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[2027强基计划]三角函数(I)

通过十五道典型例题梳理三角函数的恒等变形与综合应用，重点讲解和差化积、辅助角、万能公式、切化弦及对偶构造等解题方法。

6 分钟阅读强基计划 · 数学竞赛下载 Markdown

例4.1

（清华大学）设 α=π24\alpha=\frac{\pi}{24}，则 sin⁡αcos⁡4αcos⁡3α+sin⁡αcos⁡3αcos⁡2α+sin⁡αcos⁡2αcos⁡α+sin⁡αcos⁡α=\frac{\sin \alpha}{\cos 4 \alpha \cos 3 \alpha}+\frac{\sin \alpha}{\cos 3 \alpha \cos 2 \alpha}+\frac{\sin \alpha}{\cos 2 \alpha \cos \alpha}+\frac{\sin \alpha}{\cos \alpha}=

A. 36\frac{\sqrt{3}}{6}

B. 33\frac{\sqrt{3}}{3}

C. 32\frac{\sqrt{3}}{2}

D. 12\frac{1}{2}

考虑积化和差化简分母:

sin⁡αcos⁡4αcos⁡3α+sin⁡αcos⁡3αcos⁡2α+sin⁡αcos⁡2αcos⁡α+sin⁡αcos⁡α=2sin⁡αcos⁡7α+cos⁡α+⋯\begin{gathered} \frac{\sin \alpha}{\cos 4 \alpha \cos 3 \alpha}+\frac{\sin \alpha}{\cos 3 \alpha \cos 2 \alpha}+\frac{\sin \alpha}{\cos 2 \alpha \cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\\ =\frac{2\sin \alpha}{\cos 7\alpha+\cos\alpha}+\cdots \end{gathered}

分母更加复杂了,此路不通,考虑让分子向分母的形式凑:

sin⁡αcos⁡4αcos⁡3α+sin⁡αcos⁡3αcos⁡2α+sin⁡αcos⁡2αcos⁡α+sin⁡αcos⁡α=sin⁡(4α−3α)cos⁡4αcos⁡3α+sin⁡(3α−2α)cos⁡3αcos⁡2α+sin⁡(2α−α)cos⁡2αcos⁡α+sin⁡αcos⁡α=tan⁡4α−tan⁡3α+tan⁡3α−tan⁡2α+tan⁡2α−tan⁡α+tan⁡α=tan⁡4α=tan⁡π6=33\begin{gathered} \frac{\sin \alpha}{\cos 4 \alpha \cos 3 \alpha}+\frac{\sin \alpha}{\cos 3 \alpha \cos 2 \alpha}+\frac{\sin \alpha}{\cos 2 \alpha \cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\\ =\frac{\sin (4\alpha-3\alpha)}{\cos 4 \alpha \cos 3 \alpha}+\frac{\sin (3\alpha-2\alpha)}{\cos 3 \alpha \cos 2 \alpha}+\frac{\sin (2\alpha-\alpha)}{\cos 2 \alpha \cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\\ =\tan4\alpha-\tan3\alpha+\tan3\alpha-\tan2\alpha+\tan2\alpha-\tan\alpha+\tan\alpha\\ =\tan4\alpha=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3} \end{gathered}

例4.2

（同济大学）已知 sin⁡2(α+γ)=nsin⁡2β\sin2(\alpha + \gamma) = n \sin2\beta，则 tan⁡(α+β+γ)tan⁡(α−β+γ)=\frac{\tan(\alpha + \beta + \gamma)}{\tan(\alpha - \beta + \gamma)} =

A. n−1n+1\frac{n-1}{n+1}

B. nn+1\frac{n}{n+1}

C. nn−1\frac{n}{n-1}

D. n+1n−1\frac{n+1}{n-1}

tan⁡(α+β+γ)tan⁡(α−β+γ)=tan⁡[(α+γ)−β]tan⁡[(α+γ)+β]=tan⁡(α+γ)−tan⁡β1+tan⁡(α+γ)tan⁡βtan⁡(α+γ)+tan⁡β1−tan⁡(α+γ)tan⁡β\begin{gathered} \frac{\tan(\alpha + \beta + \gamma)}{\tan(\alpha - \beta + \gamma)}\\ =\frac{\tan[(\alpha+\gamma)-\beta]}{\tan[(\alpha+\gamma)+\beta]}\\ =\frac{\frac{\tan(\alpha+\gamma)-\tan\beta}{1+\tan(\alpha+\gamma)\tan\beta}}{\frac{\tan(\alpha+\gamma)+\tan\beta}{1-\tan(\alpha+\gamma)\tan\beta}} \end{gathered}

一筹莫展.我们看条件如何用万能公式化简:

