經(jīng)典解三角題目及答案一、選擇題（每題5分，共20分）1.若\(\sinA=\frac{1}{2}\)，且\(A\)為銳角，則\(\cosA\)的值為：A.\(\frac{\sqrt{3}}{2}\)B.\(\frac{1}{2}\)C.\(\frac{\sqrt{2}}{2}\)D.\(\frac{1}{\sqrt{3}}\)答案：A2.已知\(\tanB=2\)，求\(\sinB\cosB\)的值：A.\(\frac{1}{5}\)B.\(\frac{2}{5}\)C.\(\frac{3}{5}\)D.\(\frac{4}{5}\)答案：B3.若\(\cosC=-\frac{1}{2}\)，且\(C\)為第二象限角，則\(\sinC\)的值為：A.\(\frac{\sqrt{3}}{2}\)B.\(-\frac{\sqrt{3}}{2}\)C.\(\frac{1}{2}\)D.\(-\frac{1}{2}\)答案：A4.已知\(\sinA=\frac{\sqrt{2}}{2}\)，\(\cosA=-\frac{\sqrt{2}}{2}\)，則\(A\)所在的象限是：A.第一象限B.第二象限C.第三象限D(zhuǎn).第四象限答案：C二、填空題（每題5分，共20分）5.若\(\tan\theta=3\)，則\(\sin\theta\cos\theta\)的值為\(\_\_\_\_\_\)。答案：\(\frac{3}{10}\)6.已知\(\sin\alpha=\frac{4}{5}\)，且\(\alpha\)為銳角，求\(\cos\alpha\)的值為\(\_\_\_\_\_\)。答案：\(\frac{3}{5}\)7.若\(\cos\beta=-\frac{\sqrt{3}}{2}\)，且\(\beta\)為第三象限角，則\(\sin\beta\)的值為\(\_\_\_\_\_\)。答案：\(-\frac{1}{2}\)8.已知\(\sin\gamma=\frac{1}{2}\)，\(\cos\gamma=-\frac{\sqrt{3}}{2}\)，則\(\tan\gamma\)的值為\(\_\_\_\_\_\)。答案：\(-\frac{\sqrt{3}}{3}\)三、計算題（每題10分，共20分）9.計算\(\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ\)的值。答案：\[\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ=\frac{1}{2}\cdot\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1\]10.已知\(\tan\alpha=2\)，求\(\sin\alpha\)和\(\cos\alpha\)的值。答案：\[\sin\alpha=\frac{2}{\sqrt{1+4}}=\frac{2}{\sqrt{5}},\quad\cos\alpha=\frac{1}{\sqrt{1+4}}=\frac{1}{\sqrt{5}}\]四、證明題（每題15分，共30分）11.證明：若\(\sinA=\cosB\)，則\(A+B=90^\circ\)。證明：\[\text{已知}\sinA=\cosB,\text{根據(jù)三角函數(shù)的定義，有}\sinA=\cos(90^\circ-A),\text{因此}B=90^\circ-A.\]\[\text{所以}A+B=A+(90^\circ-A)=90^\circ.\]12.證明：\(\sin^2\theta+\cos^2\theta=1\)。證明：\[\text{根據(jù)三角函數(shù)的基本恒等式，我們知道}\sin^2\theta+\cos^2\theta=1.\]\[\text{這是因為}\sin\theta\text{和}\cos\theta\text{分