《2026年零差错高中数学必修第一册人教版》

5. 【题型三】若$\sin \alpha = \frac{1}{2}$，$\alpha \in (\frac{\pi}{2}, \pi)$，则$\cos(\pi - \alpha)$的值为$\frac{\sqrt{3}}{2}$.

答案： 5.$\frac{\sqrt{3}}{2}$【详解】由$\sin\alpha=\frac{1}{2},\alpha\in(\frac{\pi}{2},\pi)$,则$\cos(\pi-\alpha)=-\cos\alpha=-[-\sqrt{1 - (\frac{1}{2})^{2}}]=\frac{\sqrt{3}}{2}$.

6. 【题型三】若$\sin(180^{\circ} + \alpha) + \cos(90^{\circ} + \alpha) = -\frac{1}{4}$，则$\cos(270^{\circ} - \alpha) + 2\sin(360^{\circ} - \alpha) =$$-\frac{3}{8}$.

答案： 6.$-\frac{3}{8}$【详解】由$\sin(180^{\circ}+\alpha)+\cos(90^{\circ}+\alpha)=-\frac{1}{4}$,可得
$-\sin\alpha-\sin\alpha=-\frac{1}{4}$,即$\sin\alpha=\frac{1}{8}$则$\cos(270^{\circ}-\alpha)+2\sin(360^{\circ}-\alpha)=-\sin\alpha-2\sin\alpha=-3\sin\alpha=-\frac{3}{8}$.

7. 【题型一、二、三】已知$f(\alpha) = \frac{\sin(\pi - \alpha)\cos(2\pi - \alpha)\cos(-\alpha + \frac{3}{2}\pi)}{\cos(\frac{\pi}{2} - \alpha)\sin(-\pi - \alpha)}$，则$f(\frac{\pi}{3}) =$$-\frac{1}{2}$.

答案： 7.$-\frac{1}{2}$【详解】$f(\alpha)=\frac{\sin(\pi-\alpha)\cos(2\pi-\alpha)\cos(-\alpha+\frac{3}{2}\pi)}{\cos(\frac{\pi}{2}-\alpha)\sin(-\pi-\alpha)}=\frac{\sin\alpha·\cos\alpha·(-\sin\alpha)}{-\cos\alpha(-\sin\alpha)}=-\cos\alpha$,则$f(\frac{\pi}{3})=-\cos\frac{\pi}{3}=-\frac{1}{2}$.

8. 【题型三、五】若角$\alpha$的终边过点$P(-3, 4)$.
(1)求$\sin \alpha$和$\cos \alpha$的值；
(2)求$\frac{\sin(\pi - \alpha) + \cos(3\pi + \alpha)}{\cos(\frac{3\pi}{2} + \alpha) - \sin(\frac{7\pi}{2} - \alpha)}$的值.

答案： 8.解：
(1)因为角$\alpha$的终边过点$P(-3,4)$,所以$\sin\alpha=\frac{4}{\sqrt{(-3)^{2}+4^{2}}}=\frac{4}{5},\cos\alpha=\frac{-3}{\sqrt{(-3)^{2}+4^{2}}}=-\frac{3}{5}$.
(2)$\frac{\sin(\pi-\alpha)+\cos(3\pi+\alpha)}{\cos(\frac{3\pi}{2}+\alpha)-\sin(\frac{7\pi}{2}-\alpha)}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha-(-\cos\alpha)}=\frac{\frac{4}{5}-(-\frac{3}{5})}{\frac{4}{5}+\frac{3}{5}} = 7$.

9. 【题型三、四】求证：$\frac{\tan(2\pi - \alpha)\cos(\frac{3\pi}{2} - \alpha)\cos(6\pi - \alpha)}{\tan(\pi - \alpha)\sin(\alpha + \frac{3\pi}{2})\cos(\alpha + \frac{3\pi}{2})} = 1$.

答案： 9.证明：左边$=\frac{\tan(-\alpha)[-\cos(\frac{\pi}{2}-\alpha)][-\cos(\frac{\pi}{2}+\alpha)]}{(-\tan\alpha)(-\sin\alpha)\cos\alpha}=\frac{(-\tan\alpha)(-\sin\alpha)\cos\alpha}{(-\tan\alpha)(-\sin\alpha)\cos\alpha}=1$=右边,所以原式成立.

10. 【题型三、四】(1)求证：$\frac{\tan(2\pi - \alpha)\sin(-2\pi - \alpha)\cos(6\pi - \alpha)}{\sin(\alpha + \frac{3\pi}{2})\cos(\alpha + \frac{3\pi}{2})} = -\tan \alpha$；
(2)设$\tan(\alpha + \frac{8\pi}{7}) = m$，求证：$\frac{\sin(\frac{15\pi}{7} + \alpha) + 3\cos(\alpha - \frac{13\pi}{7})}{\sin(\frac{20\pi}{7} - \alpha) - \cos(\alpha + \frac{22\pi}{7})} = \frac{m + 3}{m + 1}$.

答案： 10.证明：
(1)左边$=\frac{\tan(-\alpha)\sin(-\alpha)\cos(-\alpha)}{\sin[2\pi-(\frac{\pi}{2}-\alpha)]\cos[2\pi-(\frac{\pi}{2}-\alpha)]}=\frac{(-\tan\alpha)(-\sin\alpha)\cos\alpha}{\sin[-(\frac{\pi}{2}-\alpha)]\cos[-(\frac{\pi}{2}-\alpha)]}=\frac{\sin^{2}\alpha}{\sin^{2}(\frac{\pi}{2}-\alpha)\cos(\frac{\pi}{2}-\alpha)}=\frac{\sin^{2}\alpha}{-\cos\alpha\sin\alpha}=-\tan\alpha$
右边,所以原等式成立.
(2)方法1：左边$=\sin[\pi+(\frac{8\pi}{7}+\alpha)]+3\cos[(\alpha+\frac{8\pi}{7})-3\pi]=\sin[4\pi-(\alpha+\frac{8\pi}{7})]-\cos[2\pi+(\alpha+\frac{8\pi}{7})]=-\sin(\frac{8\pi}{7}+\alpha)-3\cos(\frac{8\pi}{7}+\alpha)=\frac{-\sin(\frac{8\pi}{7}+\alpha)-3\cos(\frac{8\pi}{7}+\alpha)}{\tan(\alpha+\frac{8\pi}{7})+3}=\frac{m + 3}{m+1}$=右边,
所以原等式成立.方法2：由$\tan(\alpha+\frac{8\pi}{7})=m$,得$\tan(\alpha+\frac{\pi}{7})=m$,所以,等式左边$=\sin[2\pi+(\frac{\pi}{7}+\alpha)]+3\cos[(\alpha+\frac{\pi}{7})-2\pi]=\sin[2\pi+\pi-(\alpha+\frac{\pi}{7})]-\cos[2\pi+\pi+(\alpha+\frac{\pi}{7})]=\sin(\frac{\pi}{7}+\alpha)+3\cos(\alpha+\frac{\pi}{7})=\frac{\sin(\frac{\pi}{7}+\alpha)+3\cos(\alpha+\frac{\pi}{7})}{\tan(\alpha+\frac{\pi}{7})+3}=\frac{m + 3}{m+1}$=右边,等式成立.