Chapter Review 6 - 练习题

基础练习题

Question 1 简单

将下列各式表示为锐角的三角函数：

\(\cos 237°\)
\(\sin 312°\)
\(\tan 190°\)
\(\cos\left(\frac{11\pi}{6}\right)\)
\(\sin\left(\frac{7\pi}{6}\right)\)
\(\tan\left(\frac{11\pi}{4}\right)\)

💡 提示

使用角度关系公式将钝角、优角和大于2π的角转换为锐角。

答案

1. \(\cos 237° = -\cos 57°\)

2. \(\sin 312° = -\sin 48°\)

3. \(\tan 190° = \tan 10°\)

4. \(\cos\left(\frac{11\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)\)

5. \(\sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right)\)

6. \(\tan\left(\frac{11\pi}{4}\right) = \tan\left(\frac{3\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right)\)

Question 2 简单

不使用计算器，计算下列各式的值：

\(\cos 270°\)
\(\sin 225°\)
\(\tan 240°\)
\(\cos \pi\)
\(\tan\left(\frac{5\pi}{4}\right)\)
\(\sin\left(\frac{3\pi}{2}\right)\)

答案

1. \(\cos 270° = 0\)

2. \(\sin 225° = -\sin 45° = -\frac{\sqrt{2}}{2}\)

3. \(\tan 240° = \tan 60° = \sqrt{3}\)

4. \(\cos \pi = -1\)

5. \(\tan\left(\frac{5\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1\)

6. \(\sin\left(\frac{3\pi}{2}\right) = -1\)

中等难度练习题

Question 3 中等

已知角 \(A\) 是钝角且 \(\cos A = -\sqrt{\frac{7}{11}}\)，证明 \(\tan A = \frac{-2\sqrt{7}}{7}\)。

解题步骤：

使用毕达哥拉斯恒等式：\(\sin^2 A + \cos^2 A = 1\)
代入 \(\cos A = -\sqrt{\frac{7}{11}}\)：\(\sin^2 A + \frac{7}{11} = 1\)
解得：\(\sin^2 A = 1 - \frac{7}{11} = \frac{4}{11}\)
所以：\(\sin A = \sqrt{\frac{4}{11}} = \frac{2}{\sqrt{11}}\)（因为A是钝角，sin A > 0）
计算：\(\tan A = \frac{\sin A}{\cos A} = \frac{\frac{2}{\sqrt{11}}}{-\sqrt{\frac{7}{11}}} = \frac{-2\sqrt{7}}{7}\)

Question 4 中等

已知角 \(B\) 是优角且 \(\tan B = +\frac{\sqrt{21}}{2}\)，求 \(\sin B\) 和 \(\cos B\) 的精确值。

解题步骤：

设 \(\tan B = \frac{\sqrt{21}}{2} = \frac{\sin B}{\cos B}\)
设 \(\sin B = k\sqrt{21}\)，\(\cos B = 2k\)（k为比例常数）
使用恒等式：\((k\sqrt{21})^2 + (2k)^2 = 1\)
解得：\(21k^2 + 4k^2 = 1\)，即 \(25k^2 = 1\)，所以 \(k = \frac{1}{5}\)
因此：\(\sin B = \frac{\sqrt{21}}{5}\)，\(\cos B = \frac{2}{5}\)

Question 5 中等

化简下列表达式：

\(\cos^4\theta - \sin^4\theta\)
\(\sin^2 3\theta - \sin^2 3\theta \cos^2 3\theta\)
\(\cos^4\theta + 2\sin^2\theta \cos^2\theta + \sin^4\theta\)

解题步骤：

a) \(\cos^4\theta - \sin^4\theta = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta) = 1 \cdot \cos 2\theta = \cos 2\theta\)
b) \(\sin^2 3\theta - \sin^2 3\theta \cos^2 3\theta = \sin^2 3\theta(1 - \cos^2 3\theta) = \sin^2 3\theta \cdot \sin^2 3\theta = \sin^4 3\theta\)
c) \(\cos^4\theta + 2\sin^2\theta \cos^2\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 = 1^2 = 1\)

