1+sin+x
@雕霞3511：计算不定积分:∫ 1+sinx1+sinx+cosxdx. - 作业帮
蔺儿18432291634…… [答案] ∫ 1+sinx1+sinx+cosxdx=∫12(1+sinx+cosx)+12(sinx−cosx)+121+sinx+cosx=12∫dx−12∫cosx−sinx1+sinx+cosxdx+12∫11+sinx+cosxdx=12x−12∫d(1+sinx+cosx)1+sinx+cosx+12∫12sinx2cosx2+2cos2x2dx=12...
@雕霞3511：判断函数f(x)=根号下1+(sinx)^2+sinx - 1/根号下1+(sinx)^2+sinx+1奇偶性 -
蔺儿18432291634…… (1)f(x)定义域为(-∞,+∞)(2)f(0)=0(3)f(x)=根号下1+(sinx)^2+sinx-1/根号下1+(sinx)^2+sinx+1 f(-x)=根号下1+(sin(-x))^2+sin(-x)-1/根号下1+(sin(-x))^2+sin(-x)+1 =根号下1+(sinx)^2-sinx-1/根号下1+(sinx)^2-sinx+1 f(x)+f(-x) =根号下1+(sinx)^2+sinx-1/根号下...
@雕霞3511：(1+x)sin(1+x)的原函数 -
蔺儿18432291634…… 令1+x=t F(x)=∫(1+x)sin(1+x)=∫tsint=-tcost+∫cost+c=-tcost+sint+c=sin(1+x)-(1+x)cos(1+x)+c 满意请采纳哦
@雕霞3511：1+sin(x+y)=e - xy导数,要求有详解
蔺儿18432291634…… 两边同事求导 对1+sin(x+y)=e^(xy)求导得 cos(x+y)(1+y')=e^(xy)*(y+xy'), ∴cos(x+y)-ye^(xy)=[xe^(xy)-cos(x+y)]y', ∴y'=[cos(x+y)-ye^(xy)]/[xe^(xy)-cos(x+y)].
@雕霞3511：对任意的x1,x2属于(0,π/2) x1<x2, y1=1+sinx1/x1, y2=1+sinx2/x2 比大小
蔺儿18432291634…… 令y=1+sinx/x y'=(1+sinx/x)' =(xcosx-sinx)/x^2 =cosx(x-tanx)/x^2 因为当x属于(0,π/2)时,x<tanx, 所以 y'<0 即函数y=1+sinx/x是减函数,所以 对任意的x1,x2属于(0,π/2) x1<x2, 有 y1=1+sinx1/x1>y2=1+sinx2/x2 即 1+sinx1/x1>1+sinx2/x2
@雕霞3511：化简:(1+sin x - cos x)/(1+sin x+cos x) 谢 - 作业帮
蔺儿18432291634…… [答案] :(1+sin x-cos x)/(1+sin x+cos x) =[1+2sin(x/2)cos(x/2)-(1-2sin(x/2)^2]/[1+2sin(x/2)cos(x/2)+2cos(x/2)^2-1] =[2sin(x/2)cos(x/2)+2sin(x/2)^2]/[2sin(x/2)cos(x/2)+2cos(x/2)^2] =2sin(x/2)[cos(x/2)+sin(x/2)]/{2cosx/2*[sin(x/2)+cos(x/2)]} =sin(x/2)/cos(x/2) =tanx/2 ...
@雕霞3511：已知1+sinx÷cosx=﹣2分之1求cosx÷sinx - 1的值 求详细步骤 -
蔺儿18432291634…… sin²+cos²x=1 cos²x=1-sin²x cos²x=(1+sinx)(1-sinx) cos²x=-(1+sinx)(sinx-1) 所以cosx÷(sinx-1)=-(1+sinx)÷cos=1/2
@雕霞3511：limsinx*根号1+sin1/x -
蔺儿18432291634…… 因为sinx是无穷小,而 √1+sin1/x是有界函数,所以 原式=0
@雕霞3511：已知函数F(X)=1+sin(派/2+X) - 根号3sinx 求函数F(X)的最小正周期和值域? 解题步骤!
蔺儿18432291634…… 你好! sin(x+π/2)=cosx, ∴f(x)=1+cosx-√3sinx=1-(√3sinx-cosx)=1-2sin(x-π/6), ∴f(x)的最小正周期T=2π/1=2π; ∵2sin(x-π/6)∈[-2,2]; ∴-2sin(x-π/6)∈[-2,2]; ∴1-2sin(x-π/6)∈[-1,3]. 即函数的值域是[-1,3].
