1-cost分之sint
@吕瞿717：sint/(1 - cost)求导 - 作业帮
平疫19637668624…… [答案] f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)? =[cost(1-cost)-sintsint]/(1-cost)? =[cost-(cos?t+sin?t)]/(1-cost)? =(cost-1)/(1-cost)? =1/(cost-1)
@吕瞿717：cost(1 - cost) - sintsint化简 -
平疫19637668624…… cost(1-cost)-sint.sint =cost - [ (cost)^2 + (sint)^2 ] =cost -1
@吕瞿717：(1 - cost)2转化为sint - 作业帮
平疫19637668624…… [答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).
@吕瞿717：sint/(1 - cost)求导 -
平疫19637668624…… f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)
@吕瞿717：lim(1 - cost)/sint t趋于0 -
平疫19637668624…… 记住在t 趋于0的时候,1-cost等价于 0.5t^2 所以这里的 极限值等价于 0.5t^2 /sint,而显然此时 t/sint 趋于1,所以得到 原极限=lim(t趋于0) t/sint *0.5t =0
@吕瞿717：这个极限为什么不存在?lim(t→0)t/√(1 - cost) -
平疫19637668624…… 因为t趋于0-和0+时 极限分别是-√2和√2 x趋于0 则1-cosx~1/2x^2 所以这里应该等于t/√(1/2t²)=√2t/|t| 左右极限不相等 所以极限不存在
@吕瞿717：为什么sint/(1 - cost)=cot(t/2) - 作业帮
平疫19637668624…… [答案] sint = 2sin(t/2)cos(t/2) 1-cost = 1-(1-2sin²(t/2))=2sin²(t/2) 所以 sint/(1-cost)=2sin(t/2)cos(t/2)/2sin²(t/2)=cos(t/2)/sin(t/2)=cot(t/2)
@吕瞿717：对(1—cost)的二分之五次方进行一阶求导 - 作业帮
平疫19637668624…… [答案] [(1-cost)^(5/2)]′ =(5/2)(1-cost)^(3/2)(1-cost)′ =(5/2)sint(1-cost)^(3/2)
@吕瞿717：1/(cost+sint)的积分怎么算?
平疫19637668624…… 设x=tan(t/2),则sint=2sin(t/2)cos(t/2)=2tan(t/2)/sec²(t/2)sint=2x/(1+x²)cost=2cos²(t/2)-1=2/sec²(t/2)-1cost=(1-x²)/(1+x²)t=2arctanxdt=2dx/(1+x²)原式=∫2dx/(1-x²+2x)已经有理化了.
@吕瞿717：Sint/(1 - cost)是如何得于Cot(t/2)的?中间有个小错误是1 - cos2a=[1 - (1 - 2Sin^2a)]. - 作业帮
平疫19637668624…… [答案] 令t=2a a=t/2 则原式=sin2a/(1-cos2a) =2sinacosa/[1-(1-2sin²a)] =2sinacosa/2sin²a =cosa/sina =cota =cot(t/2)
平疫19637668624…… [答案] f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)? =[cost(1-cost)-sintsint]/(1-cost)? =[cost-(cos?t+sin?t)]/(1-cost)? =(cost-1)/(1-cost)? =1/(cost-1)
@吕瞿717：cost(1 - cost) - sintsint化简 -
平疫19637668624…… cost(1-cost)-sint.sint =cost - [ (cost)^2 + (sint)^2 ] =cost -1
@吕瞿717：(1 - cost)2转化为sint - 作业帮
平疫19637668624…… [答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).
@吕瞿717：sint/(1 - cost)求导 -
平疫19637668624…… f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)
@吕瞿717：lim(1 - cost)/sint t趋于0 -
平疫19637668624…… 记住在t 趋于0的时候,1-cost等价于 0.5t^2 所以这里的 极限值等价于 0.5t^2 /sint,而显然此时 t/sint 趋于1,所以得到 原极限=lim(t趋于0) t/sint *0.5t =0
@吕瞿717：这个极限为什么不存在?lim(t→0)t/√(1 - cost) -
平疫19637668624…… 因为t趋于0-和0+时 极限分别是-√2和√2 x趋于0 则1-cosx~1/2x^2 所以这里应该等于t/√(1/2t²)=√2t/|t| 左右极限不相等 所以极限不存在
@吕瞿717：为什么sint/(1 - cost)=cot(t/2) - 作业帮
平疫19637668624…… [答案] sint = 2sin(t/2)cos(t/2) 1-cost = 1-(1-2sin²(t/2))=2sin²(t/2) 所以 sint/(1-cost)=2sin(t/2)cos(t/2)/2sin²(t/2)=cos(t/2)/sin(t/2)=cot(t/2)
@吕瞿717：对(1—cost)的二分之五次方进行一阶求导 - 作业帮
平疫19637668624…… [答案] [(1-cost)^(5/2)]′ =(5/2)(1-cost)^(3/2)(1-cost)′ =(5/2)sint(1-cost)^(3/2)
@吕瞿717：1/(cost+sint)的积分怎么算?
平疫19637668624…… 设x=tan(t/2),则sint=2sin(t/2)cos(t/2)=2tan(t/2)/sec²(t/2)sint=2x/(1+x²)cost=2cos²(t/2)-1=2/sec²(t/2)-1cost=(1-x²)/(1+x²)t=2arctanxdt=2dx/(1+x²)原式=∫2dx/(1-x²+2x)已经有理化了.
@吕瞿717：Sint/(1 - cost)是如何得于Cot(t/2)的?中间有个小错误是1 - cos2a=[1 - (1 - 2Sin^2a)]. - 作业帮
平疫19637668624…… [答案] 令t=2a a=t/2 则原式=sin2a/(1-cos2a) =2sinacosa/[1-(1-2sin²a)] =2sinacosa/2sin²a =cosa/sina =cota =cot(t/2)
