高三数学专题 三角函数之给值求值问题

高三数学专题三角函数之给值求值问题

三角函数的“给值求值”问题一、单选题1.若$\alpha\in\left(0,\frac{\pi}{2}\right)$，且$\cos\left(\frac{\pi}{2}-\alpha\right)=2\cos^2\alpha$，则$\sin2\alpha$等于（）A.$\frac{3}{11}$B.$\frac{1}{5}$C.$\frac{1}{3}$D.$\frac{7}{15}$【答案】A2.已知$\sin\left(\alpha+\frac{\pi}{6}\right)=\frac{1}{2}$，且$\cos2\alpha-\frac{1}{2}=0$，则$\tan\alpha$的值是（）A.$-\frac{5}{9}$B.$-\frac{8}{17}$C.$\frac{5}{9}$D.$\frac{8}{17}$【答案】D【解析】因为$\sin\left(\alpha+\frac{\pi}{6}\right)=\frac{1}{2}$，所以$\alpha\in\left(0,\frac{\pi}{2}\right)$。又因为$\cos2\alpha-\frac{1}{2}=0$，所以$\cos2\alpha=\frac{1}{2}$，即$2\cos^2\alpha-1=\frac{1}{2}$，解得$\cos\alpha=\pm\frac{1}{2}$。因为$\alpha\in\left(0,\frac{\pi}{2}\right)$，所以$\cos\alpha=\frac{1}{2}$。又因为$\sin\left(\alpha+\frac{\pi}{6}\right)=\frac{1}{2}$，所以$\sin\alpha\cos\frac{\pi}{6}+\cos\alpha\sin\frac{\pi}{6}=\frac{1}{2}$，即$\frac{\sqrt{3}}{4}\sin\alpha+\frac{1}{4}=\frac{1}{2}$，解得$\sin\alpha=\frac{5}{8}$。因此，$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{5}{8}}{\frac{1}{2}}=\frac{5}{4}$。又因为$\tan\left(\alpha-\frac{\pi}{6}\right)=\frac{\tan\alpha-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}\tan\alpha}=\frac{2-\sqrt{3}}{2+\sqrt{3}}=-\frac{8}{3}$，所以$\tan\left(\alpha-\frac{\pi}{3}\right)=\frac{\tan\left(\alpha-\frac{\pi}{6}\right)-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}\tan\left(\alpha-\frac{\pi}{6}\right)}=\frac{-\frac{8}{3}-\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}\cdot\frac{8}{3}}=-\frac{8}{17}$。因此，$\tan\left(\beta-2\alpha\right)=\tan\left[\left(\alpha-\left(\alpha-\frac{\pi}{6}\right)\right)-\left(\beta-\left(\alpha+\frac{\pi}{6}\right)\right)\right]=\frac{\tan\left(\alpha-\frac{\pi}{6}\right)-\tan\left(\beta-\left(\alpha+\frac{\pi}{6}\right)\right)}{1+\tan\left(\alpha-\frac{\pi}{6}\right)\tan\left(\beta-\left(\alpha+\frac{\pi}{6}\right)\right)}=\frac{-\frac{8}{17}-\frac{5}{9}}{1+\frac{8}{17}\cdot\frac{5}{9}}=-\frac{4}{5}$。二、填空题3.已知$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac{7}{10}$，且$\alpha\in\left(0,\frac{\pi}{2}\right)$，则$\tan\alpha$等于$\underline{\hspace{2cm}}$。【答案】$\frac{24}{17}$【解析】因为$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac{7}{10}$，所以$\sin\alpha\cos\frac{\pi}{4}-\cos\alpha\sin\frac{\pi}{4}=\frac{7}{10}$，即$\frac{1}{\sqrt{2}}\sin\alpha-\frac{1}{\sqrt{2}}\cos\alpha=\frac{7}{10}$，解得$\sin\alpha-\cos\alpha=\frac{7\sqrt{2}}{10}$。因为$\alpha\in\left(0,\frac{\pi}{2}\right)$，所以$\sin\alpha>0$，$\cos\alpha>0$，所以$\sin\alpha-\cos\alpha=\sqrt{2}\left(\sin\alpha\cos\frac{\pi}{4}-\cos\alpha\sin\frac{\pi}{4}\right)=\sqrt{2}\sin\left(\alpha-\frac{\pi}{4}\right)=\frac{7\sqrt{2}}{10}$，解得$\sin\alpha=\frac{17}{20}$，$\cos\alpha=\frac{3}{20}$。因此，$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{17}{20}}{\frac{3}{20}}=\frac{17}{3}$。4.已知$\sin\alpha=\frac{4}{5}$，$\alpha\in\left(\frac{\pi}{2},\pi\right)$，则$\cos\left(\alpha-\frac{\pi}{4}\right)$等于$\underline{\hspace{2cm}}$。