(1)cos60°-sin^2 45°+sin^2 45°+1/4tan^2 30°+cos30°-sin30°(2)sin30°/(sin60°-cos45°)-(根号下(1-tan60°)^2)-tan45°第一题是这个(1)cos60°-sin^2 45°+1/4tan^2 30°+cos30°-sin30°
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![(1)cos60°-sin^2 45°+sin^2 45°+1/4tan^2 30°+cos30°-sin30°(2)sin30°/(sin60°-cos45°)-(根号下(1-tan60°)^2)-tan45°第一题是这个(1)cos60°-sin^2 45°+1/4tan^2 30°+cos30°-sin30°](/uploads/image/z/2513228-68-8.jpg?t=%EF%BC%881%EF%BC%89cos60%C2%B0-sin%5E2+45%C2%B0%2Bsin%5E2+45%C2%B0%2B1%2F4tan%5E2+30%C2%B0%2Bcos30%C2%B0-sin30%C2%B0%EF%BC%882%EF%BC%89sin30%C2%B0%2F%EF%BC%88sin60%C2%B0-cos45%C2%B0%EF%BC%89-%EF%BC%88%E6%A0%B9%E5%8F%B7%E4%B8%8B%EF%BC%881-tan60%C2%B0%EF%BC%89%5E2%EF%BC%89-tan45%C2%B0%E7%AC%AC%E4%B8%80%E9%A2%98%E6%98%AF%E8%BF%99%E4%B8%AA%EF%BC%881%EF%BC%89cos60%C2%B0-sin%5E2+45%C2%B0%2B1%2F4tan%5E2+30%C2%B0%2Bcos30%C2%B0-sin30%C2%B0)
(1)cos60°-sin^2 45°+sin^2 45°+1/4tan^2 30°+cos30°-sin30°(2)sin30°/(sin60°-cos45°)-(根号下(1-tan60°)^2)-tan45°第一题是这个(1)cos60°-sin^2 45°+1/4tan^2 30°+cos30°-sin30°
(1)cos60°-sin^2 45°+sin^2 45°+1/4tan^2 30°+cos30°-sin30°
(2)sin30°/(sin60°-cos45°)-(根号下(1-tan60°)^2)-tan45°
第一题是这个(1)cos60°-sin^2 45°+1/4tan^2 30°+cos30°-sin30°
(1)cos60°-sin^2 45°+sin^2 45°+1/4tan^2 30°+cos30°-sin30°(2)sin30°/(sin60°-cos45°)-(根号下(1-tan60°)^2)-tan45°第一题是这个(1)cos60°-sin^2 45°+1/4tan^2 30°+cos30°-sin30°
是要求结果吗?
那就直接将各三角函数对应的值带入就可得出结果了,还要我算吗?
(1)1/2-(根号下2/2)^2+(1/4)×(根号下3/3)^2+根号下3/2-1/2=(3×根号下3-4)/6
sin^2 45°-cos60°+tan60°*cos30°
3|tan30°-1|+sin平方60°-(cos60°-1)负2次方*tan45°+tan60°=?
(1)cos60°-sin^2 45°+sin^2 45°+1/4tan^2 30°+cos30°-sin30°(2)sin30°/(sin60°-cos45°)-(根号下(1-tan60°)^2)-tan45°第一题是这个(1)cos60°-sin^2 45°+1/4tan^2 30°+cos30°-sin30°
sin²30°+根号2cos45°*cos60°-tan²45°...
1/4tan²45°+1/sin²30°-3cos²30°+tan45°/cos60°-sin60°/cos60°
sin²60°-tan45°×cos60°-tan²60° 根号(cos60°-1)²+sin60°-1的绝对值急
sin²45°-cos60°- tan45°分之cos30° +2sin²60°乘tan60°
1)cos60°/sin60°-sin45°2)(1-tan60°)的绝对值+2sin²45°3)(cot60°-sin60°)/sin²45°×tan30°4)[根号(tan30°-cos30°)² ]+(cos60°-cot60°)的绝对值
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
cos²45°-1/cos60°+1/tan45°+cos²30°+sin²45°
cos60°÷si30°-tan45°+sin²45°
2*sin(60°-B)*sinB=cos(60°-2B)-cos60°
sin平方45°-tan平方30°另外还有一道 2sin平方30°×tan30°+cos60°tan60°
为什么sin(60°+18°)-sin(60°-18°)= 2cos60°sin18°
计算:2(cos^2)45°—cos60°
2分之1cos60°-根号2sin45°
sin10°sin50°为什么等于1/2(cos40-cos60)
计算:cos60°+2/根2-根8-1/2