2tan⁡(α+γ)1+tan⁡2(α+γ)=n2tan⁡(β)1+tan⁡2(β)\begin{gathered} \frac{2\tan(\alpha+\gamma)}{1+\tan^2(\alpha+\gamma)}=n\frac{2\tan(\beta)}{1+\tan^2(\beta)} \end{gathered}

条件和结论都没有得到有效的化简.前方的路不好走,考虑从结果入手:

A=α+β+γ,B=α−β+γsin⁡(A+B)=nsin⁡(A−B)sin⁡Acos⁡B+cos⁡Asin⁡B=n(sin⁡Acos⁡B−cos⁡Asin⁡B)(n+1)sin⁡Bcos⁡A=(n−1)sin⁡Acos⁡B(n+1)tan⁡B=(n−1)tan⁡Atan⁡Atan⁡B=n+1n−1\begin{gathered} A=\alpha + \beta + \gamma,B=\alpha - \beta + \gamma\\ \sin(A+B)=n\sin(A-B)\\ \sin A\cos B+\cos A\sin B=n(\sin A\cos B -\cos A\sin B)\\ (n+1)\sin B\cos A=(n-1)\sin A\cos B\\ (n+1)\tan B=(n-1)\tan A\\ \frac{\tan A}{\tan B}=\frac{n+1}{n-1} \end{gathered}

例4.3

（2025 北京大学）若 α,β\alpha, \beta 是 3cos⁡x+2sin⁡x=c3\cos x + 2\sin x = c 的两解，且 α−β≠kπ\alpha - \beta \neq k\pi (k∈Zk \in \mathbb{Z})，求 tan⁡(α+β)\tan(\alpha + \beta)。

引入辅助角φ\varphi: 13(sin⁡φcos⁡x+cos⁡φsin⁡x)=csin⁡(φ+x)=c13{sin⁡φ=313,cos⁡φ=213φ+x=(−1)narcsin⁡(c13)+nπ(n∈Z)\begin{gathered} \sqrt{13}(\sin\varphi\cos x+\cos\varphi\sin x)=c\\ \sin(\varphi+x)=\frac{c}{\sqrt{13}}\\ \begin{cases} \sin\varphi=\frac{3}{\sqrt{13}},\\ \cos\varphi=\frac{2}{\sqrt{13}} \end{cases}\\ \varphi+x=(-1)^n\arcsin(\frac{c}{\sqrt{13}})+n\pi(n\in \Z)\\ \end{gathered}

由于α−β≠kπ\alpha - \beta \neq k\pi (k∈Zk \in \mathbb{Z}),所以α,β\alpha,\beta对应的n奇偶性不同,不妨设:

φ+α=arcsin⁡(c13)φ+β=π−arcsin⁡(c13)tan⁡(α+β)=tan⁡(π−2φ)=−tan⁡2φ=−2tan⁡φ1−tan⁡2φ=2×32(32)2−1=125\begin{gathered} \varphi+\alpha=\arcsin(\frac{c}{\sqrt{13}})\\ \varphi+\beta=\pi-\arcsin(\frac{c}{\sqrt{13}})\\ \tan(\alpha+\beta)=\tan(\pi-2\varphi)\\ =-\tan2\varphi=-\frac{2\tan\varphi}{1-\tan^2\varphi}\\ =\frac{2\times\frac{3}{2}}{(\frac{3}{2})^2-1}\\ =\frac{12}{5} \end{gathered}

例4.4

2026 北京大学）在 △ABC\triangle ABC 中，已知 sin⁡A+3cos⁡Acos⁡A−3sin⁡A=tan⁡7π12\frac{\sin A + \sqrt{3} \cos A}{\cos A - \sqrt{3} \sin A} = \tan \frac{7\pi}{12}，则 sin⁡2B+2cos⁡C\sin 2B + 2 \cos C 的取值范围为__________。

