Question

计算：1.tan30°*cos60°+sin^245°

2.sin^245°+sin^260°+根号6cos30°·cos45°
3.sin60°-cos45°分之sin30°-根号（1-tan60°）^2-tan45°
4.（3分之2）^-1+（π-3.14）^0-2sin60°-根号12+|1-3倍根号3|
5.2sin30°-（2分之1）^-1·|1-tan60°|+根号3+1分之2
6.cos45°-cos60°分之sin30°-2sin45°+根号3tan30°

Answer 1

权威答案 觉得保证步骤和答案正确 望采纳

Answer 2

1. tan30°*cos60°+sin^245°

   =√3/3 ×1/2 +(√2/2)方

   =√6/6 +1/2

2. sin^245°+sin^260°+根号6cos30°·cos45°

   =(√2/2)^2+(√3/2)^2+√6×√3/2 ×√2/2

   =1/2 +3/4+3/2

   =11/4
   3.sin60°-cos45°分之sin30°-根号（1-tan60°）^2-tan45°

   =√3/2 -(√2/2)分之(1/2) -tan60°+1-1

   =√3/2 -√2/2 -√3

   =-√2/2-√3/2
   4.（3分之2）^-1+（π-3.14）^0-2sin60°-根号12+|1-3倍根号3|

   =2分之3 +1-2×√3/2-2√3 +3√3-1

   =2分之5 -√3-2√3+3√3-1

   =2分之3
   5.2sin30°-（2分之1）^-1·|1-tan60°|+根号3+1分之2

   =2×1/2 -2×(√3-1) +2(√3-1)/[(√3)^2-1^2]

   =1 -2√3 +2 -√3+1

   =4-3√3
   6.cos45°-cos60°分之sin30°-2sin45°+根号3tan30°                    

   =√2/2 -(1/2)分之(1/2) -2×√2/2 +√3×√3/3

   =（√2/2 -√2 ） -1+1

   =-√2/2
   

追答

第五题楼下是正确的。

Answer 3

1.tan30°*cos60°+sin^245°
= √3/3*1/2 +(√2/2)²
= √3/6+1/2
= (3+√3)/6

2.sin^245°+sin^260°+根号6cos30°·cos45°
= (√2/2)²+(√3/2)²+√6*√3/2*√2/2
=1/2+3/4+6/4
= 11/4

3.sin60°-cos45°分之sin30°-根号（1-tan60°）^2-tan45°
= √3/2 - (1/2)/(√2/2) - √(1-√3)² - 1
= √3/2-√2/2-√3+1-1
= -(√3+√2)/2

4.（3分之2）^-1+（π-3.14）^0-2sin60°-根号12+|1-3倍根号3|
= 3/2 + 1 - 2*√3/2 -2√3 + 3√3-1
= 3/2 + 1 - √3 -2√3 + 3√3-1
= 3/2

5.2sin30°-（2分之1）^-1·|1-tan60°|+根号3+1分之2
= 2*1/2 - 2 *|1-√3| + 2/(√3+1)
= 1 -2(√3-1) + 2(√3-1)/{(√3+1)(√3-1)}
= 1 -2(√3-1) + 2(√3-1)/(3-1)
= 1 - √3 + 1
= 2-√3

6.cos45°-cos60°分之sin30°-2sin45°+根号3tan30°
= √2/2 - (1/2)/(1/2) - 2*√2/2 + √3*√3/3
= √2/2 - 1 - √2 + 1
= -√2/2

Answer 4

1.tan30°*cos60°+sin^245°=三分之根号3*二分之一-（sin65）

=六分之根号3-sin65
同学你好，如果问题已解决，记得右上角采纳哦~~~您的采纳是对我的肯定~谢谢哦