齐次式的求值问题-三角函数的拓展与综合知识点考前进阶单选题自测题答案-广东省等高一数学必修,平均正确率55.99999999999999%

1. 已知 $$\sin \left( \alpha-\frac{\pi}{4} \right)=\frac{3\sqrt{2}}{10} (0 < \alpha < \pi)$$，则 $$\frac{\sin (\pi-2\alpha)}{\sin \alpha+\cos \alpha}=$$

解析：

由 $$\sin (\pi-2\alpha) = \sin 2\alpha = 2\sin \alpha \cos \alpha$$

分母 $$\sin \alpha + \cos \alpha$$

原式 $$= \frac{2\sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha}$$

设 $$\sin \alpha + \cos \alpha = t$$，则 $$t^2 = 1 + \sin 2\alpha$$

由已知 $$\sin \left( \alpha-\frac{\pi}{4} \right) = \frac{3\sqrt{2}}{10}$$

展开：$$\sin \alpha \cos \frac{\pi}{4} - \cos \alpha \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}(\sin \alpha - \cos \alpha) = \frac{3\sqrt{2}}{10}$$

得 $$\sin \alpha - \cos \alpha = \frac{3}{5}$$

又 $$(\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2 = 2$$

代入：$$t^2 + \left( \frac{3}{5} \right)^2 = 2$$

$$t^2 = 2 - \frac{9}{25} = \frac{41}{25}$$

$$\sin 2\alpha = t^2 - 1 = \frac{41}{25} - 1 = \frac{16}{25}$$

原式 $$= \frac{\sin 2\alpha}{t} = \frac{16/25}{\pm \sqrt{41}/5} = \pm \frac{16\sqrt{41}}{205}$$

由 $$0 < \alpha < \pi$$，$$\sin \alpha - \cos \alpha = \frac{3}{5} > 0$$，故 $$\sin \alpha > \cos \alpha$$

当 $$\alpha \in (0, \pi)$$ 时，$$\sin \alpha + \cos \alpha$$ 符号不确定

但 $$\sin 2\alpha = \frac{16}{25} > 0$$，故 $$\alpha \in (0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi)$$

若 $$\alpha \in (0, \frac{\pi}{2})$$，则 $$\sin \alpha + \cos \alpha > 0$$

若 $$\alpha \in (\frac{\pi}{2}, \pi)$$，则 $$\sin \alpha > 0$$，$$\cos \alpha < 0$$，但 $$\sin \alpha - \cos \alpha = \frac{3}{5}$$ 较小，可能 $$\sin \alpha + \cos \alpha < 0$$

计算 $$\sin \alpha = \frac{(\sin \alpha + \cos \alpha) + (\sin \alpha - \cos \alpha)}{2} = \frac{t + \frac{3}{5}}{2}$$

$$\cos \alpha = \frac{t - \frac{3}{5}}{2}$$

由 $$\sin^2 \alpha + \cos^2 \alpha = 1$$ 验证，$$t = \pm \frac{\sqrt{41}}{5}$$ 均满足

但原式分母 $$\sin \alpha + \cos \alpha = t$$，故结果有正负两种可能

选项中有负值，故选 B

答案：B

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

解析：

$$\tan \left( \alpha+\frac{\pi}{4} \right) = \frac{\tan \alpha + 1}{1 - \tan \alpha} = -3$$

设 $$\tan \alpha = t$$，则 $$\frac{t+1}{1-t} = -3$$

$$t+1 = -3(1-t) = -3 + 3t$$

$$t - 3t = -3 - 1$$

$$-2t = -4$$

$$t = 2$$

原式 $$= \frac{2\sin \alpha+\cos \alpha}{\cos \alpha-\sin \alpha} = \frac{2\tan \alpha+1}{1-\tan \alpha} = \frac{2 \times 2 + 1}{1 - 2} = \frac{5}{-1} = -5$$

答案：D

3. 已知 $$\tan \left( \frac{\theta}{2}-\frac{\pi}{6} \right)=2$$，则 $$\cos \left( \theta-\frac{\pi}{3} \right)=$$