考虑对条件齐次化: tan⁡A+tan⁡π31−tan⁡Atan⁡π3=tan⁡7π12tan⁡(A+π3)=tan⁡7π12A+π3=7π12+kπ(k∈Z)A∈(0,π)A=π4B+C=π−A=3π4sin⁡[2(3π4−C)]+2cos⁡C=sin⁡(3π2−2C)+2cos⁡C=−sin⁡(π2−2C)+2cos⁡C=2cos⁡C−cos⁡2C=2cos⁡C−(2cos⁡2C−1)=−2cos⁡2C+2cos⁡C+1=−2(cos⁡C−12)2+32C∈(0,3π4)cos⁡C∈(−22,1)−2(cos⁡C−12)2+32∈(−2,32]\begin{gathered} \frac{\tan A+\tan\frac{\pi}{3}}{1-\tan A\tan\frac{\pi}{3}}=\tan\frac{7\pi}{12}\\ \tan(A+\frac{\pi}{3})=\tan\frac{7\pi}{12}\\ A+\frac{\pi}{3}=\frac{7\pi}{12}+k\pi(k\in\Z)\\ A\in(0,\pi)\\ A=\frac{\pi}{4}\\ B+C=\pi-A=\frac{3\pi}{4}\\ \sin[2(\frac{3\pi}{4}-C)]+2\cos C\\ =\sin(\frac{3\pi}{2}-2C)+2\cos C\\ =-\sin(\frac{\pi}{2}-2C)+2\cos C\\ =2\cos C-\cos2C\\ =2\cos C-(2\cos^2C-1)\\ =-2\cos^2C+2\cos C+1=-2(\cos C-\frac{1}{2})^2+\frac{3}{2}\\ C\in(0,\frac{3\pi}{4})\\ \cos C\in(-\frac{\sqrt{2}}{2},1)\\ -2(\cos C-\frac{1}{2})^2+\frac{3}{2}\in(-\sqrt{2},\frac{3}{2}] \end{gathered}

例4.5

（清华大学）已知 x,yx, y 满足 sin⁡x+sin⁡y=13\sin x + \sin y = \frac{1}{3}，cos⁡x−cos⁡y=15\cos x - \cos y = \frac{1}{5}，则 cos⁡(x+y)+sin⁡(x−y)\cos(x + y) + \sin(x - y) 的值为

A. 32765\frac{32}{765}
B. 1613825\frac{161}{3825} C. 18425\frac{18}{425}
D. 1633825\frac{163}{3825}

使用和差化积: 2sin⁡x+y2cos⁡x−y2=13,(1)−2sin⁡x+y2sin⁡x−y2=15(2)(2)÷(1):tan⁡x−y2=−35\begin{gathered} 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=\frac{1}{3}, (1)\\ -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}=\frac{1}{5} (2)\\ (2)\div(1):\tan\frac{x-y}{2}=-\frac{3}{5}\\ \end{gathered}

使用万能公式计算sin⁡(x−y)\sin(x - y): sin⁡(x−y)=2tan⁡x−y21+tan⁡2x−y2=−1517\begin{gathered} \sin(x-y)=\frac{2\tan\frac{x-y}{2}}{1+\tan^2\frac{x-y}{2}}=-\frac{15}{17} \end{gathered}

然后,考虑条件平方相加: 2−2cos⁡(x+y)=34225cos⁡(x+y)=208225\begin{gathered} 2-2\cos(x+y)=\frac{34}{225}\\ \cos(x+y)=\frac{208}{225}\\ \end{gathered}

最后的计算结果:208225−1517=1613825\frac{208}{225}-\frac{15}{17}=\frac{161}{3825}

例4.6

（复旦大学）已知 sin⁡α+cos⁡β=32\sin \alpha + \cos \beta = \frac{\sqrt{3}}{2}，cos⁡α+sin⁡β=2\cos \alpha + \sin \beta = \sqrt{2}，求 tan⁡α⋅cot⁡β\tan \alpha \cdot \cot \beta 的值。

审视一下所求式: tan⁡α⋅cot⁡β=sin⁡αcos⁡βcos⁡αsin⁡β\begin{gathered} \tan \alpha \cdot \cot \beta=\frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta} \end{gathered}

条件平方相加可以凑出分子加分母.

2+2(sin⁡αcos⁡β+cos⁡αsin⁡β)=2+34sin⁡αcos⁡β+cos⁡αsin⁡β=38\begin{gathered} 2+2(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=2+\frac{3}{4}\\ \sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{3}{8} \end{gathered}

如果可以凑出分子减分母,问题便迎刃而解.

我们通过条件平方相减实现: 2(sin⁡αcos⁡β−cos⁡αsin⁡β)+sin⁡2α+cos⁡2β−cos⁡2α−sin⁡2β=−542(sin⁡αcos⁡β−cos⁡αsin⁡β)−cos⁡2α+cos⁡2β=−542(sin⁡αcos⁡β−cos⁡αsin⁡β)−2sin⁡(β+α)sin⁡(β−α)=−54(sin⁡αcos⁡β−cos⁡αsin⁡β)−38sin⁡(β−α)=−58(sin⁡αcos⁡β−cos⁡αsin⁡β)+38(sin⁡αcos⁡β−cos⁡αsin⁡β)=−58sin⁡αcos⁡β−cos⁡αsin⁡β=−511\begin{gathered} 2(\sin\alpha\cos\beta-\cos\alpha\sin\beta)+\sin^2\alpha+\cos^2\beta-\cos^2\alpha-\sin^2\beta=-\frac{5}{4}\\ 2(\sin\alpha\cos\beta-\cos\alpha\sin\beta)-\cos2\alpha+\cos2\beta=-\frac{5}{4}\\ 2(\sin\alpha\cos\beta-\cos\alpha\sin\beta)-2\sin(\beta+\alpha)\sin(\beta-\alpha)=-\frac{5}{4}\\ (\sin\alpha\cos\beta-\cos\alpha\sin\beta)-\frac{3}{8}\sin(\beta-\alpha)=-\frac{5}{8}\\ (\sin\alpha\cos\beta-\cos\alpha\sin\beta)+\frac{3}{8}(\sin\alpha\cos\beta-\cos\alpha\sin\beta)=-\frac{5}{8}\\ \sin\alpha\cos\beta-\cos\alpha\sin\beta=-\frac{5}{11} \end{gathered}

