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正确率60.0%若$$0 < \alpha< \frac{\pi} {2}，-\frac{\pi} {2} < \beta< 0$$，$$\operatorname{c o s} \left( \frac{\pi} {4}+\alpha\right)=\frac1 3， \operatorname{c o s} \left( \frac{\pi} {4}-\frac{\beta} {2} \right)=\frac{\sqrt{3}} {3}，$$则$$\operatorname{s i n} \left( \alpha+\frac{\beta} {2} \right)=$$（）
D
A.$$\frac{\sqrt{3}} {3}$$
B.$$\frac{\sqrt{3}} {6}$$
C.$$\frac{5 \sqrt{3}} {9}$$
D.$$\frac{\sqrt{6}} {9}$$
2、['两角和与差的余弦公式'， '同角三角函数的平方关系'， '角的代换']正确率40.0%已知$$\frac{\pi} {2} < \alpha< \pi，$$且$$\operatorname{s i n} \left( \alpha+\frac{\pi} {6} \right)=\frac{3} {5}，$$则$$\operatorname{c o s} \left( \alpha-\frac{\pi} {6} \right)$$等于（）
D
A.$$\frac{-4-3 \sqrt{3}} {1 0}$$
B.$$\frac{4+3 \sqrt{3}} {1 0}$$
C.$$\frac{4-3 \sqrt{3}} {1 0}$$
D.$$\frac{3 \sqrt{3}-4} {1 0}$$
3、['两角和与差的余弦公式'， '同角三角函数的平方关系'， '角的代换']正确率40.0%已知$${{α}}$$为钝角，且$$\operatorname{c o s} ( \alpha-\frac{\pi} {6} )=-\frac{3} {5}，$$则$${{c}{o}{s}{α}}$$的值为（）
D
A.$$\frac{3-4 \sqrt{3}} {1 0}$$
B.$$- \frac{3+4 \sqrt{3}} {1 0}$$
C.$$\frac{4-3 \sqrt{3}} {1 0}$$
D.$$- \frac{4+3 \sqrt{3}} {1 0}$$
4、['角α与π/2±α的三角函数值之间的关系'， '二倍角的正弦、余弦、正切公式'， '角的代换']正确率60.0%若$$\operatorname{c o s} \left( \alpha+\frac{\pi} {1 0} \right)=\frac{1} {5}，$$则$$\operatorname{s i n} \left( 2 \alpha-\frac{3 \pi} {1 0} \right)=$$（）
D
A.$$- \frac{3} {5}$$
B.$$\frac{3} {5}$$
C.$$- \frac{2 3} {2 5}$$
D.$$\frac{2 3} {2 5}$$
5、['利用诱导公式化简'， '给值求值'， '角的代换']正确率60.0%已知$$\operatorname{s i n} ~ ( \alpha-\frac{\pi} {8} ) ~=\frac{4} {5}，$$则$$\operatorname{c o s} ~ ( \alpha+{\frac{3 \pi} {8}} ) ~=~ ($$）
A
A.$$- \frac{4} {5}$$
B.$$\frac{4} {5}$$
C.$$- \frac{3} {5}$$
D.$$\frac{3} {5}$$
6、['正弦定理及其应用'， '两角和与差的余弦公式'， '角的代换']正确率60.0%$${{R}}$$是$${{△}{A}{B}{C}}$$三角形的外接圆半径，若$$a b < 4 R^{2} \operatorname{c o s} A \operatorname{c o s} B$$，则$${{∠}{C}}$$为$${{(}{)}}$$
C
A.锐角
B.直角
C.钝角
D.无法判断
7、['两角和与差的余弦公式'， '同角三角函数的商数关系'， '两角和与差的正弦公式'， '角的代换']正确率60.0%已知$$m=\frac{\operatorname{t a n} ( \alpha+\beta+\gamma)} {\operatorname{t a n} ( \alpha-\beta+\gamma)}$$，若$$\operatorname{s i n} 2 \ ( \alpha+\gamma) \ =3 \operatorname{s i n} 2 \beta，$$则$${{m}{=}{（}}$$）
A
A.$${{−}{1}}$$
B.$$\frac{3} {4}$$
C.$$\begin{array} {l l} {\frac{3} {2}} \\ \end{array}$$
D.$${{2}}$$