解析：

设 $$u = \frac{\theta}{2} - \frac{\pi}{6}$$，则 $$\theta = 2u + \frac{\pi}{3}$$

$$\theta - \frac{\pi}{3} = 2u$$

$$\cos (\theta - \frac{\pi}{3}) = \cos 2u = \frac{1 - \tan^2 u}{1 + \tan^2 u}$$

已知 $$\tan u = 2$$

代入：$$\cos 2u = \frac{1 - 4}{1 + 4} = \frac{-3}{5} = -\frac{3}{5}$$

答案：B

4. 已知 $$\tan (\pi-\theta)=3$$，则 $$\frac{\sin \left( \frac{\pi}{2}+\theta\right)-\cos \left( \pi-\theta\right)}{\sin \left( \frac{\pi}{2}-\theta\right)-\sin \left( \pi-\theta\right)}=$$

解析：

$$\tan (\pi-\theta) = -\tan \theta = 3$$，故 $$\tan \theta = -3$$

化简分子：$$\sin \left( \frac{\pi}{2}+\theta\right) - \cos (\pi-\theta) = \cos \theta - (-\cos \theta) = 2\cos \theta$$

分母：$$\sin \left( \frac{\pi}{2}-\theta\right) - \sin (\pi-\theta) = \cos \theta - \sin \theta$$

原式 $$= \frac{2\cos \theta}{\cos \theta - \sin \theta} = \frac{2}{1 - \tan \theta} = \frac{2}{1 - (-3)} = \frac{2}{4} = \frac{1}{2}$$

答案：D

5. 已知 $$\alpha$$ 为锐角，且 $$\tan \alpha=1+\sqrt{2}$$，且 $$f(x)=\sin (\omega x+\varphi) (\omega>0, |\varphi| < \frac{\pi}{2})$$ 相邻对称轴之间的距离为 $$\frac{\pi}{2}$$，且 $$f(0)=-\frac{\sqrt{2}}{2}$$，则求值

解析：

相邻对称轴距离为半周期，故 $$\frac{T}{2} = \frac{\pi}{2}$$，$$T = \pi$$，$$\omega = \frac{2\pi}{T} = 2$$

$$f(0) = \sin \varphi = -\frac{\sqrt{2}}{2}$$，且 $$|\varphi| < \frac{\pi}{2}$$，故 $$\varphi = -\frac{\pi}{4}$$

$$f(x) = \sin (2x - \frac{\pi}{4})$$

$$\tan \alpha = 1 + \sqrt{2}$$，为锐角，可求 $$\sin \alpha$$ 和 $$\cos \alpha$$

但问题中未明确要求表达式，可能求 $$f(\alpha)$$ 或其他

由选项猜测可能求 $$\sin (2\alpha - \frac{\pi}{4})$$ 或类似

计算 $$\sin (2\alpha - \frac{\pi}{4}) = \sin 2\alpha \cos \frac{\pi}{4} - \cos 2\alpha \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}(\sin 2\alpha - \cos 2\alpha)$$

由 $$\tan \alpha = 1 + \sqrt{2}$$，可求 $$\sin 2\alpha$$ 和 $$\cos 2\alpha$$

$$\sin 2\alpha = \frac{2\tan \alpha}{1+\tan^2 \alpha} = \frac{2(1+\sqrt{2})}{1+(1+\sqrt{2})^2} = \frac{2+2\sqrt{2}}{1+3+2\sqrt{2}} = \frac{2+2\sqrt{2}}{4+2\sqrt{2}} = \frac{2(1+\sqrt{2})}{2(2+\sqrt{2})} = \frac{1+\sqrt{2}}{2+\sqrt{2}}$$