联立和与差,得: sin⁡αcos⁡β=−7176cos⁡αsin⁡β=73176tan⁡α⋅cot⁡β=sin⁡αcos⁡βcos⁡αsin⁡β=−773\begin{gathered} \sin\alpha\cos\beta=-\frac{7}{176}\\ \cos\alpha\sin\beta=\frac{73}{176}\\ \tan \alpha \cdot \cot \beta=\frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta}=-\frac{7}{73} \end{gathered}

例4.7

（复旦大学）解方程：cos⁡3x⋅tan⁡5x‾=sin⁡7x\cos 3x \cdot \underline{\tan 5x} = \sin 7x。

tan⁡5x\tan5x是不和谐之处,应该同乘cos⁡5x\cos5x实现切化弦,同时考虑增根问题: cos⁡3xsin⁡5x=sin⁡7xcos⁡5xsin⁡8x+sin⁡2x=sin⁡12x+sin⁡2xsin⁡8x=sin⁡12x12x=8x+2kπ(k∈Z) or 12x=(π−8x)+2kπ(k∈Z)x=kπ2(k∈Z) or x=(2k+1)π20(k∈Z)\begin{gathered} \cos3x\sin5x=\sin7x\cos5x\\ \sin8x+\sin2x=\sin12x+\sin2x\\ \sin8x=\sin12x\\ 12x=8x+2k\pi(k\in\Z)\text{ or }12x=(\pi-8x)+2k\pi(k\in\Z)\\ x=\frac{k\pi}{2}(k\in\Z)\text{ or }x=\frac{(2k+1)\pi}{20}(k\in\Z) \end{gathered}

舍去定义域外的根: 5x≠π2+kπ(k∈Z)x≠(2k+1)π10\begin{gathered} 5x\ne\frac{\pi}{2}+k\pi(k\in\Z)\\ x\ne\frac{(2k+1)\pi}{10} \end{gathered}

对于x=kπ2(k∈Z)x=\frac{k\pi}{2}(k\in\Z),kk不能取奇数,故化为x=kπ(k∈Z)x=k\pi(k\in\Z)

对于x=(2k+1)π20(k∈Z)x=\frac{(2k+1)\pi}{20}(k\in\Z),k∈Zk\in\Z均符合条件.

{x∣x=kπ 或 x=(2k+1)π20,k∈Z}\boxed{\{x|x=k\pi\text{ 或 }x=\frac{(2k+1)\pi}{20},k\in\Z\}}

例4.8

（北京大学）已知 sin⁡x,sin⁡y,sin⁡z\sin x, \sin y, \sin z 是递增的等差数列，求证：cos⁡x,cos⁡y,cos⁡z\cos x, \cos y, \cos z 不是等差数列。

采取反证法:假设cos⁡x,cos⁡y,cos⁡z\cos x, \cos y, \cos z 是等差数列. sin⁡x+sin⁡z=2sin⁡y(1)cos⁡x+cos⁡z=2cos⁡y(2)(1)2+(2)2:2+2cos⁡(x−z)=4cos⁡(x−z)=+1\begin{gathered} \sin x+\sin z=2\sin y(1)\\ \cos x+\cos z=2\cos y(2)\\ (1)^2+(2)^2:2+2\cos(x-z)=4\\ \cos(x-z)=+1 \end{gathered}

cos⁡(x−z)=1⟺z−x=2kπ(k∈Z)\cos(x-z)=1\Longleftrightarrow z-x=2k\pi(k\in\Z),故sin⁡x=sin⁡z\sin x=\sin z,这与递增的条件矛盾.

