《2025年全效核心素养测评高中数学必修第一册人教版》

13. 化简：(1)$\frac{\cos(\alpha-\pi)}{\sin(\pi-\alpha)}\cdot\sin(\alpha-\frac{\pi}{2})\cos(\frac{\pi}{2}+\alpha)$；
(2)$\frac{\tan(2\pi-\alpha)\cos(\frac{3\pi}{2}-\alpha)\cos(6\pi-\alpha)}{\tan(\pi-\alpha)\sin(\alpha+\frac{3\pi}{2})\cos(\alpha+\frac{3\pi}{2})}$.

答案： 13. 解：
(1) 原式$ = \frac{\cos[-(\pi - \alpha)]}{\sin \alpha} \cdot \sin\left[-\left(\frac{\pi}{2} - \alpha\right)\right] \cdot$
$(- \sin \alpha) = \frac{\cos(\pi - \alpha)}{\sin \alpha} \cdot \left[ - \sin\left(\frac{\pi}{2} - \alpha\right) \right] \cdot (- \sin \alpha) =$
$\frac{- \cos \alpha}{\sin \alpha} \cdot (- \cos \alpha) (- \sin \alpha) = - \cos^2 \alpha.$
(2)原式$= \frac{\tan(-\alpha)\left[ - \cos\left( \frac{\pi}{2} - \alpha \right) \right] \cos(-\alpha)}{( - \tan \alpha)\left[ - \sin\left( \frac{\pi}{2} + \alpha \right) \right]\left[ - \cos\left( \frac{\pi}{2} + \alpha \right) \right]}$
$\frac{( - \tan \alpha)( - \sin \alpha)\cos\alpha}{( - \tan \alpha)( - \cos \alpha)\sin\alpha} = 1.$

14. （2024·山东济宁高一期末）在平面直角坐标系$xOy$中，角$\alpha$与$\beta$的顶点均为坐标原点$O$，始边均为$x$轴的非负半轴. 若角$\alpha$的终边$OP$与单位圆交于点$P(\frac{\sqrt{6}}{3},y_{0})(y_{0}\gt0)$，将$OP$绕原点$O$按逆时针方向旋转$\frac{\pi}{2}$后与角$\beta$的终边$OQ$重合.
(1)求$\tan\beta$的值；
(2)求$\frac{\sin(-\beta)-\cos(\pi-\beta)}{\sin(\pi+\beta)-\sin(\frac{\pi}{2}-\beta)}$的值.

答案： 14. 解：
(1)由题意可知$ \sin \alpha = y_0, \cos \alpha = \frac{\sqrt{6}}{3}, \because \sin^2 \alpha + \cos^2 \alpha = 1,$
$\therefore y_0^2 + \left( \frac{\sqrt{6}}{3} \right)^2 = 1, $又$ y_0 ＞ 0, \therefore $可得$ y_0 = \frac{\sqrt{3}}{3}, $即$ \sin \alpha = \frac{\sqrt{3}}{3}.$
又$ \beta = \alpha + \frac{\pi}{2} + 2k\pi (k \in \mathbf{Z}), \therefore \sin \beta = \sin\left( \alpha + \frac{\pi}{2} + 2k\pi \right) =$
$\cos \alpha = \frac{\sqrt{6}}{3} (k \in \mathbf{Z}), \cos \beta = \cos\left( \alpha + \frac{\pi}{2} + 2k\pi \right) = - \sin \alpha =$
$\frac{\sqrt{3}}{3} (k \in \mathbf{Z}). \therefore \tan \beta = \frac{\sin \beta}{\cos \beta} = - \sqrt{2}.$
(2)由
(1)知$ \tan \beta = - \sqrt{2},$
$\therefore \frac{\sin( - \beta) - \cos(\pi - \beta)}{\sin(\pi + \beta) - \sin\left( \frac{\pi}{2} - \beta \right)} = \frac{- \sin \beta + \cos \beta}{- \sin \beta - \cos \left( \frac{\pi}{2} - \beta \right)} = \frac{- \tan \beta + 1}{- \tan \beta - 1} =$
$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = 3 + 2\sqrt{2}.$

15. $\sin^{2}1^{\circ}+\sin^{2}2^{\circ}+\sin^{2}3^{\circ}+\cdots+\sin^{2}88^{\circ}+\sin^{2}89^{\circ}+\sin^{2}90^{\circ}=$________.

答案： $15. \frac{91}{2}$

16. 三角比内容丰富，公式很多. 仔细观察、大胆猜想、科学求证，你也能发现其中的一些奥秘. 现有如下两个恒等式：
(1)$\frac{\cos2^{\circ}}{\sin47^{\circ}}+\frac{\cos88^{\circ}}{\sin133^{\circ}}=\sqrt{2}$；
(2)$\frac{\cos5^{\circ}}{\sin50^{\circ}}+\frac{\cos85^{\circ}}{\sin130^{\circ}}=\sqrt{2}$.
根据以上恒等式，请你猜想出一个一般性的结论并证明（参考公式：$\sin(45^{\circ}+\alpha)=\sin\alpha\cos45^{\circ}+\cos\alpha\sin45^{\circ}$）.

答案： 16. 解：猜想$ \frac{\cos \alpha}{\sin(45° + \alpha)} + \frac{\cos(90° - \alpha)}{\sin(135° - \alpha)} = \sqrt{2}.$
证明：由诱导公式可得$ \cos(90° - \alpha) = \sin \alpha, \sin(135° - \alpha) =$
$\sin(45° + \alpha), \therefore \frac{\cos \alpha}{\sin(45° + \alpha)} + \frac{\cos(90° - \alpha)}{\sin(135° - \alpha)} = \frac{\sin \alpha + \cos \alpha}{\sin(45° + \alpha)} =$
$\frac{\sin \alpha + \cos \alpha}{\sin \alpha \cos 45° + \cos \alpha \sin 45°} = \sqrt{2}.$