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$\therefore \sin^2 \alpha + 4 \sin \alpha \cos \alpha + 4 \cos^2 \alpha = \frac{5}{2}.$

$\therefore \frac{\tan^2 \alpha + 4 \tan \alpha + 4}{\tan^2 \alpha + 1} = \frac{5}{2}, \text{ \解得 } \tan \alpha = -\frac{1}{3} \text{ \或 } \tan \alpha = 3.$

$\text{当 } \tan \alpha = -\frac{1}{3} \text{ \时, } \tan 2\alpha = \frac{2 \times (-\frac{1}{3})}{1 - (-\frac{1}{3})^2} = -\frac{3}{4}.$

$\text{当 } \tan \alpha = 3 \text{ \时, } \tan 2\alpha = \frac{6}{1-9} = -\frac{3}{4}, \therefore \tan 2\alpha = -\frac{3}{4}, \text{ \故选 B.}$

$8. \text{选 D } \cos\left(\frac{\pi}{3} + 2\alpha\right) = -\cos\left(\frac{2\pi}{3} - 2\alpha\right) = -1 + 2\sin^2\left(\frac{\pi}{3} - \alpha\right) = -1 + \frac{2}{9} = -\frac{7}{9}, \text{ \故选 D.}$

$9. \text{选 C } 4\cos 50^\circ - \tan 40^\circ$

$= 4\cos 50^\circ - \frac{\sin 40^\circ}{\cos 40^\circ}$

$= \frac{4\sin 40^\circ \cos 40^\circ - \sin 40^\circ}{\cos 40^\circ}$

$= \frac{2\sin 80^\circ - \sin 40^\circ}{\cos 40^\circ}$

$= \frac{2\sin(50^\circ + 30^\circ) - \cos 50^\circ}{\cos 40^\circ}$

$= \frac{\sqrt{3}\sin 50^\circ}{\cos 40^\circ}$

$= \sqrt{3}, \text{ \故选 C.}$

$10. \text{选 A \原式} = \frac{\sin \frac{\pi}{9} \cos \frac{\pi}{9} - \cos \frac{2\pi}{9} \cos \frac{5\pi}{9}}{\sin \frac{\pi}{9}}$

$= \frac{\frac{1}{2} \sin \frac{2\pi}{9} \cos \frac{2\pi}{9} - \cos \frac{2\pi}{9} \cos \frac{5\pi}{9}}{\sin \frac{\pi}{9}}$

$= \frac{-\frac{1}{4} \sin \frac{4\pi}{9} \cos \frac{4\pi}{9}}{\sin \frac{\pi}{9}}$

$= \frac{-\frac{1}{8} \sin \frac{8\pi}{9}}{\sin \frac{\pi}{9}}$

$= -\frac{1}{8}.$

$11. \text{解析: \原式} = \frac{2\sin \alpha + 2\sin \alpha \cos \alpha}{1 + \cos \alpha} = \frac{2\sin \alpha(1 + \cos \alpha)}{1 + \cos \alpha} = 2\sin \alpha.$

\答案: $2\sin \alpha$

$12. \text{解析: } \because f(x) = 2\tan x + \frac{1-2\sin^2 \frac{x}{2}}{\frac{1}{2} \sin x} = 2\tan x + \frac{2\cos x}{\sin x} = \frac{2\sin x}{\cos x} +$

$\frac{2\cos x}{\sin x} = \frac{2}{\sin x \cos x} = \frac{4}{\sin 2x}.$

$\therefore f\left(\frac{\pi}{12}\right) = \frac{4}{\sin \frac{\pi}{6}} = 8.$

\答案: 8

\高考真题

$13. \text{选 B } \cos 2\alpha = 1 - 2\sin^2 \alpha = 1 - \frac{2}{9} = \frac{7}{9}.$

$14. \text{选 B \由二倍角公式可知, } 4\sin \alpha \cos \alpha = 2\cos^2 \alpha.$

$\therefore \alpha \in \left(0, \frac{\pi}{2}\right), \therefore \cos \alpha \neq 0.$

$\therefore 2\sin \alpha = \cos \alpha, \text{ \又 } \sin^2 \alpha + \cos^2 \alpha = 1.$

$\therefore \sin \alpha = \frac{\sqrt{5}}{5}, \text{ \故选 B.}$

$15. \text{选 D } \sin 2\alpha = \cos\left(\frac{\pi}{2} - 2\alpha\right) = 2\cos^2\left(\frac{\pi}{4} - \alpha\right) - 1 = -\frac{7}{25}, \text{ \故}$

\选 D.