例4.9

(2024 清华大学) 已知 {sin⁡θ,sin⁡2θ,sin⁡3θ}={cos⁡θ,cos⁡2θ,cos⁡3θ}\{\sin \theta, \sin 2\theta, \sin 3\theta\} = \{\cos \theta, \cos 2\theta, \cos 3\theta\}，则 θ\theta 的可能值是______。

元素配对种类繁多,考虑整体条件或为简便: sin⁡θ+sin⁡2θ+sin⁡3θ=cos⁡θ+cos⁡2θ+cos⁡3θ2sin⁡2θcos⁡θ+sin⁡2θ=2cos⁡2θcos⁡θ+cos⁡2θsin⁡2θ(2cos⁡θ+1)=cos⁡2θ(2cos⁡θ+1)\begin{gathered} \sin\theta+\sin2\theta+\sin3\theta=\cos\theta+\cos2\theta+\cos3\theta\\ 2\sin2\theta\cos\theta+\sin2\theta=2\cos2\theta\cos\theta+\cos2\theta\\ \sin2\theta(2\cos\theta+1)=\cos2\theta(2\cos\theta+1) \end{gathered}

考虑两条岔路,先难后易: 2cos⁡θ+1=0cos⁡θ=−12cos⁡2θ=2cos⁡2θ−1=−12=cos⁡θ\begin{gathered} 2\cos\theta+1=0\\ \cos\theta=-\frac{1}{2}\\ \cos2\theta=2\cos^2\theta-1=-\frac{1}{2}=\cos\theta \end{gathered}

这与集合的互异性矛盾. sin⁡2θ=cos⁡2θsin⁡(2θ−π4)=02θ−π4=kπ(k∈Z)θ=π8+kπ2(k∈Z)\begin{gathered} \sin2\theta=\cos2\theta\\ \sin(2\theta-\frac{\pi}{4})=0\\ 2\theta-\frac{\pi}{4}=k\pi(k\in\Z)\\ \theta=\frac{\pi}{8}+\frac{k\pi}{2}(k\in\Z) \end{gathered}

接下来,应该检验θ=π8+kπ2(k∈Z)\theta=\frac{\pi}{8}+\frac{k\pi}{2}(k\in\Z):

{sin⁡θ,sin⁡3θ}={cos⁡θ,cos⁡3θ}sin⁡θsin⁡3θ=cos⁡θcos⁡3θ−(cos⁡4θ−cos⁡θ)=cos⁡4θ+cos⁡θcos⁡4θ=0\begin{gathered} \{\sin\theta,\sin3\theta\}=\{\cos\theta,\cos3\theta\}\\ \sin\theta\sin3\theta=\cos\theta\cos3\theta\\ -(\cos4\theta-\cos\theta)=\cos4\theta+\cos\theta\\ \cos4\theta=0 \end{gathered} θ=π8+kπ2(k∈Z)\theta=\frac{\pi}{8}+\frac{k\pi}{2}(k\in\Z)满足cos⁡4θ=0\cos4\theta=0,进一步考虑元素的互异性: sin⁡θ≠sin⁡3θ3θ≠θ+2kπ and 3θ≠(π−θ)+2kπ(k∈Z)θ≠kπ and θ≠π4+kπ2\begin{gathered} \sin\theta\ne\sin3\theta\\ 3\theta\ne \theta+2k\pi\text{ and }3\theta\ne(\pi-\theta)+2k\pi(k\in\Z)\\ \theta\ne k\pi\text{ and }\theta\ne\frac{\pi}{4}+\frac{k\pi}{2} \end{gathered} 那么,所有的θ=π8+kπ2(k∈Z)\theta=\frac{\pi}{8}+\frac{k\pi}{2}(k\in\Z)都能使得{sin⁡θ,sin⁡3θ}={cos⁡θ,cos⁡3θ}\{\sin\theta,\sin3\theta\}=\{\cos\theta,\cos3\theta\}(因为元素的和/积对应相等,且满足元素的互异性).

更进一步,检验整体集合的元素互异性: sin⁡θ≠sin⁡2θ2θ≠θ+2kπ and 2θ≠(π−θ)+2kπ(k∈Z)θ≠2kπ and θ≠π3+2kπ3(k∈Z)sin⁡2θ≠sin⁡3θ3θ≠2θ+2kπ and 3θ≠(π−2θ)+2kπ(k∈Z)θ≠2kπ and θ≠π5+2kπ5(k∈Z)\begin{gathered} \sin\theta\ne\sin2\theta\\ 2\theta\ne\theta+2k\pi\text{ and }2\theta\ne(\pi-\theta)+2k\pi(k\in\Z)\\ \theta\ne2k\pi\text{ and }\theta\ne\frac{\pi}{3}+\frac{2k\pi}{3}(k\in\Z)\\ \sin2\theta\ne\sin3\theta\\ 3\theta\ne2\theta+2k\pi\text{ and }3\theta\ne(\pi-2\theta)+2k\pi(k\in\Z)\\ \theta\ne2k\pi\text{ and }\theta\ne\frac{\pi}{5}+\frac{2k\pi}{5}(k\in\Z) \end{gathered} 显然θ=π8+kπ2(k∈Z)\theta=\frac{\pi}{8}+\frac{k\pi}{2}(k\in\Z)可以胜任这些要求,故为最终结果.

