求曲线r=sinΘ和r=根号3*cosΘ所围成的图形的面积
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![求曲线r=sinΘ和r=根号3*cosΘ所围成的图形的面积](/uploads/image/f/5743922-50-2.jpg?t=%E6%B1%82%E6%9B%B2%E7%BA%BFr%3Dsin%CE%98%E5%92%8Cr%3D%E6%A0%B9%E5%8F%B73%2Acos%CE%98%E6%89%80%E5%9B%B4%E6%88%90%E7%9A%84%E5%9B%BE%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF)
2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB有正弦定理2RsinA=a,2RsinC=c所以(a+c)(sinA-sinC)=(√3a-b)sinBsinA=a/2R,si
(R)={,,,,},s(R)={,,,,},t(R)={,,,,}
正弦定理这一定理对于任意三角形ABC,都有a/sinA=b/sinB=c/sinC=2RR为三角形外接圆半径a/sinA=b/sinB=c/sinC=2R带入原式=2R[(a/2R)^2-(c-R)^
就是说R=丨r×h丨
很久很久以前的教材曾经规定ρ可以取负数,极坐标点(-2,1)表示在射线θ=1反向延长线上距离原点2的那个点,弄得很复杂.现在规定ρ取非负实数,就简单多了,即现在的极坐标点的第1个坐标是不可以为负数的,
f(x)=2√3sinxcosx+2sin^2x-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=π,单调递增区间:2kπ-π/2
m(2rcosθ,2rsinθ)圆心轨迹是以原点为圆心,2r为半径的圆内切的定圆就是以原点为圆心,3r为半径的圆外切的定圆就是以原点为圆心,r为半径的圆
y=sin^2+cos^2+cos^2+根号3sinxcosx=1+(1+cos2x)/2+根号3/2sin2x=3/2+sin(2x+π/6)T=2π/w=π递增-π/2+2kπ≤(2x+π/6)≤
1=sinθr2=3cosθ两曲线在θ(0,π/2),交点θt=tg-1(1/3),r=1/10^0.5S=S1+S2S1=∫1/2*r1^2*dθθ(0,θt)S2=∫1/2*r2^2*dθθ(θt
所给极坐标方程已经是最简表达形式;两边同乘以ρ,何以看出不是两条相交直线?再问:我知道了、我给它混到圆的方程里了。三克油~
因为f(x)=根号3sin(2x-π/6)+2sin的平方(x-π/12)=根号3sin(2x-π/6)-(1-2sin的平方(x-π/12))+1=根号3sin(2x-π/6)-cos(2x-π/6
f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin^2x和差角公式展开f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x=2sinx*cosx+根号3cosx
2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB有正弦定理2RsinA=a,2RsinC=c所以(a+c)(sinA-sinC)=(√3a-b)sinBsinA=a/2R,si
(R)={,,,,,};s(R)={,,,,,};t(R)={,,,,,,,,}
(1)f(x)=0.5(1-cos2x)+√3sin2x=0.5√3sin2x-0.5cos2x+1/2=sin(2x-π/6)+1/2∴T=2π/2=π∵-π/2+2kπ
郭敦顒回答:r=a(sin(θ/3))^3的全长,这题应是求定积分,积分区间为[0,6π],即积分下限为0,上限为6π,于是,r=∫a(sin(θ/3))^3dθ,积分下限为0,上限为6π,换元,令α
解析:由正弦定理a/sinA=b/sinB=c/sinC=2R∴a^2-c^2=(√3a-b)b=√3ab-b^2即(a^2+b^2-c^2)/2ab=√3/2∴cosC=√3/2∠C=30°∵a^2
2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB有正弦定理2RsinA=a,2RsinC=c所以(a+c)(sinA-sinC)=(√3a-b)sinBsinA=a/2R,si
f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x
f(x)=sin²x+√3sinxcosx+2cos²x=(1-cos2x)/2+(√3/2)sin2x+(1+cos2x)=3/2+(√3/2)sin2x+(1/2)cos2x=