竞赛类题目，不要直接套公式，而是要考虑角度直接的关系，构造最合适的角度加减法

Q1

Given the function \(\boldsymbol{f(x)=2\sin\left(\frac{1}{3}x-\frac{\pi}{6}\right)}\),
(1) Find the value of \(\boldsymbol{f\left(\frac{5\pi}{4}\right)}\).
(2) Let \(\boldsymbol{\alpha,\beta\in\left[0,\frac{\pi}{2}\right]}\), \(\boldsymbol{f\left(3\alpha+\frac{\pi}{2}\right)=\frac{10}{13}}\), \(\boldsymbol{f(3\beta+2\pi)=\frac{6}{5}}\). Find the values of \(\boldsymbol{\cos(\alpha+\beta)}\) and \(\boldsymbol{\sin(\alpha+\beta)}\).

已知函数 \(\boldsymbol{f(x)=2\sin\left(\dfrac{1}{3}x-\dfrac{\pi}{6}\right)}\)，
(1) 求 \(\boldsymbol{f\left(\dfrac{5\pi}{4}\right)}\) 的值；
(2) 设 \(\boldsymbol{\alpha,\beta\in\left[0,\dfrac{\pi}{2}\right]}\)，\(\boldsymbol{f\left(3\alpha+\dfrac{\pi}{2}\right)=\dfrac{10}{13}}\)，\(\boldsymbol{f(3\beta+2\pi)=\dfrac{6}{5}}\)，求 \(\boldsymbol{\cos(\alpha+\beta)}\) 及 \(\boldsymbol{\sin(\alpha+\beta)}\)

Q1

Solution (English)

(1) Evaluate \(f\left(\frac{5\pi}{4}\right)\)

Substitute \(x=\frac{5\pi}{4}\) into the function: \(\begin{align*} f\left(\frac{5\pi}{4}\right)&=2\sin\left(\frac{1}{3}\cdot\frac{5\pi}{4}-\frac{\pi}{6}\right)\\ &=2\sin\left(\frac{5\pi}{12}-\frac{2\pi}{12}\right)\\ &=2\sin\frac{\pi}{4}\\ &=2\cdot\frac{\sqrt{2}}{2}=\boldsymbol{\sqrt{2}} \end{align*}\)

(2) Calculate \(\cos(\alpha+\beta)\) and \(\sin(\alpha+\beta)\)

1. Simplify \(f\left(3\alpha+\frac{\pi}{2}\right)\)

\(\begin{align*} f\left(3\alpha+\frac{\pi}{2}\right)&=2\sin\left[\frac{1}{3}\left(3\alpha+\frac{\pi}{2}\right)-\frac{\pi}{6}\right]\\ &=2\sin\left(\alpha+\frac{\pi}{6}-\frac{\pi}{6}\right)\\ &=2\sin\alpha=\frac{10}{13} \end{align*}\)

Thus \(\boldsymbol{\sin\alpha=\frac{5}{13}}\).

Since \(\alpha\in\left[0,\frac{\pi}{2}\right]\), we have \(\boldsymbol{\cos\alpha=\sqrt{1-\sin^2\alpha}=\frac{12}{13}}\).

2. Simplify \(f(3\beta+2\pi)\)

\(\begin{align*} f(3\beta+2\pi)&=2\sin\left[\frac{1}{3}(3\beta+2\pi)-\frac{\pi}{6}\right]\\ &=2\sin\left(\beta+\frac{2\pi}{3}-\frac{\pi}{6}\right)\\ &=2\sin\left(\beta+\frac{\pi}{2}\right)=2\cos\beta=\frac{6}{5} \end{align*}\)

Thus \(\boldsymbol{\cos\beta=\frac{3}{5}}\).

Since \(\beta\in\left[0,\frac{\pi}{2}\right]\), we have \(\boldsymbol{\sin\beta=\sqrt{1-\cos^2\beta}=\frac{4}{5}}\).