例4.10

求 cos⁡π7⋅cos⁡2π7⋅cos⁡3π7\cos \frac{\pi}{7} \cdot \cos \frac{2\pi}{7} \cdot \cos \frac{3\pi}{7} 的值。 sin⁡π7cos⁡π7⋅cos⁡2π7⋅(−cos⁡4π7)=−sin⁡2π7cos⁡2π7⋅cos⁡4π72=−sin⁡4π7cos⁡4π74=−sin⁡8π78=sin⁡π78\begin{gathered} \sin \frac{\pi}{7}\cos \frac{\pi}{7} \cdot \cos \frac{2\pi}{7} \cdot (-\cos \frac{4\pi}{7})\\ =-\frac{\sin\frac{2\pi}{7}\cos \frac{2\pi}{7} \cdot \cos \frac{4\pi}{7}}{2}\\ =-\frac{\sin \frac{4\pi}{7}\cos \frac{4\pi}{7}}{4}\\ =-\frac{\sin\frac{8\pi}{7}}{8}\\ =\frac{\sin\frac{\pi}{7}}{8} \end{gathered}

得到cos⁡π7⋅cos⁡2π7⋅cos⁡3π7=18\cos \frac{\pi}{7} \cdot \cos \frac{2\pi}{7} \cdot \cos \frac{3\pi}{7}=\frac{1}{8}

或者,构造对偶式: A=cos⁡π7⋅cos⁡2π7⋅cos⁡3π7,B=sin⁡π7⋅sin⁡2π7⋅sin⁡3π7AB=18sin⁡2π7sin⁡4π7sin⁡6π7=18sin⁡π7⋅sin⁡2π7⋅sin⁡3π7=18B⟹A=18\begin{gathered} A=\cos \frac{\pi}{7} \cdot \cos \frac{2\pi}{7} \cdot \cos \frac{3\pi}{7},B=\sin \frac{\pi}{7} \cdot \sin \frac{2\pi}{7} \cdot \sin \frac{3\pi}{7}\\ AB=\frac{1}{8}\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}\sin\frac{6\pi}{7}\\ =\frac{1}{8}\sin \frac{\pi}{7} \cdot \sin \frac{2\pi}{7} \cdot \sin \frac{3\pi}{7}=\frac{1}{8}B\\ \Longrightarrow A=\frac{1}{8} \end{gathered}

例4.11

求 cos⁡π11⋅cos⁡2π11⋯cos⁡10π11\cos\frac{\pi}{11}\cdot\cos\frac{2\pi}{11}\cdots\cos\frac{10\pi}{11} 的值。

不难发现,乘数中出现了周期性: A=cos⁡π11cos⁡2π11cos⁡3π11cos⁡4π11cos⁡5π11cos⁡π11⋅cos⁡2π11⋯cos⁡10π11=−A2\begin{gathered} A=\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}\\ \cos\frac{\pi}{11}\cdot\cos\frac{2\pi}{11}\cdots\cos\frac{10\pi}{11}=-A^2 \end{gathered} 照猫画虎,引入对偶式: B=sin⁡π11sin⁡2π11sin⁡3π11sin⁡4π11sin⁡5π11AB=125sin⁡2π11sin⁡4π11sin⁡6π11sin⁡8π11sin⁡10π11=125sin⁡2π11sin⁡4π11sin⁡5π11sin⁡3π11sin⁡1π11=125B⟹A=125cos⁡π11⋅cos⁡2π11⋯cos⁡10π11=−A2=−1210=−11024\begin{gathered} B=\sin\frac{\pi}{11}\sin\frac{2\pi}{11}\sin\frac{3\pi}{11}\sin\frac{4\pi}{11}\sin\frac{5\pi}{11}\\ AB=\frac{1}{2^5}\sin\frac{2\pi}{11}\sin\frac{4\pi}{11}\sin\frac{6\pi}{11}\sin\frac{8\pi}{11}\sin\frac{10\pi}{11}\\ =\frac{1}{2^5}\sin\frac{2\pi}{11}\sin\frac{4\pi}{11}\sin\frac{5\pi}{11}\sin\frac{3\pi}{11}\sin\frac{1\pi}{11}=\frac{1}{2^5}B\\ \Longrightarrow A=\frac{1}{2^5}\\ \cos\frac{\pi}{11}\cdot\cos\frac{2\pi}{11}\cdots\cos\frac{10\pi}{11}=-A^2=-\frac{1}{2^{10}}=-\frac{1}{1024} \end{gathered}