$16. \text{解析: \由已知得, } \frac{\tan \alpha + 1}{1 - \tan \alpha} = -\frac{2}{3}.$

$\text{解得 } \tan \alpha = 2 \text{ \或 } \tan \alpha = -\frac{1}{3}.$

$\text{又因为 } \sin\left(2\alpha + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}(\sin 2\alpha + \cos 2\alpha)$

$= \frac{\sqrt{2}}{2}(2\sin \alpha \cos \alpha + 2\cos^2 \alpha - 1)$

$= \sqrt{2}\sin \alpha \cos \alpha + \sqrt{2}\cos^2 \alpha - \frac{\sqrt{2}}{2}$

$= \frac{\sqrt{2}\sin \alpha \cos \alpha + \sqrt{2}\cos^2 \alpha - \frac{\sqrt{2}}{2}}{\sin^2 \alpha + \cos^2 \alpha}$

$= \frac{\sqrt{2}\tan \alpha + \sqrt{2} - \frac{\sqrt{2}}{2}}{\tan^2 \alpha + 1}$

$\text{所以当 } \tan \alpha = 2 \text{ \时, \原式} = \frac{\sqrt{2}}{10}.$

$\text{当 } \tan \alpha = -\frac{1}{3}, \text{ \原式} = \frac{\sqrt{2}}{10}.$

\答案: $\frac{\sqrt{2}}{10}$

\实战演练

$17. \text{选 A } \frac{\sqrt{2}}{2} \cos 15^\circ - \frac{\sqrt{2}}{2} \sin 195^\circ$

$= \sin 45^\circ \cos 15^\circ + \cos 45^\circ \sin 15^\circ$

$= \sin(45^\circ + 15^\circ)$

$= \sin 60^\circ$

$= \frac{\sqrt{3}}{2}, \text{ \故选 A.}$

$18. \text{选 C } \cos\left(\frac{5\pi}{3} + 2\alpha\right) = \cos\left(\frac{\pi}{3} - 2\alpha\right) = 2\cos^2\left(\frac{\pi}{6} - \alpha\right) - 1 = 2 \times$

$\frac{4}{9} - 1 = -\frac{1}{9}, \text{ \故选 C.}$

$19. \text{选 B } \because \frac{12}{\sin \alpha} + \frac{12}{\cos \alpha} = 35, \alpha \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right).$

$\therefore 12(\sin \alpha + \cos \alpha) = 35 \sin \alpha \cos \alpha.$

$\text{令 } \sin \alpha + \cos \alpha = t, t \in (1, \sqrt{2}), \text{ \则 } \sin \alpha \cos \alpha = \frac{t^2 - 1}{2}.$

$\therefore 12t = 35 \cdot \frac{t^2 - 1}{2},$

$\therefore 35t^2 - 24t - 35 = 0,$

$\therefore t = \frac{7}{5} \text{ \或 } t = -\frac{5}{7} \text{ \(舍).}$

$\therefore \sin \alpha + \cos \alpha = \frac{7}{5}, \text{ \两边平方得 } 2\sin \alpha \cos \alpha = \frac{24}{25}.$

$\therefore \sin 2\alpha = \frac{24}{25}, \text{ \又 } \because 2\alpha \in \left(\frac{\pi}{2}, \pi\right), \therefore \cos 2\alpha = -\frac{7}{25}.$

$\therefore \tan 2\alpha = -\frac{24}{7}, \text{ \故选 B.}$

$20. \text{选 D } \because \frac{\sin \alpha - 2\cos \alpha}{\sin \alpha + \cos \alpha} = 2, \therefore \frac{\tan \alpha - 2}{\tan \alpha + 1} = 2,$

$\therefore \tan \alpha = -4.$

$\therefore \sin \alpha = -4\cos \alpha.$

$\therefore \cos^2 \alpha = \frac{1}{17}.$

$\therefore \sin 2\alpha = 2\sin \alpha \cos \alpha = -8\cos^2 \alpha = -\frac{8}{17}, \text{ \故选 D.}$