3. Apply angle addition formulas

\(\begin{align*} \boldsymbol{\cos(\alpha+\beta)}&=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{12}{13}\cdot\frac{3}{5}-\frac{5}{13}\cdot\frac{4}{5}=\boldsymbol{\frac{16}{65}}\\ \boldsymbol{\sin(\alpha+\beta)}&=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{5}{13}\cdot\frac{3}{5}+\frac{12}{13}\cdot\frac{4}{5}=\boldsymbol{\frac{63}{65}} \end{align*}\)

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Let \(\boldsymbol{\alpha}\) be an acute angle, and \(\boldsymbol{\cos\left(\alpha+\frac{\pi}{6}\right)=\frac{5}{13}}\), then \(\boldsymbol{\cos\alpha=}\).

设\(\boldsymbol{\alpha}\)为锐角，且\(\boldsymbol{\cos\left(\alpha+\frac{\pi}{6}\right)=\frac{5}{13}}\)，则\(\boldsymbol{\cos\alpha=}\)；

If \(\boldsymbol{\cos(\alpha+\beta)=\frac{1}{5}}\) and \(\boldsymbol{\cos(\alpha-\beta)=\frac{3}{5}}\), then \(\boldsymbol{\tan\alpha\cdot\tan\beta=}\).

若\(\boldsymbol{\cos(\alpha+\beta)=\frac{1}{5},\cos(\alpha-\beta)=\frac{3}{5}}\)，则\(\boldsymbol{\tan\alpha\cdot\tan\beta=}\)；

Given \(\boldsymbol{\sin\alpha+\sin\beta=\frac{4}{5}}\) and \(\boldsymbol{\cos\alpha+\cos\beta=\frac{3}{5}}\), then \(\boldsymbol{\cos(\alpha-\beta)=}\).

已知\(\boldsymbol{\sin\alpha+\sin\beta=\frac{4}{5},\cos\alpha+\cos\beta=\frac{3}{5}}\)，则\(\boldsymbol{\cos(\alpha-\beta)=}\)；

代入展开： \(\begin{align*} \cos\alpha&=\cos\left(\alpha+\frac{\pi}{6}\right)\cos\frac{\pi}{6}+\sin\left(\alpha+\frac{\pi}{6}\right)\sin\frac{\pi}{6}\\ &=\frac{5}{13}\cdot\frac{\sqrt{3}}{2}+\frac{12}{13}\cdot\frac{1}{2}\\ &=\boldsymbol{\frac{12+5\sqrt{3}}{26}} \end{align*}\)

展开余弦和、差公式： \(\begin{cases} \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\dfrac{1}{5}\quad①\\ \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\dfrac{3}{5}\quad② \end{cases}\)

\(①+②\)：\(\boldsymbol{2\cos\alpha\cos\beta=\frac{4}{5}\implies\cos\alpha\cos\beta=\frac{2}{5}}\)

\(②-①\)：\(\boldsymbol{2\sin\alpha\sin\beta=\frac{2}{5}\implies\sin\alpha\sin\beta=\frac{1}{5}}\)

正切定义\(\boldsymbol{\tan x=\frac{\sin x}{\cos x}}\)，因此：

\(\tan\alpha\cdot\tan\beta=\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}=\frac{\frac{1}{5}}{\frac{2}{5}}=\boldsymbol{\frac{1}{2}}\)

对已知等式两边平方后相加： \(\begin{align*} (\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2&=\left(\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2\\ (\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta)+(\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta)&=1 \end{align*}\)

由\(\boldsymbol{\sin^2x+\cos^2x=1}\)化简： \(\begin{align*} (\sin^2\alpha+\cos^2\alpha)+(\sin^2\beta+\cos^2\beta)+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)&=1\\ 1+1+2\cos(\alpha-\beta)&=1 \end{align*}\)

解得：

\(\boldsymbol{\cos(\alpha-\beta)=-\frac{1}{2}}\)

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Given \(\boldsymbol{\sin2(\alpha+\gamma)=n\sin2\beta}\), find the value of \(\boldsymbol{\dfrac{\tan(\alpha+\beta+\gamma)}{\tan(\alpha-\beta+\gamma)}}\).