例4.12

求 cos⁡π7−cos⁡2π7+cos⁡3π7\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} 的值。 继续构造对偶式: A=cos⁡π7−cos⁡2π7+cos⁡3π7B=sin⁡π7−sin⁡2π7+sin⁡3π7A2+B2=3−2cos⁡π7−2cos⁡π7+2cos⁡2π7=3−4cos⁡π7+2cos⁡2π7A2−B2=cos⁡2π7+cos⁡4π7+cos⁡8π7−2cos⁡3π7−2cos⁡5π7+2cos⁡4π7=cos⁡2π7−cos⁡3π7−cos⁡π7−4cos⁡3π7+2cos⁡2π7=−cos⁡π7+3cos⁡2π7−5cos⁡3π7(A2+B2)+(A2−B2)=3−5A=2A22A2+5A−3=0⟹A=−3(discard),12\begin{gathered} A=\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7}\\ B=\sin\frac{\pi}{7} - \sin\frac{2\pi}{7} + \sin\frac{3\pi}{7}\\ A^2+B^2=3-2\cos\frac{\pi}{7}-2\cos\frac{\pi}{7}+2\cos\frac{2\pi}{7}\\ =3-4\cos\frac{\pi}{7}+2\cos\frac{2\pi}{7}\\ A^2-B^2=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}-2\cos\frac{3\pi}{7}-2\cos\frac{5\pi}{7}+2\cos\frac{4\pi}{7}\\ =\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}-\cos\frac{\pi}{7}-4\cos\frac{3\pi}{7}+2\cos\frac{2\pi}{7}\\ =-\cos\frac{\pi}{7}+3\cos\frac{2\pi}{7}-5\cos\frac{3\pi}{7}\\ (A^2+B^2)+(A^2-B^2)=3-5A=2A^2\\ 2A^2+5A-3=0\\ \Longrightarrow A=-3(\text{discard}),\frac{1}{2} \end{gathered} cos⁡π7−cos⁡2π7+cos⁡3π7=12\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7}=\frac{1}{2} 或者,考虑用诱导公式去掉讨厌的负号: cos⁡π7−cos⁡2π7+cos⁡3π7=cos⁡π7+cos⁡3π7+cos⁡5π7\cos\frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7}=\cos\frac{\pi}{7} + \cos\frac{3\pi}{7} + \cos\frac{5\pi}{7} 我们发现,这正是之前讨论过的经典问题,剩余两种处理思路(单位根/构造裂项)不加赘述.

例4.13

(北京大学) (1+cos⁡π5)(1+cos⁡3π5)\left(1+\cos \frac{\pi}{5}\right)\left(1+\cos \frac{3\pi}{5}\right) 的值为

A. 1+551+\frac{\sqrt{5}}{5}

B. 54\frac{5}{4}

C. 1+331+\frac{\sqrt{3}}{3}

D. 前三个答案都不对

(1+cos⁡π5)(1+cos⁡3π5)=1+cos⁡π5cos⁡3π5+cos⁡π5+cos⁡3π5=1+12(cos⁡4π5+cos⁡2π5)+cos⁡π5+cos⁡3π5=1+12(cos⁡π5+cos⁡3π5)=1+cos⁡2π5cos⁡π5=1+sin⁡π5cos⁡π5cos⁡2π5sin⁡π5=1+sin⁡2π5cos⁡2π52sin⁡π5=1+sin⁡4π54sin⁡π5=54\begin{gathered} \left(1+\cos \frac{\pi}{5}\right)\left(1+\cos \frac{3\pi}{5}\right)\\ =1+\cos\frac{\pi}{5}\cos\frac{3\pi}{5}+\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}\\ =1+\frac{1}{2}(\cos\frac{4\pi}{5}+\cos\frac{2\pi}{5})+\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}\\ =1+\frac{1}{2}(\cos\frac{\pi}{5}+\cos\frac{3\pi}{5})\\ =1+\cos\frac{2\pi}{5}\cos\frac{\pi}{5}\\ =1+\frac{\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}}{\sin\frac{\pi}{5}}\\ =1+\frac{\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}}{2\sin\frac{\pi}{5}}\\ =1+\frac{\sin\frac{4\pi}{5}}{4\sin\frac{\pi}{5}}=\frac{5}{4} \end{gathered}

对于cos⁡2π5cos⁡π5\cos\frac{2\pi}{5}\cos\frac{\pi}{5},仍可以构造对偶式: A=cos⁡2π5cos⁡π5,B=sin⁡2π5sin⁡π5AB=14sin⁡4π5sin⁡2π5=14B⟹A=14\begin{gathered} A=\cos\frac{2\pi}{5}\cos\frac{\pi}{5},\\ B=\sin\frac{2\pi}{5}\sin\frac{\pi}{5}\\ AB=\frac{1}{4}\sin\frac{4\pi}{5}\sin\frac{2\pi}{5}=\frac{1}{4}B\\ \Longrightarrow A=\frac{1}{4} \end{gathered}