$21. \text{选 D } \because \frac{1 + \cos 2\alpha}{\sin 2\alpha} = \frac{2\cos^2 \alpha}{2\sin \alpha \cos \alpha} = \frac{\cos \alpha}{\sin \alpha} = \frac{1}{2},$

$\therefore \tan \alpha = 2.$

$\therefore \tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha} = -\frac{4}{3}, \text{ \故选 D.}$

$22. \text{选 D } \frac{1 + \cos \alpha}{\sin \alpha} = \frac{\sin \alpha}{1 - \cos \alpha} = 5, \text{ \故选 D.}$

$23. \text{选 A } \frac{\tan 20^\circ + \tan 40^\circ + \tan 120^\circ}{\tan 20^\circ \tan 40^\circ}$

$= \frac{\tan 60^\circ (1 - \tan 20^\circ \tan 40^\circ) + \tan 120^\circ}{\tan 20^\circ \tan 40^\circ}$

$= -\sqrt{3}, \text{ \故选 A.}$

$24. \text{解析: \解法一: } \sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) \cos\left(\frac{\alpha}{2} + \frac{\pi}{4}\right) = -\frac{3}{4}.$

$\therefore \left(\frac{\sqrt{2}}{2} \sin \frac{\alpha}{2} - \frac{\sqrt{2}}{2} \cos \frac{\alpha}{2}\right) \left(\frac{\sqrt{2}}{2} \cos \frac{\alpha}{2} - \frac{\sqrt{2}}{2} \sin \frac{\alpha}{2}\right) =$

$-\frac{3}{4}, \text{ \化简得, } \left(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}\right)^2 = \frac{3}{2}.$

$\therefore 1 - 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = \frac{3}{2}, \therefore 1 - \sin \alpha = \frac{3}{2}.$

$\therefore \sin \alpha = -\frac{1}{2}.$

$\text{解法二: } \because \sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) \cos\left(\frac{\alpha}{2} + \frac{\pi}{4}\right) = -\frac{3}{4}.$

$\therefore \sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) \cos\left(\frac{\pi}{2} + \frac{\alpha}{2} - \frac{\pi}{4}\right) = -\frac{3}{4}.$

$\therefore -\sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) \sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) = -\frac{3}{4}.$

$\therefore \sin^2\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) = \frac{3}{4}.$

$\therefore \frac{1 - \cos\left(\alpha - \frac{\pi}{2}\right)}{2} = \frac{3}{4}.$

$\therefore \frac{1 - \sin \alpha}{2} = \frac{3}{4}, \therefore \sin \alpha = -\frac{1}{2}.$

\答案: $-\frac{1}{2}$

\大题冲关——提升能力

\高考大题

$25. \text{解: (1) \由正弦定理及已知 } 3c \sin B = 4a \sin C, \text{ \得 } 3bc = 4ac.$

$\therefore 3b = 4a,$

$\text{又 } b+c=2a, \therefore b=\frac{4}{3}a, c=\frac{2}{3}a.$

$\therefore \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{a^2 + \frac{4}{9}a^2 - \frac{16}{9}a^2}{2 \cdot a \cdot \frac{2}{3}a} = -\frac{1}{4}.$

$(2) \text{由(1)知, } \cos B = -\frac{1}{4}, B \in (0, \pi), \therefore \sin B = \frac{\sqrt{15}}{4}.$

$\therefore \sin 2B = -\frac{\sqrt{15}}{8}, \cos 2B = -\frac{7}{8}.$

$\therefore \sin\left(2B + \frac{\pi}{6}\right) = \sin 2B \cos \frac{\pi}{6} + \cos 2B \sin \frac{\pi}{6} = -\frac{3\sqrt{5} + 7}{16}.$

\大题精练

$26. \text{解: (1) } \because f(x) = \cos 2x + \frac{\sqrt{3}}{2} \sin 2x - \frac{1}{2} \cos 2x$

$= \frac{\sqrt{3}}{2} \sin 2x + \frac{1}{2} \cos 2x = \sin\left(2x + \frac{\pi}{6}\right).$

$\therefore \text{函数 } f(x) \text{ \的最小正周期 } T = \pi.$

$(2) \text{由 } f(\alpha) = \frac{1}{3}, \text{ \可得 } \sin\left(2\alpha + \frac{\pi}{6}\right) = \frac{1}{3}.$

$\because \alpha \in \left(0, \frac{\pi}{2}\right), \therefore 2\alpha + \frac{\pi}{6} \in \left(\frac{\pi}{6}, \frac{7\pi}{6}\right).$

