16个诱导公式题目及答案解析

16个诱导公式题目及答案解析一、选择题（共8分）1.（2分）已知\sin(\frac{\pi}{6}-\theta)=\frac{1}{2}，求\cos(\frac{\pi}{3}+\theta)的值。A.\frac{1}{2}B.\frac{\sqrt{3}}{2}C.-\frac{1}{2}D.-\frac{\sqrt{3}}{2}答案：C解析：根据诱导公式，\cos(\frac{\pi}{3}+\theta)=\sin(\frac{\pi}{2}-(\frac{\pi}{3}+\theta))=\sin(\frac{\pi}{6}-\theta)=\frac{1}{2}。因此，选项C正确。2.（2分）若\sin(\theta)=\frac{1}{2}，求\cos(\frac{5\pi}{6}-\theta)的值。A.\frac{\sqrt{3}}{2}B.-\frac{\sqrt{3}}{2}C.\frac{1}{2}D.-\frac{1}{2}答案：A解析：根据诱导公式，\cos(\frac{5\pi}{6}-\theta)=\sin(\frac{\pi}{2}-(\frac{5\pi}{6}-\theta))=\sin(\theta-\frac{\pi}{3})=\sin(\theta)\cos(\frac{\pi}{3})-\cos(\theta)\sin(\frac{\pi}{3})=\frac{1}{2}\times\frac{1}{2}-\sqrt{1-\left(\frac{1}{2}\right)^2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}。因此，选项A正确。二、填空题（共6分）3.（3分）计算\sin(\frac{2\pi}{3}+\theta)的值。答案：\sqrt{3}\sin(\theta)-\cos(\theta)解析：根据诱导公式，\sin(\frac{2\pi}{3}+\theta)=\sin(\frac{\pi}{2}+\frac{\pi}{6}+\theta)=\cos(\frac{\pi}{6}+\theta)=\cos(\frac{\pi}{6})\cos(\theta)-\sin(\frac{\pi}{6})\sin(\theta)=\frac{\sqrt{3}}{2}\cos(\theta)-\frac{1}{2}\sin(\theta)=\sqrt{3}\sin(\theta)-\cos(\theta)。4.（3分）计算\cos(\frac{7\pi}{4}-\theta)的值。答案：-\sin(\theta)解析：根据诱导公式，\cos(\frac{7\pi}{4}-\theta)=\cos(2\pi-(\frac{\pi}{4}+\theta))=\cos(\frac{\pi}{4}+\theta)=\cos(\frac{\pi}{4})\cos(\theta)-\sin(\frac{\pi}{4})\sin(\theta)=\frac{\sqrt{2}}{2}\cos(\theta)-\frac{\sqrt{2}}{2}\sin(\theta)=-\sin(\theta)。三、简答题（共6分）5.（3分）已知\tan(\theta)=2，求\tan(\frac{\pi}{2}-\theta)的值。答案：\frac{1}{2}解析：根据诱导公式，\tan(\frac{\pi}{2}-\theta)=\frac{1}{\tan(\theta)}=\frac{1}{2}。6.（3分）已知\cos(\theta)=\frac{3}{5}，求\sin(\frac{3\pi}{2}-\theta)的值。答案：-\frac{4}{5}解析：根据诱导公式，\sin(\frac{3\pi}{2}-\theta)=-\cos(\theta)=-\frac{3}{5}。由于\cos(\theta)=\frac{3}{5}，我们可以利用勾股定理求出\sin(\theta)=\sqrt{1-\cos^2(\theta)}=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}。因此，\sin(\frac{3\pi}{2}-\theta)=-\sin(\theta)=-\frac{4}{5}。四、计算题（共10分）7.（5分）计算\sin(\frac{\pi}{4}+\theta)\cos(\frac{\pi}{4}-\theta)的值。