有理化：$$\frac{1+\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{(1+\sqrt{2})(2-\sqrt{2})}{4-2} = \frac{2 - \sqrt{2} + 2\sqrt{2} - 2}{2} = \frac{\sqrt{2}}{2}$$

$$\cos 2\alpha = \frac{1-\tan^2 \alpha}{1+\tan^2 \alpha} = \frac{1-(3+2\sqrt{2})}{4+2\sqrt{2}} = \frac{-2-2\sqrt{2}}{4+2\sqrt{2}} = \frac{-2(1+\sqrt{2})}{2(2+\sqrt{2})} = -\frac{1+\sqrt{2}}{2+\sqrt{2}} = -\frac{\sqrt{2}}{2}$$

代入：$$\sin (2\alpha - \frac{\pi}{4}) = \frac{\sqrt{2}}{2} \left( \frac{\sqrt{2}}{2} - (-\frac{\sqrt{2}}{2}) \right) = \frac{\sqrt{2}}{2} \times \sqrt{2} = 1$$

答案：C

6. 如果 $$\tan \theta=2$$，那么 $$1+\sin \theta \cos \theta$$ 的值是

解析：

$$1 + \sin \theta \cos \theta = 1 + \frac{1}{2} \sin 2\theta$$

$$\sin 2\theta = \frac{2\tan \theta}{1+\tan^2 \theta} = \frac{4}{1+4} = \frac{4}{5}$$

故 $$1 + \frac{1}{2} \times \frac{4}{5} = 1 + \frac{2}{5} = \frac{7}{5}$$

答案：B

7. 已知角 $$\alpha$$ 的顶点在坐标原点 $$O$$，始边与 $$x$$ 轴的非负半轴重合，将 $$\alpha$$ 的终边按顺时针方向旋转 $$\frac{\pi}{4}$$ 后经过点 $$(3,4)$$，则 $$\sin 2\alpha=$$

解析：

旋转后终边经过点 (3,4)，该点对应角 $$\beta$$ 满足 $$\tan \beta = \frac{4}{3}$$

顺时针旋转 $$\frac{\pi}{4}$$，故 $$\alpha = \beta + \frac{\pi}{4}$$

$$\sin 2\alpha = \sin (2\beta + \frac{\pi}{2}) = \cos 2\beta$$

由 $$\tan \beta = \frac{4}{3}$$，可求 $$\cos 2\beta = \frac{1-\tan^2 \beta}{1+\tan^2 \beta} = \frac{1-16/9}{1+16/9} = \frac{-7/9}{25/9} = -\frac{7}{25}$$

答案：B

8. 若 $$\tan \alpha=2$$，则 $$\cos \left( \frac{\pi}{2}-2\alpha \right)=$$

解析：

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

$$\sin 2\alpha = \frac{2\tan \alpha}{1+\tan^2 \alpha} = \frac{4}{1+4} = \frac{4}{5}$$

答案：D

9. 若 $$\tan \theta=-2$$，则 $$\frac{\sin \theta(1+\sin 2\theta)}{\sin \theta+\cos \theta}=$$

解析：

分子：$$\sin \theta (1+\sin 2\theta) = \sin \theta + \sin \theta \sin 2\theta$$

分母：$$\sin \theta + \cos \theta$$

原式 $$= \frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta \sin 2\theta}{\sin \theta + \cos \theta}$$

$$= \frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{2\sin^2 \theta \cos \theta}{\sin \theta + \cos \theta}$$

$$= \frac{\sin \theta (1 + 2\sin \theta \cos \theta)}{\sin \theta + \cos \theta}$$

由 $$\tan \theta = -2$$，设 $$\sin \theta = -2k$$，$$\cos \theta = k$$，则 $$\sin^2 \theta + \cos^2 \theta = 4k^2 + k^2 = 5k^2 = 1$$，$$k = \pm \frac{1}{\sqrt{5}}$$

代入：原式 $$= \frac{-2k (1 + 2(-2k)(k))}{-2k + k} = \frac{-2k (1 - 4k^2)}{-k} = 2(1-4k^2)$$

$$k^2 = \frac{1}{5}$$，故 $$1-4k^2 = 1 - \frac{4}{5} = \frac{1}{5}$$

原式 $$= 2 \times \frac{1}{5} = \frac{2}{5}$$

答案：C