计算4sin(2π/11)+tan(3π/11)要解法和答案

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计算4sin(2π/11)+tan(3π/11)要解法和答案

计算4sin(2π/11)+tan(3π/11)要解法和答案
计算4sin(2π/11)+tan(3π/11)
要解法和答案

计算4sin(2π/11)+tan(3π/11)要解法和答案
4sin(2π/11)+tan(3π/11)=根号11
证明过程如图所示

计算4sin(2π/11)+tan(3π/11)要解法和答案 高一三角函数计算求 具体方法!(1)cos310*tan(-108)(2)sin(5π/4)*cos(4π/5)*tan(11π/6) 计算:√3sin(-20/3∏)-cos13/4π×tan(-37/4π) 计算(1)cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos(-210°)cos420°]/cot(-600°)sin(-1050°)cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπt 计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos(-210°)cos420°]/cot(-600°)sin(-1050°)cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπtan10°t 已知tan(π/4+α)=3,计算下列各式的值:(1)tanα (2)2sinαcosα+cos2α已知tan(π/4+α)=3,计算下列各式的值:(1)tanα (2)2sinαcosα+cos2α 高一必修4三角函数sin420*cos750+sin(-330)*cos(-660)tan675+tan765-tan(-330)+tan(-690)sin(25π/6)+cos(25π/3)+tan(-25π/4)2.已知sin(π+α)=-1/2 计算(1)sin(5π-α) (2)sin(π/2+α) (3)cos(α-3π/2) (4)tan(π/2-α) 计算sin(-α-5π)*cos(α-2分之π)-tan(α-2分之3π)*tan(2π-α) 2cos π/2-tan π/4+3/4tan的平方 π/6-sin π/6+cos的平方 π/6 +sin 3π/2 求计算过程及...2cos π/2-tan π/4+3/4tan的平方 π/6-sin π/6+cos的平方 π/6 +sin 3π/2 计算2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2同上 写出过程 ①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)②计算(1)tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5) (2)sin(-60°)+cos(225°)+tan135°(3)cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)(4)tan10°+tan170°+sin1866°-sin(-606°) sin(π+a)+tan π/4 已知sin(π+a)=-1/2,计算 (1) sin(5π-a) (2)sin(π+a)(3)cos(a-π/2) (4)tan(π/2-a) 计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)], 计算:sin25π/6+cos25π/4+tan(-25π/4)+sin(-26π/3) 计算sin(-π+π/3)+tan(3π/2-π/6)+sin15, 高中数学三角函数计算题计算:计算:(1)cosπ/3-tanπ/4+(3/4)tan平方(π/6)-sinπ/6+cos平方(π/6)(2)[(2tan平方30°-1)cos平方30°]/(2sin平方45°+1)题就是那么出的啊... 求计算sin²π/2+cos⁴3π/2-tan²π/3