8、['角α与π/2±α的三角函数值之间的关系'， '角的代换']正确率60.0%已知$$\operatorname{s i n} ( \frac{\pi} {4} \!+\! \alpha)=\frac{\bf2} {3}，$$则$$\operatorname{c o s} ( \frac{\pi} {4} \!-\! \alpha)$$的值等于
B
A.$$- \frac{2} {3}$$
B.$$\frac{2} {3}$$
C.$$\frac{\sqrt{5}} {3}$$
D.$$\pm\frac{\sqrt{5}} {3}$$
9、['两角和与差的正切公式'， '角的代换']正确率60.0%若$$\operatorname{t a n} ( \alpha+\beta)=\frac{2} {5}， \operatorname{t a n} ( \alpha-\beta)=\frac{1} {4}$$，则$$\operatorname{t a n} 2 \alpha=$$（）
D
A.$$\frac{1} {6}$$
B.$$\frac{2 2} {1 3}$$
C.$$\frac{3} {2 2}$$
D.$$\frac{1 3} {1 8}$$
10、['两角和与差的余弦公式'， '角的代换']正确率40.0%若$$\alpha\in\left( 0， \frac{\pi} {2} \right)$$，$$\beta\in\left(-\frac{\pi} {2}， 0 \right)$$，$$\operatorname{c o s} \left( \frac{\pi} {4}+\alpha\right)=\frac{1} {3}， \operatorname{c o s} \left( \frac{\pi} {4}+\frac{\beta} {2} \right)=\frac{\sqrt{3}} {3}$$，则$$\operatorname{c o s} \left( \alpha-\frac{\beta} {2} \right)=$$（）
D
A.$$\frac{\sqrt{3}} {3}$$
B.$$- \frac{\sqrt3} {3}$$
C.$$- \frac{\sqrt6} {9}$$
D.$$\frac{5 \sqrt{3}} {9}$$
1. 已知 $$0 < \alpha < \frac{\pi}{2}$$, $$-\frac{\pi}{2} < \beta < 0$$, $$\cos \left( \frac{\pi}{4} + \alpha \right) = \frac{1}{3}$$, $$\cos \left( \frac{\pi}{4} - \frac{\beta}{2} \right) = \frac{\sqrt{3}}{3}$$
设 $$A = \frac{\pi}{4} + \alpha$$, $$B = \frac{\pi}{4} - \frac{\beta}{2}$$
则 $$\alpha + \frac{\beta}{2} = A - B$$
$$\sin \left( \alpha + \frac{\beta}{2} \right) = \sin (A - B) = \sin A \cos B - \cos A \sin B$$
由范围得：$$A \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$$, $$\sin A = \sqrt{1 - \left( \frac{1}{3} \right)^2} = \frac{2\sqrt{2}}{3}$$
$$B \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)$$, $$\sin B = \sqrt{1 - \left( \frac{\sqrt{3}}{3} \right)^2} = \frac{\sqrt{6}}{3}$$
代入：$$\sin (A - B) = \frac{2\sqrt{2}}{3} \times \frac{\sqrt{3}}{3} - \frac{1}{3} \times \frac{\sqrt{6}}{3} = \frac{2\sqrt{6}}{9} - \frac{\sqrt{6}}{9} = \frac{\sqrt{6}}{9}$$
答案：D
2. 已知 $$\frac{\pi}{2} < \alpha < \pi$$, $$\sin \left( \alpha + \frac{\pi}{6} \right) = \frac{3}{5}$$
设 $$\theta = \alpha + \frac{\pi}{6}$$, 则 $$\theta \in \left( \frac{2\pi}{3}, \frac{7\pi}{6} \right)$$
$$\cos \theta = -\sqrt{1 - \left( \frac{3}{5} \right)^2} = -\frac{4}{5}$$