中文：已知\(\boldsymbol{\sin2(\alpha+\gamma)=n\sin2\beta}\)，求\(\boldsymbol{\dfrac{\tan(\alpha+\beta+\gamma)}{\tan(\alpha-\beta+\gamma)}}\)的值。

解题过程

x=alpha+beta+gama y=alpha-beta+gama

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最终答案

\(\boldsymbol{\dfrac{n+1}{n-1}}\)

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Find the value of \(\boldsymbol{\left(\tan10^\circ-\sqrt{3}\right)\cdot\dfrac{\cos10^\circ}{\sin50^\circ}}\).

中文：求\(\boldsymbol{\left(\tan10^\circ-\sqrt{3}\right)\cdot\dfrac{\cos10^\circ}{\sin50^\circ}}\)的值。

解题过程

步骤 1：将正切化为正弦、余弦

\(\begin{align*} \left(\tan10^\circ-\sqrt{3}\right)\cdot\frac{\cos10^\circ}{\sin50^\circ} &=\left(\frac{\sin10^\circ}{\cos10^\circ}-\sqrt{3}\right)\cdot\frac{\cos10^\circ}{\sin50^\circ}\\ &=\frac{\sin10^\circ-\sqrt{3}\cos10^\circ}{\cos10^\circ}\cdot\frac{\cos10^\circ}{\sin50^\circ}\\ &=\frac{\boldsymbol{\sin10^\circ-\sqrt{3}\cos10^\circ}}{\sin50^\circ} \end{align*}\)

步骤 2：利用辅助角公式化简分子

提取系数2，结合两角差的正弦公式： \(\begin{align*} \sin10^\circ-\sqrt{3}\cos10^\circ &=2\left(\frac12\sin10^\circ-\frac{\sqrt3}{2}\cos10^\circ\right)\\ &=2\sin\left(10^\circ-60^\circ\right)\\ &=2\sin(-50^\circ)\\ &=\boldsymbol{-2\sin50^\circ} \end{align*}\)

步骤 3：代入约分计算

\(\frac{-2\sin50^\circ}{\sin50^\circ}=\boldsymbol{-2}\)

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Find the value of \(\boldsymbol{(1+\tan1^\circ)(1+\tan2^\circ)\dots(1+\tan44^\circ)(1+\tan45^\circ)}\).

中文：求\(\boldsymbol{(1+\tan1^\circ)(1+\tan2^\circ)\dots(1+\tan44^\circ)(1+\tan45^\circ)}\)的值。

解题过程

步骤 1：推导核心恒等式

若\(\boldsymbol{A+B=45^\circ}\)，则\(\tan(A+B)=\tan45^\circ=1\)。

由正切和角公式： \(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=1\)

整理得：\(\boldsymbol{\tan A+\tan B=1-\tan A\tan B}\)。

因此： \(\begin{align*} (1+\tan A)(1+\tan B)&=1+\tan A+\tan B+\tan A\tan B\\ &=1+(1-\tan A\tan B)+\tan A\tan B\\ &=\boldsymbol{2} \end{align*}\)

步骤 2：角度配对

观察角度关系：

\(1^\circ+44^\circ=45^\circ,\ 2^\circ+43^\circ=45^\circ,\ \dots,\ 22^\circ+23^\circ=45^\circ\)

共22 组，每组满足\((1+\tan k^\circ)(1+\tan(45^\circ-k)^\circ)=2\)。

剩余项：\(\boldsymbol{1+\tan45^\circ=1+1=2}\)。

步骤 3：计算连乘积

\(\begin{align*} \text{原式}&=\underbrace{2\times2\times\dots\times2}_{22\text{个}}\times2\\ &=2^{22}\times2\\ &=\boldsymbol{2^{23}} \end{align*}\)