此外,利用sin⁡π5=5−14\sin\frac{\pi}{5}=\frac{\sqrt{5}-1}{4}(黄金分割率的一半)也可行

例4.14

(2024 北京大学) 求 sin⁡36∘−sin⁡3114∘+sin⁡3126∘\sin^3 6^\circ - \sin^3 114^\circ + \sin^3 126^\circ。

注意到114∘=120∘−6∘,126∘=120∘+6∘114\degree=120\degree-6\degree,126\degree=120\degree+6\degree.

逆用正弦三倍角公式降幂升角:sin⁡3x=3sin⁡x−4sin⁡3x⟺sin⁡3x=3sin⁡x−sin⁡3x4\sin3x=3\sin x-4\sin^3x\Longleftrightarrow \sin^3x=\frac{3\sin x-\sin3x}{4} sin⁡36∘−sin⁡3114∘+sin⁡3126∘=3sin⁡6∘−sin⁡18∘4−3sin⁡114∘−sin⁡342∘4+3sin⁡126∘−sin⁡378∘4=3sin⁡6∘−sin⁡18∘4−3sin⁡114∘+sin⁡18∘4+3sin⁡126∘−sin⁡18∘4=34(sin⁡6∘−sin⁡18∘−sin⁡114∘+sin⁡126∘)=34(2sin⁡66∘cos⁡60∘−sin⁡114∘−sin⁡18∘)=−34sin⁡18∘=−345−14=−3(5−1)16\begin{gathered} \sin^3 6^\circ - \sin^3 114^\circ + \sin^3 126^\circ\\ =\frac{3\sin6\degree-\sin18\degree}{4}-\frac{3\sin114\degree-\sin342\degree}{4}+\frac{3\sin126\degree-\sin378\degree}{4}\\ =\frac{3\sin6\degree-\sin18\degree}{4}-\frac{3\sin114\degree+\sin18\degree}{4}+\frac{3\sin126\degree-\sin18\degree}{4}\\ =\frac{3}{4}(\sin6\degree-\sin18\degree-\sin114\degree+\sin126\degree)\\ =\frac{3}{4}(2\sin66\degree\cos60\degree-\sin114\degree-\sin18\degree)\\ =-\frac{3}{4}\sin18\degree\\ =-\frac{3}{4}\frac{\sqrt{5}-1}{4}=-\frac{3(\sqrt{5}-1)}{16} \end{gathered}

例4.15

求值：sin⁡210∘+cos⁡240∘+sin⁡10∘⋅cos⁡40∘\sin^2 10^\circ + \cos^2 40^\circ + \sin 10^\circ \cdot \cos 40^\circ

经典题目:背景为余弦定理.

sin⁡210∘+cos⁡240∘+sin⁡10∘⋅cos⁡40∘=sin⁡210∘+sin⁡250∘−2cos⁡120∘sin⁡10∘⋅sin⁡50∘=sin⁡2120∘=34\begin{gathered} \sin^2 10^\circ + \cos^2 40^\circ + \sin 10^\circ \cdot \cos 40^\circ\\ =\sin^2 10^\circ + \sin^2 50^\circ -2 \cos120\degree\sin 10^\circ \cdot \sin 50^\circ\\ =\sin^2120\degree=\frac{3}{4} \end{gathered}

思路打开:沿用对偶式 A=sin⁡210∘+cos⁡240∘+sin⁡10∘⋅cos⁡40∘B=cos⁡210∘+sin⁡240∘+cos⁡10∘⋅sin⁡40∘A+B=2+sin⁡50∘A−B=cos⁡80∘−sin⁡20∘−sin⁡30∘2A=32+sin⁡50∘+sin⁡10∘−sin⁡20∘=32+2sin⁡30∘sin⁡20∘−sin⁡20∘=32⟹A=34\begin{gathered} A=\sin^2 10^\circ + \cos^2 40^\circ + \sin 10^\circ \cdot \cos 40^\circ\\ B=\cos^2 10^\circ + \sin^2 40^\circ + \cos 10^\circ \cdot \sin 40^\circ\\ A+B=2+\sin50\degree\\ A-B=\cos80\degree-\sin20\degree-\sin30\degree\\ 2A=\frac{3}{2}+\sin50\degree+\sin10\degree-\sin20\degree\\ =\frac{3}{2}+2\sin30\degree\sin20\degree-\sin20\degree=\frac{3}{2}\\ \Longrightarrow A=\frac{3}{4} \end{gathered}