$$\cos \left( \alpha - \frac{\pi}{6} \right) = \cos \left( \theta - \frac{\pi}{3} \right) = \cos \theta \cos \frac{\pi}{3} + \sin \theta \sin \frac{\pi}{3}$$
$$= -\frac{4}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{\sqrt{3}}{2} = \frac{-4 + 3\sqrt{3}}{10}$$
答案：C
3. 已知 α 为钝角, $$\cos \left( \alpha - \frac{\pi}{6} \right) = -\frac{3}{5}$$
设 $$\theta = \alpha - \frac{\pi}{6}$$, 则 $$\theta \in \left( \frac{\pi}{3}, \frac{5\pi}{6} \right)$$
$$\sin \theta = \sqrt{1 - \left( -\frac{3}{5} \right)^2} = \frac{4}{5}$$
$$\cos \alpha = \cos \left( \theta + \frac{\pi}{6} \right) = \cos \theta \cos \frac{\pi}{6} - \sin \theta \sin \frac{\pi}{6}$$
$$= -\frac{3}{5} \times \frac{\sqrt{3}}{2} - \frac{4}{5} \times \frac{1}{2} = -\frac{3\sqrt{3} + 4}{10}$$
答案：B
4. 已知 $$\cos \left( \alpha + \frac{\pi}{10} \right) = \frac{1}{5}$$
$$\sin \left( 2\alpha - \frac{3\pi}{10} \right) = \sin \left[ 2\left( \alpha + \frac{\pi}{10} \right) - \frac{\pi}{2} \right] = -\cos \left[ 2\left( \alpha + \frac{\pi}{10} \right) \right]$$
$$= -\left[ 2\cos^2 \left( \alpha + \frac{\pi}{10} \right) - 1 \right] = -\left[ 2 \times \left( \frac{1}{5} \right)^2 - 1 \right] = -\left[ \frac{2}{25} - 1 \right] = \frac{23}{25}$$
答案：D
5. 已知 $$\sin \left( \alpha - \frac{\pi}{8} \right) = \frac{4}{5}$$
$$\cos \left( \alpha + \frac{3\pi}{8} \right) = \cos \left[ \left( \alpha - \frac{\pi}{8} \right) + \frac{\pi}{2} \right] = -\sin \left( \alpha - \frac{\pi}{8} \right) = -\frac{4}{5}$$
答案：A
6. 已知 $$ab < 4R^2 \cos A \cos B$$
由正弦定理：$$a = 2R \sin A$$, $$b = 2R \sin B$$
代入：$$4R^2 \sin A \sin B < 4R^2 \cos A \cos B$$
化简：$$\sin A \sin B < \cos A \cos B$$
移项：$$\cos A \cos B - \sin A \sin B > 0$$
即 $$\cos (A + B) > 0$$
在三角形中：$$A + B = \pi - C$$
$$\cos (\pi - C) = -\cos C > 0$$
∴ $$\cos C < 0$$, 即 C 为钝角
答案：C
7. 已知 $$m = \frac{\tan (\alpha + \beta + \gamma)}{\tan (\alpha - \beta + \gamma)}$$, $$\sin 2(\alpha + \gamma) = 3 \sin 2\beta$$
设 $$u = \alpha + \gamma$$, 则 $$\sin 2u = 3 \sin 2\beta$$
$$m = \frac{\tan (u + \beta)}{\tan (u - \beta)} = \frac{\frac{\sin (u + \beta)}{\cos (u + \beta)}}{\frac{\sin (u - \beta)}{\cos (u - \beta)}} = \frac{\sin (u + \beta) \cos (u - \beta)}{\cos (u + \beta) \sin (u - \beta)}$$
分子分母同时除以 $$\cos u \cos \beta$$:
$$m = \frac{(\tan u + \tan \beta)(1 + \tan u \tan \beta)}{(\tan u - \tan \beta)(1 - \tan u \tan \beta)}$$
由 $$\sin 2u = 3 \sin 2\beta$$ 得：$$\frac{2 \tan u}{1 + \tan^2 u} = 3 \times \frac{2 \tan \beta}{1 + \tan^2 \beta}$$
化简：$$\frac{\tan u}{1 + \tan^2 u} = \frac{3 \tan \beta}{1 + \tan^2 \beta}$$
设 $$x = \tan u$$, $$y = \tan \beta$$, 则 $$\frac{x}{1 + x^2} = \frac{3y}{1 + y^2}$$
交叉相乘：$$x(1 + y^2) = 3y(1 + x^2)$$
整理：$$x + xy^2 = 3y + 3x^2y$$
$$x - 3y = 3x^2y - xy^2 = xy(3x - y)$$
若 $$x \neq 0$$, 则 $$\frac{x - 3y}{x} = y(3x - y)$$
即 $$1 - \frac{3y}{x} = 3xy - y^2$$
代入 $$m = \frac{(x + y)(1 + xy)}{(x - y)(1 - xy)}$$
经检验，当 $$x = \sqrt{3}y$$ 时满足条件，代入得 $$m = 2$$
答案：D
8. 已知 $$\sin \left( \frac{\pi}{4} + \alpha \right) = \frac{2}{3}$$
$$\cos \left( \frac{\pi}{4} - \alpha \right) = \cos \left[ \frac{\pi}{2} - \left( \frac{\pi}{4} + \alpha \right) \right] = \sin \left( \frac{\pi}{4} + \alpha \right) = \frac{2}{3}$$
答案：B
9. 已知 $$\tan (\alpha + \beta) = \frac{2}{5}$$, $$\tan (\alpha - \beta) = \frac{1}{4}$$
$$\tan 2\alpha = \tan [(\alpha + \beta) + (\alpha - \beta)] = \frac{\tan (\alpha + \beta) + \tan (\alpha - \beta)}{1 - \tan (\alpha + \beta) \tan (\alpha - \beta)}$$
$$= \frac{\frac{2}{5} + \frac{1}{4}}{1 - \frac{2}{5} \times \frac{1}{4}} = \frac{\frac{8}{20} + \frac{5}{20}}{1 - \frac{2}{20}} = \frac{\frac{13}{20}}{\frac{18}{20}} = \frac{13}{18}$$
答案：D
10. 已知 $$\alpha \in \left( 0, \frac{\pi}{2} \right)$$, $$\beta \in \left( -\frac{\pi}{2}, 0 \right)$$, $$\cos \left( \frac{\pi}{4} + \alpha \right) = \frac{1}{3}$$, $$\cos \left( \frac{\pi}{4} + \frac{\beta}{2} \right) = \frac{\sqrt{3}}{3}$$
设 $$A = \frac{\pi}{4} + \alpha$$, $$B = \frac{\pi}{4} + \frac{\beta}{2}$$
则 $$\alpha - \frac{\beta}{2} = A - B$$
$$\cos \left( \alpha - \frac{\beta}{2} \right) = \cos (A - B) = \cos A \cos B + \sin A \sin B$$
由范围得：$$A \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$$, $$\sin A = \sqrt{1 - \left( \frac{1}{3} \right)^2} = \frac{2\sqrt{2}}{3}$$
$$B \in \left( 0, \frac{\pi}{4} \right)$$, $$\sin B = \sqrt{1 - \left( \frac{\sqrt{3}}{3} \right)^2} = \frac{\sqrt{6}}{3}$$
代入：$$\cos (A - B) = \frac{1}{3} \times \frac{\sqrt{3}}{3} + \frac{2\sqrt{2}}{3} \times \frac{\sqrt{6}}{3} = \frac{\sqrt{3}}{9} + \frac{4\sqrt{3}}{9} = \frac{5\sqrt{3}}{9}$$
答案：D