高难度练习题

Question 6 困难

a) 已知 \(2(\sin x + 2\cos x) = \sin x + 5\cos x\)，求 \(\tan x\) 的精确值。

b) 已知 \(\sin x\cos y + 3\cos x\sin y = 2\sin x\sin y - 4\cos x\cos y\)，用 \(\tan x\) 表示 \(\tan y\)。

解题步骤：

a) 展开左边：\(2\sin x + 4\cos x = \sin x + 5\cos x\)
整理得：\(\sin x = \cos x\)
所以：\(\tan x = \frac{\sin x}{\cos x} = 1\)
b) 将方程重新排列：\(\sin x\cos y + 3\cos x\sin y + 4\cos x\cos y = 2\sin x\sin y\)
左边：\(\sin x\cos y + \cos x(3\sin y + 4\cos y)\)
右边：\(2\sin x\sin y\)
除以 \(\cos x\cos y\)：\(\tan x + 3\tan y + 4 = 2\tan x\tan y\)
整理得：\(\tan y = \frac{\tan x + 4}{2\tan x - 3}\)

Question 7 困难

证明对于所有 \(\theta\) 值：

\((1 + \sin\theta)^2 + \cos^2\theta \equiv 2(1 + \sin\theta)\)
\(\cos^4\theta + \sin^2\theta \equiv \sin^4\theta + \cos^2\theta\)

证明过程：

a) 左边：\((1 + \sin\theta)^2 + \cos^2\theta\)

\(= 1 + 2\sin\theta + \sin^2\theta + \cos^2\theta\)

\(= 1 + 2\sin\theta + (\sin^2\theta + \cos^2\theta)\)

\(= 1 + 2\sin\theta + 1 = 2 + 2\sin\theta = 2(1 + \sin\theta)\) = 右边

b) 左边：\(\cos^4\theta + \sin^2\theta\)

\(= \cos^2\theta(\cos^2\theta) + \sin^2\theta\)

\(= \cos^2\theta(1 - \sin^2\theta) + \sin^2\theta\)

\(= \cos^2\theta - \cos^2\theta\sin^2\theta + \sin^2\theta\)

\(= \cos^2\theta + \sin^2\theta(1 - \cos^2\theta)\)

\(= \cos^2\theta + \sin^2\theta \cdot \sin^2\theta = \cos^2\theta + \sin^4\theta\) = 右边

综合应用题

Question 8 困难

不尝试求解，说明下列方程在区间 \(0 \leq \theta \leq 360°\) 内有多少个解，并给出简要理由：

\(2\sin\theta = 3\)
\(\sin\theta = -\cos\theta\)
\(2\sin\theta + 3\cos\theta + 6 = 0\)
\(\tan\theta + \frac{1}{\tan\theta} = 0\)

分析过程：

a) \(\sin\theta = \frac{3}{2} > 1\)，无解（因为 \(\sin\theta\) 的最大值为1）
b) \(\sin\theta = -\cos\theta\)，即 \(\tan\theta = -1\)，在给定区间内有2个解
c) \(2\sin\theta + 3\cos\theta = -6\)，左边最大值约为3.6，小于6，无解
d) \(\tan\theta + \frac{1}{\tan\theta} = 0\)，即 \(\tan^2\theta + 1 = 0\)，无实数解

Question 9 困难

a) 因式分解 \(4xy - y^2 + 4x - y\)。

b) 求解方程 \(4\sin\theta\cos\theta - \cos^2\theta + 4\sin\theta - \cos\theta = 0\)，在区间 \(0 \leq \theta \leq 2\pi\) 内。

解题步骤：

a) \(4xy - y^2 + 4x - y = y(4x - y) + (4x - y) = (4x - y)(y + 1)\)
b) 设 \(x = \sin\theta\)，\(y = \cos\theta\)，方程变为：\(4xy - y^2 + 4x - y = 0\)
即 \((4x - y)(y + 1) = 0\)
情况1：\(4\sin\theta - \cos\theta = 0\)，即 \(\tan\theta = \frac{1}{4}\)
情况2：\(\cos\theta + 1 = 0\)，即 \(\cos\theta = -1\)
求解得：\(\theta = \arctan\frac{1}{4}, \pi\)