$\text{又 } 0 < \sin\left(2\alpha + \frac{\pi}{6}\right) = \frac{1}{3} < \frac{1}{2}, \therefore 2\alpha + \frac{\pi}{6} \in \left(\frac{\pi}{2}, \pi\right).$

$\therefore \cos\left(2\alpha + \frac{\pi}{6}\right) = -\frac{2\sqrt{2}}{3}.$

$\therefore \cos 2\alpha = \cos\left[\left(2\alpha + \frac{\pi}{6}\right) - \frac{\pi}{6}\right]$

$= \cos\left(2\alpha + \frac{\pi}{6}\right) \cos \frac{\pi}{6} + \sin\left(2\alpha + \frac{\pi}{6}\right) \sin \frac{\pi}{6} = \frac{1-2\sqrt{6}}{6}.$

$27. \text{解: (1) \因为 } \alpha \in \left(\frac{\pi}{2}, \pi\right), \cos \alpha = -\frac{3}{5}, \text{ \所以 } \sin \alpha = \sqrt{1-\cos^2 \alpha} =$

$\sqrt{1-\left(-\frac{3}{5}\right)^2} = \frac{4}{5}. \text{ \所以 } \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{4}{3}.$

$(2) \text{由(1)知, } \sin \alpha = \frac{4}{5}, \cos \alpha = -\frac{3}{5},$

$\text{所以 } \sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \times \frac{4}{5} \times \left(-\frac{3}{5}\right) = -\frac{24}{25}.$

$\cos 2\alpha = 2\cos^2 \alpha - 1 = 2 \times \left(-\frac{3}{5}\right)^2 - 1 = -\frac{7}{25}.$

$\text{所以 } \frac{\cos 2\alpha}{\sin 2\alpha + 1} = \frac{-\frac{7}{25}}{-\frac{24}{25} + 1} = -7.$

$28. \text{解: \因为 } \tan \alpha = 2 > 0, \alpha \in (0, \pi), \text{ \所以 } \alpha \in \left(0, \frac{\pi}{2}\right).$

$\text{因为 } \cos \beta = -\frac{7\sqrt{2}}{10}, \beta \in (0, \pi).$

$\text{所以 } \beta \in \left(\frac{\pi}{2}, \pi\right), \text{ \且 } \tan \beta = -\frac{1}{7}.$

$\text{所以 } \alpha - \beta \in (-\pi, 0), \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = 3 > 0.$

$\text{所以 } \alpha - \beta \in \left(-\pi, -\frac{\pi}{2}\right), \text{ \所以 } 2\alpha - \beta \in (-\pi, 0).$

$\text{因为 } \tan(2\alpha - \beta) = \tan[\alpha + (\alpha - \beta)] = \frac{\tan \alpha + \tan(\alpha - \beta)}{1 - \tan \alpha \tan(\alpha - \beta)} = -1.$

$\text{所以 } 2\alpha - \beta = -\frac{\pi}{4}.$

\考点测试19 \三角函数的性质

\小题集训——夯实基础

\基础巩固

$1. \text{选 A } \because y = \tan x \text{ \在 } \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text{ \上单调递增, } -\frac{\pi}{5} > -\frac{3\pi}{7},$

$\therefore \tan\left(-\frac{\pi}{5}\right) > \tan\left(-\frac{3\pi}{7}\right), \text{ \故选 A.}$

$2. \text{选 B \当 } f(x) \text{ \是奇函数时, } \varphi = \frac{\pi}{2} + k\pi (k \in \mathbb{Z}), \text{ \充分性不成立; \当 } \varphi = \frac{\pi}{2} \text{ \时, } f(x) = A\cos\left(\omega x + \frac{\pi}{2}\right) = -A\sin \omega x, \text{ \为奇函数, \必要性成立, \故选 B.}$

$3. \text{选 B \由 } -\frac{\pi}{2} + k\pi < 2x - \frac{\pi}{3} < \frac{\pi}{2} + k\pi (k \in \mathbb{Z}), \text{ \得 } -\frac{\pi}{12} + \frac{k\pi}{2} < x < \frac{5}{12}\pi + \frac{k\pi}{2} (k \in \mathbb{Z}). \therefore f(x) \text{ \的单调递增区间为}$

