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1. $\sin 30^{\circ}· \cos 45^{\circ}-\tan 60^{\circ}+3\tan 30^{\circ}$.
答案：
1. 解:原式$=\frac {1}{2}×\frac {\sqrt {2}}{2}-\sqrt {3}+3×\frac {\sqrt {3}}{3}=\frac {\sqrt {2}}{4}-\sqrt {3}+\sqrt {3}=\frac {\sqrt {2}}{4}$.
2. $4\sin ^{2}30^{\circ}+\tan 60^{\circ}-2\cos 30^{\circ}$.
答案：
2. 解:原式$=4×(\frac {1}{2})^{2}+\sqrt {3}-2×\frac {\sqrt {3}}{2}=1+\sqrt {3}-\sqrt {3}=1$.
3. $2\sin 60^{\circ}· \tan 30^{\circ}+\cos ^{2}30^{\circ}-\tan 45^{\circ}$.
答案：
3. 解:原式$=2×\frac {\sqrt {3}}{2}×\frac {\sqrt {3}}{3}+(\frac {\sqrt {3}}{2})^{2}-1=1+\frac {3}{4}-1=\frac {3}{4}$.
4. $\tan 45^{\circ}· \sin 45^{\circ}+\cos ^{2}30^{\circ}$.
答案：
4. 解:原式$=1×\frac {\sqrt {2}}{2}+(\frac {\sqrt {3}}{2})^{2}=\frac {\sqrt {2}}{2}+\frac {3}{4}=\frac {2\sqrt {2}+3}{4}$.
5. $4\cos 30^{\circ}+\tan ^{2}45^{\circ}-2\tan 60^{\circ}$.
答案：
5. 解:原式$=4×\frac {\sqrt {3}}{2}+1^{2}-2×\sqrt {3}=2\sqrt {3}+1-2\sqrt {3}=1$.
6. $\tan 30^{\circ}\sin 60^{\circ}-\cos 45^{\circ}\sin 45^{\circ}$.
答案：
6. 解:原式$=\frac {\sqrt {3}}{3}×\frac {\sqrt {3}}{2}-\frac {\sqrt {2}}{2}×\frac {\sqrt {2}}{2}=\frac {1}{2}-\frac {1}{2}=0$.
7. $8\sin ^{2}60^{\circ}+\tan 45^{\circ}-4\cos 30^{\circ}$.
答案：
7. 解:原式$=8×(\frac {\sqrt {3}}{2})^{2}+1-4×\frac {\sqrt {3}}{2}=8×\frac {3}{4}+1-2\sqrt {3}=6+1-2\sqrt {3}=7-2\sqrt {3}$.
8. $2\cos ^{2}45^{\circ}+\tan 60^{\circ}· \tan 30^{\circ}-\cos 60^{\circ}$.
答案：
8. 解:原式$=2×(\frac {\sqrt {2}}{2})^{2}+\sqrt {3}×\frac {\sqrt {3}}{3}-\frac {1}{2}=1+1-\frac {1}{2}=\frac {3}{2}$.
9. (2024·惠山区期末)$\frac{\cos 45^{\circ}}{\tan ^{2}30^{\circ}}+\sin 45^{\circ}$.
答案：
9. 解:原式$=\frac {\frac {\sqrt {2}}{2}}{(\frac {\sqrt {3}}{3})^{2}}+\frac {\sqrt {2}}{2}=\frac {\frac {\sqrt {2}}{2}}{\frac {1}{3}}+\frac {\sqrt {2}}{2}=\frac {3\sqrt {2}}{2}+\frac {\sqrt {2}}{2}=2\sqrt {2}$.
10. (2024·溧阳期末)$\frac{\sin 60^{\circ}-\sin 30^{\circ}}{2\cos 45^{\circ}}$.
答案：
10. 解:原式$=\frac {\frac {\sqrt {3}}{2}-\frac {1}{2}}{2×\frac {\sqrt {2}}{2}}=\frac {\sqrt {3}-1}{2\sqrt {2}}=\frac {\sqrt {6}-\sqrt {2}}{4}$.
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