tan(-2010°)+cos(-79π/6)

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/23 22:45:04
tan(-2010°)+cos(-79π/6)tan(-2010°)+cos(-79π/6)tan(-2010°)+cos(-79π/6)【1】tan(-2010º)=-tan2010

tan(-2010°)+cos(-79π/6)
tan(-2010°)+cos(-79π/6)

tan(-2010°)+cos(-79π/6)
【1】
tan(-2010º)=-tan2010º=-tan[6×360º-150]=tan150º=tan(180º-30º)=-tan30º=-(√3)/3
【2】
cos(-79π/6)=cos(79π/6)=cos[12π+(7π/6)]=cos(7π/6)=cos[π+(π/6)]=-cos(π/6)=-(√3)/2
【3】
原式=-(√3)[(1/3)+(1/2)]=-(√3)×(5/6)=-(5√3)/6

tan(-2010°)+cos(-79π/6)
=-tan30°+(-cosπ/6)
=-√3/3-√3/2
=-5√3/6

原式=-tan2010°+cos(79π/6)
=-tan(1800°+210°)+cos[12π+(7π/6)]
=-tan210°+cos(7π/6)
=-tan(180°+30°)+cos(π+π/6)
=-tan30°-cos(π/6)
=-√3/3-√3/2
=-(5√3)/6

tan(-2010°)+cos(-79π/6)
=tan(-6x360°+150°)+cos(-6x2π-7π/6)
=-√3/3-√3/2=-5√3/6

tan(-2010°)+cos(-79π/6)
=tan(1980-2010)°+cos(14π-79π/6)
=-tan30°-cos30°
=-√3/3-√3/2
=-5√3/6

-2010°=-1080°-30°,所以tan(-2010°)=-tan30°=-根3/3
-79π/6=-13π-π/6,所以cos(-79π/6)=cosπ/6=根3/2
所以原式=根3/6

急看不清!!

tan(-2010°)+cos(-79π/6) sinπ+cos+(-π)+tanπ过程. 化简:sin²(α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)1.cos(27°+α)cos(33°-α)-sin(27°+α)sin(33°-α) 2.sin²(α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π) 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 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 化简:(1)sin(-α+180°)-tan(-α)+tan(-α+360°)/tan(α+180°)+cos(-α)+cos(α+180°)(2)sin^2(α+π)cos(-α+π)/tan(α+π)+tan(α+2π)cos^3(-α-π) tan(-150°)cos(-210°)cos(-420°)tan(-600°)/sin(-1050°) 计算 tan 30°+cos 30°/tan 45°-cos 60° 等于多少 cos^(50°+α)+cos^(40°-α)-tan(30°-α)*tan(60°+α) 计算 tan 30°+cos 30°/tan 45°-cos 60= Sinπ/3Tanπ/3+tanπ/6COSπ/6-Tanπ/4COSπ/2 求证cos(720°+α)(2/cosα+tanα)(1/cosα-2tanα)=2cosα-3tanα 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°) 若α∈(0,π/6) 比较tan(sinα),tan(cosα),tan(tanα)的大小 【sin330°*tan(-13π/3)】/【cos(-19π/6)*cos690°】 [sin330°-tan(-13π/3)]/[cos(-19π/6)-cos690°] 求值sin17/6π,cos(-43π/6),tan(-1560°) sin(-1920°)=?cos(-1560°)=?tan(-17π/6)=?