$\left(-\frac{\pi}{12} + \frac{k\pi}{2}, \frac{5}{12}\pi + \frac{k\pi}{2}\right) (k \in \mathbb{Z}).$

$4. \text{选 C } \because y = 2\sin^2 x + \sin 2x = 1 - \cos 2x + \sin 2x = \sqrt{2}\sin\left(2x - \frac{\pi}{4}\right) +$

$1, \therefore T = \frac{2\pi}{\omega} = \pi, \text{ \函数 } y = 2\sin^2 x + \sin 2x \text{ \的最小正周期是 } \pi, \text{ \故选 C.}$

$5. \text{选 B } \because f(x) = 8\sin\left(\omega x - \frac{\pi}{3}\right) (\omega > 0) \text{ \的最小正周期为 } \pi,$

$\therefore \omega = 2,$

$\therefore f(x) = 8\sin \left(2x - \frac{\pi}{3}\right).$

$\therefore -\frac{\pi}{2} \le x \le \frac{2\pi}{3},$

$\therefore -\frac{5\pi}{12} \le 2x - \frac{\pi}{3} \le \pi.$

$\therefore \text{当 } -\frac{5\pi}{12} \le 2x - \frac{\pi}{3} \le \frac{\pi}{2} \text{ \时, } f(x) \text{ \单调递增,}$

$\text{当 } \frac{\pi}{2} \le 2x - \frac{\pi}{3} \le \pi \text{ \时, } f(x) \text{ \单调递减,}$

$\therefore \begin{cases} \frac{2m}{3} - \frac{\pi}{3} \le \frac{\pi}{2}, \\ m - \frac{\pi}{3} \ge \frac{\pi}{2}, \end{cases} \therefore \begin{cases} \frac{5\pi}{12} \le m \le \frac{2\pi}{3}, \\ \text{故选 B.} \end{cases}$

$6. \text{选 D } \because f(x) = \sin(x + \varphi) \text{ \在 } \left(\frac{\pi}{3}, \frac{2\pi}{3}\right) \text{ \上单调递增,}$

$\therefore \frac{\pi}{3} + \varphi \ge -\frac{\pi}{2} + 2k\pi.$

$\text{解得 } 2k\pi - \frac{5\pi}{6} \le \varphi \le 2k\pi - \frac{\pi}{6}.$

$\text{当 } k=1 \text{ \时, } \frac{7\pi}{6} \le \varphi \le \frac{11\pi}{6}, \therefore \varphi \text{ \的值可能是 } \frac{3\pi}{2}, \text{ \故选 D.}$

$7. \text{选 D } \because y = \sin 2x \text{ \在 } \left(\frac{\pi}{2}, \pi\right) \text{ \上不单调, \排除 A;}$

$y = 2|\cos x| \text{ \在 } \left(\frac{\pi}{2}, \pi\right) \text{ \上单调递增, \排除 B;}$

$y = \cos \frac{x}{2} \text{ \的最小正周期为 } 4\pi, \text{ \排除 C;}$

$y = \tan(-x) \text{ \的最小正周期为 } \pi, \text{ \且在区间 } \left(\frac{\pi}{2}, \pi\right) \text{ \上为减函数, \故选 D.}$

$8. \text{选 A } \because 0 \le x \le 9, \therefore -\frac{\pi}{3} \le \frac{\pi}{6}x - \frac{\pi}{3} \le \frac{7\pi}{6},$

$\therefore -\frac{\sqrt{3}}{2} \le \sin\left(\frac{\pi}{6}x - \frac{\pi}{3}\right) \le 1,$

$\therefore -\sqrt{3} \le 2\sin\left(\frac{\pi}{6}x - \frac{\pi}{3}\right) \le 2.$

$\therefore \text{函数 } y = 2\sin\left(\frac{\pi}{6}x - \frac{\pi}{3}\right) \text{ \的最大值为 } 2, \text{ \最小值为 } -\sqrt{3}.$

$\therefore \text{和为 } 2 - \sqrt{3}, \text{ \故选 A.}$

$9. \text{选 D } \because f(x) = \cos\left(x + \frac{\pi}{4}\right) \sin x = \frac{1}{2} \sin\left(2x + \frac{\pi}{4}\right) - \frac{\sqrt{2}}{4}, \therefore T = \frac{2\pi}{2} = \pi, \text{ A \错;}$

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\高考总复习金考卷

\数学(\文科)

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