【答案】$-\frac{2}{5}$【解析】因为$\sin\alpha=\frac{4}{5}$，$\alpha\in\left(\frac{\pi}{2},\pi\right)$，所以$\cos\alpha<0$。又因为$\cos^2\alpha+\sin^2\alpha=1$，所以$\cos\alpha=-\frac{3}{5}$。因此，$\cos\left(\alpha-\frac{\pi}{4}\right)=\cos\alpha\cos\frac{\pi}{4}+\sin\alpha\sin\frac{\pi}{4}=-\frac{3}{5}\cdot\frac{1}{\sqrt{2}}+\frac{4}{5}\cdot\frac{1}{\sqrt{2}}=-\frac{2}{5}$。5.已知锐角$\alpha$，$\beta$满足$\left(\tan\alpha-1\right)\left(\tan\beta-1\right)=2$，则$\alpha+\beta$的值为$\underline{\hspace{2cm}}$。【答案】$\frac{3\pi}{4}$【解析】因为$\left(\tan\alpha-1\right)\left(\tan\beta-1\right)=2$，所以$\tan\alpha+\tan\beta=\tan\alpha\tan\beta-1$，即$\tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\tan\alpha+\tan\beta}{\tan\alpha\tan\beta-1}=\frac{\frac{\tan\alpha-1}{\tan\alpha}\cdot\frac{\tan\beta-1}{\tan\beta}}{\frac{\tan\alpha-1}{\tan\alpha}+\frac{\tan\beta-1}{\tan\beta}}=\frac{\left(\tan\alpha-1\right)\left(\tan\beta-1\right)}{\tan\alpha+\tan\beta-2}=\frac{2}{\tan\alpha+\tan\beta-2}$。因为$\alpha$，$\beta$都是锐角，所以$\tan\alpha>0$，$\tan\beta>0$，所以$\tan\alpha-1>0$，$\tan\beta-1>0$，所以$\tan\alpha>1$，$\tan\beta>1$。因此，$\tan\left(\alpha+\beta\right)=\frac{2}{\tan\alpha+\tan\beta-2}<0$，所以$\alpha+\beta\in\left(\frac{\pi}{2},\pi\right)$。又因为$\tan\left(\alpha+\beta\right)=\tan\left(\pi-\left(\alpha+\beta\right)\right)$，所以$\alpha+\beta\in\left(\frac{\pi}{2},\frac{3\pi}{4}\right)$。因为$\alpha$，$\beta$都是锐角，所以$\alpha+\beta\in\left(0,\pi\right)$，所以$\alpha+\beta\in\left(\frac{\pi}{2},\frac{3\pi}{4}\right)$。又因为$\left(\tan\alpha-1\right)\left(\tan\beta-1\right)=2$，所以$\tan\alpha-1=\frac{2}{\tan\beta-1}$，即$\tan\alpha=\frac{2+\tan\beta}{\tan\beta-1}$。因此，$\tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac{2+\tan\beta}{\tan\beta-1}+\tan\beta}{1-\frac{2+\tan\beta}{\tan\beta-1}\cdot\tan\beta}=\frac{3\tan\beta-2}{\tan\beta+3}$，解得$\tan\beta=\frac{2}{3}$，$\tan\alpha=\frac{5}{2}$。因此，$\alpha+\beta=\arctan\left(\frac{5}{2}\right)+\arctan\frac{2}{3}=\arctan\frac{13}{4}=\frac{3\pi}{4}$。三、解答题6.已知$\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}=3$，$\tan\left(\alpha-\beta\right)=2$，则$\tan\left(\beta-2\alpha\right)$的值为多少？【解答】因为$\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}=3$，所以$\sin\alpha+\cos\alpha=3\sin\alpha-3\cos\alpha$，即$\sin\alpha+3\cos\alpha=0$，所以$\sin\alpha=-3\cos\alpha$。因为$\tan\left(\alpha-\beta\right)=2$，所以$\tan\alpha-\tan\beta=\frac{2}{1+\tan\alpha\tan\beta}$，即$\tan\alpha-2=\frac{2}{1+\tan\alpha\tan\beta}$，所以$\tan\alpha\tan\beta=-\frac{1}{2}$。因此，$\sin\beta=\frac{\sin\left(\alpha-\beta\right)}{\cos\alpha-\sin\alpha}=\frac{2}{5}$，$\cos\beta=\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}=-\frac{4}{5}$。因此，$\tan\left(\beta-2\alpha\right)=\tan\left[\left(\alpha-\left(\alpha-\beta\right)\right)-2\alpha\right]=\tan\left(\beta-\alpha\right)\cdot\tan\left(-2\alpha\right