1-cost等于多少sint
@郗骂3393：cost(1 - cost) - sintsint化简 -
益削18579646649…… cost(1-cost)-sint.sint =cost - [ (cost)^2 + (sint)^2 ] =cost -1
@郗骂3393：(1 - cost)2转化为sint - 作业帮
益削18579646649…… [答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).
@郗骂3393：求极限,t趋于0 lim t/根号下1 - cost 等于多少?? -
益削18579646649…… 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.
@郗骂3393：求极限,t趋于0 lim t/根号下1 - cost 等于多少? - 作业帮
益削18579646649…… [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....
@郗骂3393：lim(1 - cost)/sint t趋于0 -
益削18579646649…… 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
@郗骂3393：1/cost等于根号5,t为 -
益削18579646649…… 解:1/cost=√5 cost=1/√5 t=arccos(1/√5) t=63.4349488°
@郗骂3393：lim(1 - cost)=t^2/2求解答,详细. -
益削18579646649…… 因为1-cost=2sin^2 t/2 当t-->0时,因为sint-->t,故sin^2 t/2-->(t/2)^2=t^2/4.所以lim(1-cost)=lim2sin^2 t/2=lim2 *t^2/4=t^2/2
@郗骂3393：lim(1 - cost)=t^2/2求解答, - 作业帮
益削18579646649…… [答案] 因为1-cost=2sin^2 t/2 当t-->0时,因为sint-->t, 故sin^2 t/2-->(t/2)^2=t^2/4. 所以lim(1-cost)=lim2sin^2 t/2=lim2 *t^2/4=t^2/2
@郗骂3393：sint/(1 - cost)求导 -
益削18579646649…… f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)² =[cost(1-cost)-sintsint]/(1-cost)² =[cost-(cos²t+sin²t)]/(1-cost)² =(cost-1)/(1-cost)² =1/(cost-1)
@郗骂3393：为什么sint/(1 - cost)=cot(t/2) - 作业帮
益削18579646649…… [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)
益削18579646649…… cost(1-cost)-sint.sint =cost - [ (cost)^2 + (sint)^2 ] =cost -1
@郗骂3393：(1 - cost)2转化为sint - 作业帮
益削18579646649…… [答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).
@郗骂3393：求极限,t趋于0 lim t/根号下1 - cost 等于多少?? -
益削18579646649…… 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.
@郗骂3393：求极限,t趋于0 lim t/根号下1 - cost 等于多少? - 作业帮
益削18579646649…… [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....
@郗骂3393：lim(1 - cost)/sint t趋于0 -
益削18579646649…… 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
@郗骂3393：1/cost等于根号5,t为 -
益削18579646649…… 解:1/cost=√5 cost=1/√5 t=arccos(1/√5) t=63.4349488°
@郗骂3393：lim(1 - cost)=t^2/2求解答,详细. -
益削18579646649…… 因为1-cost=2sin^2 t/2 当t-->0时,因为sint-->t,故sin^2 t/2-->(t/2)^2=t^2/4.所以lim(1-cost)=lim2sin^2 t/2=lim2 *t^2/4=t^2/2
@郗骂3393：lim(1 - cost)=t^2/2求解答, - 作业帮
益削18579646649…… [答案] 因为1-cost=2sin^2 t/2 当t-->0时,因为sint-->t, 故sin^2 t/2-->(t/2)^2=t^2/4. 所以lim(1-cost)=lim2sin^2 t/2=lim2 *t^2/4=t^2/2
@郗骂3393：sint/(1 - cost)求导 -
益削18579646649…… f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)² =[cost(1-cost)-sintsint]/(1-cost)² =[cost-(cos²t+sin²t)]/(1-cost)² =(cost-1)/(1-cost)² =1/(cost-1)
@郗骂3393：为什么sint/(1 - cost)=cot(t/2) - 作业帮
益削18579646649…… [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)
