}{2})，化简求值：(sin^4x+cos^4x)

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})

例6化简求值：(cfrac{sin47^{circ}-sin17^{circ}cos30^{circ}}{cos17^{circ}})

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})。

例7求值(cfrac{1+cos20^{circ}}{2sin20^{circ}}-sin10^{circ}(cfrac{1}{tan5^{circ}}-tan5^{circ}))

分析：原式(=cfrac{2cos^210^{circ}}{2cdot 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}+2cdot cfrac{sqrt{3}}{2}sin10^{circ}}{2sin10^{circ}})

(=cfrac{sqrt{3}}{2})。

例8求值(cfrac{cos10^{circ}-sqrt{3}cos(-100^{circ})}{sqrt{1-sin10^{circ}}})

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})。

例9化简求值(cfrac{cos40^{circ}}{cos25^{circ}cdot sqrt{1-sin40^{circ}}})

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}).

例10化简求值(cfrac{sqrt{3}tan12^{circ}-3}{(4cos^212^{circ}-2)sin12^{circ}})

分析：原式=(cfrac{sqrt{3}cfrac{sin12^{circ}}{cos12^{circ}}-3cfrac{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{2sqrt{3}sin(-48^{circ})}{sin24^{circ}cos24^{circ}}=-4sqrt{3})。

例11【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(2sin18^{circ})，则(cfrac{2cos^{2}27^{circ}-1}{tsqrt{4-t^{2}}}=)【】

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

分析：由于(t=2sin18^{circ})，故有

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

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

对应练习

练1化简求值(cfrac{sqrt{3}-tan12^{circ}}{(2cos^212^{circ}-1)sin12^{circ}}=8)

练2化简求值(sin50^{circ}(1+sqrt{3}tan10^{circ})=1)