@雕霞3511：若(1+sinx)/cosx= - 1/2,则cos/sinx - 1= -
蔺儿18432291634…… (1+sinx)/cosx=-1/2 (1+sinx)cosx/(cosx*cosx)=-1/2 (1+sinx)cosx/(cosx)^2=-1/2 (1+sinx)cosx/[1-(sinx)^2]=-1/2 (1+sinx)cosx/[(1-sinx)(1+sinx)]=-1/2 cosx/(1-sinx)=-1/2 cosx/(sinx-1)=1/2
蔺儿18432291634…… [答案] ∫ 1+sinx1+sinx+cosxdx=∫12(1+sinx+cosx)+12(sinx−cosx)+121+sinx+cosx=12∫dx−12∫cosx−sinx1+sinx+cosxdx+12∫11+sinx+cosxdx=12x−12∫d(1+sinx+cosx)1+sinx+cosx+12∫12sinx2cosx2+2cos2x2dx=12...
@雕霞3511：判断函数f(x)=根号下1+(sinx)^2+sinx - 1/根号下1+(sinx)^2+sinx+1奇偶性 -
蔺儿18432291634…… (1)f(x)定义域为(-∞,+∞)(2)f(0)=0(3)f(x)=根号下1+(sinx)^2+sinx-1/根号下1+(sinx)^2+sinx+1 f(-x)=根号下1+(sin(-x))^2+sin(-x)-1/根号下1+(sin(-x))^2+sin(-x)+1 =根号下1+(sinx)^2-sinx-1/根号下1+(sinx)^2-sinx+1 f(x)+f(-x) =根号下1+(sinx)^2+sinx-1/根号下...
@雕霞3511：(1+x)sin(1+x)的原函数 -
蔺儿18432291634…… 令1+x=t F(x)=∫(1+x)sin(1+x)=∫tsint=-tcost+∫cost+c=-tcost+sint+c=sin(1+x)-(1+x)cos(1+x)+c 满意请采纳哦
@雕霞3511：1+sin(x+y)=e - xy导数,要求有详解
蔺儿18432291634…… 两边同事求导 对1+sin(x+y)=e^(xy)求导得 cos(x+y)(1+y')=e^(xy)*(y+xy'), ∴cos(x+y)-ye^(xy)=[xe^(xy)-cos(x+y)]y', ∴y'=[cos(x+y)-ye^(xy)]/[xe^(xy)-cos(x+y)].
@雕霞3511：对任意的x1,x2属于(0,π/2) x1<x2, y1=1+sinx1/x1, y2=1+sinx2/x2 比大小
蔺儿18432291634…… 令y=1+sinx/x y'=(1+sinx/x)' =(xcosx-sinx)/x^2 =cosx(x-tanx)/x^2 因为当x属于(0,π/2)时,x<tanx, 所以 y'<0 即函数y=1+sinx/x是减函数,所以 对任意的x1,x2属于(0,π/2) x1<x2, 有 y1=1+sinx1/x1>y2=1+sinx2/x2 即 1+sinx1/x1>1+sinx2/x2
@雕霞3511：化简:(1+sin x - cos x)/(1+sin x+cos x) 谢 - 作业帮
蔺儿18432291634…… [答案] :(1+sin x-cos x)/(1+sin x+cos x) =[1+2sin(x/2)cos(x/2)-(1-2sin(x/2)^2]/[1+2sin(x/2)cos(x/2)+2cos(x/2)^2-1] =[2sin(x/2)cos(x/2)+2sin(x/2)^2]/[2sin(x/2)cos(x/2)+2cos(x/2)^2] =2sin(x/2)[cos(x/2)+sin(x/2)]/{2cosx/2*[sin(x/2)+cos(x/2)]} =sin(x/2)/cos(x/2) =tanx/2 ...
@雕霞3511：已知1+sinx÷cosx=﹣2分之1求cosx÷sinx - 1的值 求详细步骤 -
蔺儿18432291634…… sin²+cos²x=1 cos²x=1-sin²x cos²x=(1+sinx)(1-sinx) cos²x=-(1+sinx)(sinx-1) 所以cosx÷(sinx-1)=-(1+sinx)÷cos=1/2
@雕霞3511：limsinx*根号1+sin1/x -
蔺儿18432291634…… 因为sinx是无穷小,而 √1+sin1/x是有界函数,所以 原式=0
@雕霞3511：已知函数F(X)=1+sin(派/2+X) - 根号3sinx 求函数F(X)的最小正周期和值域? 解题步骤!
蔺儿18432291634…… 你好! sin(x+π/2)=cosx, ∴f(x)=1+cosx-√3sinx=1-(√3sinx-cosx)=1-2sin(x-π/6), ∴f(x)的最小正周期T=2π/1=2π; ∵2sin(x-π/6)∈[-2,2]; ∴-2sin(x-π/6)∈[-2,2]; ∴1-2sin(x-π/6)∈[-1,3]. 即函数的值域是[-1,3].
@雕霞3511：若(1+sinx)/cosx= - 1/2,则cos/sinx - 1= -
蔺儿18432291634…… (1+sinx)/cosx=-1/2 (1+sinx)cosx/(cosx*cosx)=-1/2 (1+sinx)cosx/(cosx)^2=-1/2 (1+sinx)cosx/[1-(sinx)^2]=-1/2 (1+sinx)cosx/[(1-sinx)(1+sinx)]=-1/2 cosx/(1-sinx)=-1/2 cosx/(sinx-1)=1/2
