1-cost+2
@廖玉3690：(1 - cost)2转化为sint - 作业帮
施崔13817005417…… [答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).
@廖玉3690：设t属于R,则复数Z=(1 - cost)+(2+sint)i的对应点所成的图形是? - 作业帮
施崔13817005417…… [答案] x=1-cost cost=1-x y=2+sint sint=y-2 两个式子平方相加 (1-x)^2+(y-2)^2=1 (x-1)^2+(y-2)^2=1 对应点所成的图形是以(1,2)为圆心,1为半径的圆
@廖玉3690：简单的三角变形cos2t如何变形成1 - cost^2 - 作业帮
施崔13817005417…… [答案] cos2t=cost^2-sint^2=cost^2-(1-cost^2)=2cost^2-1 cos2t应该等于2cost^2-1而不是1-cost^2
@廖玉3690：三角函数cos2t如何变形成1 - cost^2 -
施崔13817005417…… cos2t=cos(t+t)=costcost-sintsint=cos^2t-sin^2t=cos^2t-(1-cos^2t)=2cos^2t-1 (cos^2t=(cost)平方)
@廖玉3690：高数 求弧长 参数方程 x=a(t - sint) y=a(1 - cost) t[0,2π] 求过程 谢谢 -
施崔13817005417…… dx/dt=a(1-cost), dy/dt=asint 由公式: 弧长S=∫√[(dx/dt)^2+(dy/dt)^2] dt 积分从0到2π =∫√a^2[1-2cost+(cost)^2]+(asint)^2] dt =a∫√(2-2cost) dt =a∫2|sin(t/2)| dt =8πa
@廖玉3690：化简1+cos(t)+cos(2t)+...+cos(nt) -
施崔13817005417…… 解 法一 cost+cos2t+...+cosnt=1/2[(cost+cosnt)+(cos2t+cos(n-1)t)+...+(cosnt+cost)]=[cos(n+1)t/2][cos((n-1)t/2)+cos(((n-3)t/2)+...+cos((n-(2n-1))t/2)=[cos(n+1)t/2/sin(t/2)]*[sin(t/2)*cos((n-1)t/2)+sin(t/2)*cos(((n-3)t/2)+...+sin(t/2)*cos((n-(2n-1))t/2)=1/2[cos...
@廖玉3690：高数 定积分 求旋转面面积x=a(t - sint) y=a(1 - cost),绕x轴 -
施崔13817005417…… x = a (t-sint) ,y = a (1-cost),0<t<2π dx = a (1-cost)dt, dy = a sint, y' = dy/dx = (dy/dt) / (dx/dt) = sint / (1 - cost) 所求旋转面面积为: S = ∫(t=0 , t=2π) 2π y (1+y'^2)^(1/2) dx = ∫(t=0 , t=2π) 2π a (1 - cost) (1+(sint / (1 - cost))^2)^(1/2) a (1-cost) dt = 64/3 π a^2
@廖玉3690：定积分(1 - cost)^(2/5)dt.t的范围是从0到2派 - 作业帮
施崔13817005417…… [答案] ∫[0,2π](1-cost)^(5/2) dt =∫[0,2π][2sin(t/2)^2]^(5/2)dt =8√2∫[0,2π]sin(t/2)^5 d(t/2) =-8√2∫[0,2π]sin(t/2)^4d(cos(t/2)) =-8√2∫[0,2π](1-cos(t/2)^2)^2dcos(t/2) =-8√2∫[0,2π] [1-2cos(t/2)+cos(t/2)^4 ] dcos(t/2) =-8√2[(cosπ-(cosπ)^2+(1/5)(cosπ)^5 )-(cos0-cos0^2...
@廖玉3690：求极限,t趋于0 lim t/根号下1 - cost 等于多少? - 作业帮
施崔13817005417…… [答案] lim(t→0)t/√(1-cost)=lim(t→0)1/{[1/2]*(1-cost)^(-1/2)*sint}=lim(t→0)[2(1-cost)^1/2]/sint=lim(t→0)[(1-cost)^(-1/2)]/cost=lim(t→0)1/[cost(1-cost)^(1/2)].当t→0的时候,1-cost→0,所以说极限不存在....
@廖玉3690：∫(T - sinT)(1 - cosT)2 dT 怎么做? -
施崔13817005417…… ∫(T-sinT)(1-cosT)^2dT =∫T[1-2cost+cos^2(T) ]dT-∫sinT(1-cosT)^ dT =∫T[1-2cost+(1+cos2T)/2]dT-∫(1-cosT)^2 d(1-cosT) =3T/2-2∫TdsinT+1/4∫Tdsin2T-(1-cosT)^3/3 =3T/2-(1-cosT)^3/3-2TsinT+2∫sinTdT+Tsin2T/4-1/4∫sin2TdT =3T/2-(1-cosT)^3/3-2TsinT-2cosT+Tsin2T/4+1/8cos2T+C
施崔13817005417…… [答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).