答案：\frac{1}{2}解析：根据诱导公式，\sin(\frac{\pi}{4}+\theta)\cos(\frac{\pi}{4}-\theta)=\frac{\sqrt{2}}{2}\cos(\theta)+\frac{\sqrt{2}}{2}\sin(\theta)\times\frac{\sqrt{2}}{2}\cos(\theta)+\frac{\sqrt{2}}{2}\sin(\theta)=\frac{1}{2}(\cos^2(\theta)+\sin^2(\theta))=\frac{1}{2}。8.（5分）计算\cos(\frac{\pi}{6}+\theta)\sin(\frac{\pi}{3}-\theta)的值。答案：\frac{1}{4}解析：根据诱导公式，\cos(\frac{\pi}{6}+\theta)\sin(\frac{\pi}{3}-\theta)=\frac{\sqrt{3}}{2}\cos(\theta)-\frac{1}{2}\sin(\theta)\times\frac{\sqrt{3}}{2}\cos(\theta)-\frac{1}{2}\sin(\theta)=\frac{3}{4}\cos^2(\theta)-\frac{1}{4}\sin^2(\theta)=\frac{3}{4}-\frac{1}{4}=\frac{1}{4}。五、证明题（共8分）9.（4分）证明\sin(\frac{\pi}{2}+\theta)=\cos(\theta)。证明：根据诱导公式，\sin(\frac{\pi}{2}+\theta)=\cos(\frac{\pi}{2}-(\frac{\pi}{2}+\theta))=\cos(-\theta)=\cos(\theta)。因此，\sin(\frac{\pi}{2}+\theta)=\cos(\theta)。10.（4分）证明\cos(\pi-\theta)=-\cos(\theta)。证明：根据诱导公式，\cos(\pi-\theta)=\cos(\pi)\cos(\theta)+\sin(\pi)\sin(\theta)=-1\times\cos(\theta)+0\times\sin(\theta)=-\cos(\theta)。因此，\cos(\pi-\theta)=-\cos(\theta)。六、综合题（共10分）11.（5分）已知\sin(\theta)=\frac{1}{3}，求\sin(\frac{5\pi}{6}+\theta)的值。答案：-\frac{2\sqrt{2}+1}{6}解析：根据诱导公式，\sin(\frac{5\pi}{6}+\theta)=\sin(\pi-(\frac{\pi}{6}-\theta))=\sin(\frac{\pi}{6}-\theta)=\sin(\frac{\pi}{6})\cos(\theta)-\cos(\frac{\pi}{6})\sin(\theta)=\frac{1}{2}\cos(\theta)-\frac{\sqrt{3}}{2}\sin(\theta)。已知\sin(\theta)=\frac{1}{3}，我们可以利用勾股定理求出\cos(\theta)=\sqrt{1-\sin^2(\theta)}=\sqrt{1-\left(\frac{1}{3}\right)^2}=\frac{2\sqrt{2}}{3}。代入公式得\sin(\frac{5\pi}{6}+\theta)=\frac{1}{2}\times\frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{2}\times\frac{1}{3}=\frac{\sqrt{2}}{3}-\frac{\sqrt{3}}{6}=-\frac{2\sqrt{2}+1}{6}。12.（5分）已知\cos(\theta)=\frac{4}{5}，求\cos(\frac{7\pi}{4}+\theta)的值。答案：\frac{7}{25}解析：根据诱导公式，\cos(\frac{7\pi}{4}+\theta)=\cos(2\pi-(\frac{\pi}{4}-\theta))=\cos(\frac{\pi}{4}-\theta)=\cos(\frac{\pi}{4})\cos(\theta)+\sin(\frac{\pi}{4})\sin(\theta)=\frac{\sqrt{2}}{2}\cos(\theta)+\frac{\sqrt{2}}{2}\sin(\theta)。已知\cos(\theta)=\frac{4}{5}，我们可以利用勾股定理求出\sin(\theta)=\sqrt{1-\cos^2(\theta)}=\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}。代入公式得\cos(\