@廖玉3690：设t属于R,则复数Z=(1 - cost)+(2+sint)i的对应点所成的图形是? - 作业帮
施崔13817005417…… [答案] x=1-cost cost=1-x y=2+sint sint=y-2 两个式子平方相加 (1-x)^2+(y-2)^2=1 (x-1)^2+(y-2)^2=1 对应点所成的图形是以(1,2)为圆心,1为半径的圆
@廖玉3690：简单的三角变形cos2t如何变形成1 - cost^2 - 作业帮
施崔13817005417…… [答案] cos2t=cost^2-sint^2=cost^2-(1-cost^2)=2cost^2-1 cos2t应该等于2cost^2-1而不是1-cost^2
@廖玉3690：三角函数cos2t如何变形成1 - cost^2 -
施崔13817005417…… cos2t=cos(t+t)=costcost-sintsint=cos^2t-sin^2t=cos^2t-(1-cos^2t)=2cos^2t-1 (cos^2t=(cost)平方)
@廖玉3690：高数 求弧长 参数方程 x=a(t - sint) y=a(1 - cost) t[0,2π] 求过程 谢谢 -
施崔13817005417…… dx/dt=a(1-cost), dy/dt=asint 由公式: 弧长S=∫√[(dx/dt)^2+(dy/dt)^2] dt 积分从0到2π =∫√a^2[1-2cost+(cost)^2]+(asint)^2] dt =a∫√(2-2cost) dt =a∫2|sin(t/2)| dt =8πa
@廖玉3690：化简1+cos(t)+cos(2t)+...+cos(nt) -
施崔13817005417…… 解 法一 cost+cos2t+...+cosnt=1/2[(cost+cosnt)+(cos2t+cos(n-1)t)+...+(cosnt+cost)]=[cos(n+1)t/2][cos((n-1)t/2)+cos(((n-3)t/2)+...+cos((n-(2n-1))t/2)=[cos(n+1)t/2/sin(t/2)]*[sin(t/2)*cos((n-1)t/2)+sin(t/2)*cos(((n-3)t/2)+...+sin(t/2)*cos((n-(2n-1))t/2)=1/2[cos...
@廖玉3690：高数 定积分 求旋转面面积x=a(t - sint) y=a(1 - cost),绕x轴 -
施崔13817005417…… x = a (t-sint) ,y = a (1-cost),0<t<2π dx = a (1-cost)dt, dy = a sint, y' = dy/dx = (dy/dt) / (dx/dt) = sint / (1 - cost) 所求旋转面面积为: S = ∫(t=0 , t=2π) 2π y (1+y'^2)^(1/2) dx = ∫(t=0 , t=2π) 2π a (1 - cost) (1+(sint / (1 - cost))^2)^(1/2) a (1-cost) dt = 64/3 π a^2
@廖玉3690：定积分(1 - cost)^(2/5)dt.t的范围是从0到2派 - 作业帮
施崔13817005417…… [答案] ∫[0,2π](1-cost)^(5/2) dt =∫[0,2π][2sin(t/2)^2]^(5/2)dt =8√2∫[0,2π]sin(t/2)^5 d(t/2) =-8√2∫[0,2π]sin(t/2)^4d(cos(t/2)) =-8√2∫[0,2π](1-cos(t/2)^2)^2dcos(t/2) =-8√2∫[0,2π] [1-2cos(t/2)+cos(t/2)^4 ] dcos(t/2) =-8√2[(cosπ-(cosπ)^2+(1/5)(cosπ)^5 )-(cos0-cos0^2...
@廖玉3690：求极限,t趋于0 lim t/根号下1 - cost 等于多少? - 作业帮
施崔13817005417…… [答案] lim(t→0)t/√(1-cost)=lim(t→0)1/{[1/2]*(1-cost)^(-1/2)*sint}=lim(t→0)[2(1-cost)^1/2]/sint=lim(t→0)[(1-cost)^(-1/2)]/cost=lim(t→0)1/[cost(1-cost)^(1/2)].当t→0的时候,1-cost→0,所以说极限不存在....
@廖玉3690：∫(T - sinT)(1 - cosT)2 dT 怎么做? -
施崔13817005417…… ∫(T-sinT)(1-cosT)^2dT =∫T[1-2cost+cos^2(T) ]dT-∫sinT(1-cosT)^ dT =∫T[1-2cost+(1+cos2T)/2]dT-∫(1-cosT)^2 d(1-cosT) =3T/2-2∫TdsinT+1/4∫Tdsin2T-(1-cosT)^3/3 =3T/2-(1-cosT)^3/3-2TsinT+2∫sinTdT+Tsin2T/4-1/4∫sin2TdT =3T/2-(1-cosT)^3/3-2TsinT-2cosT+Tsin2T/4+1/8cos2T+C
