sin+2x与sinx+2的区别
甘先婷3026证明成立:[cos(3x) - sin(3x)]/(cosx+sinx)=1 - 2sin(2x). -
水钞胜13199303784 ______ cos(3x)-sin(3x)= cos(2x+x)-sin(2x+x)= cos2xcosx-sin2xsinx-sin2xcosx-cos2xsinx= (cos^2x-sin^2x)cosx-2sinxcosxsinx-2sinxcosxcosx-(cos^2x-sin^2x)sinx= cos^3x-sin^2xcosx-2sin^2xcosx-2sinxcos^2x-cos^2xsinx+sin^3x= cos^3x+sin^3x-3cos^2...
甘先婷3026当x趋近于零时,无穷小量sin(2sinx+x^2)为什么与x不等价? -
水钞胜13199303784 ______[答案] 可以利用洛必达法则 lim【x→0】[sin(2sinx+x²)]/x =lim【x→0】cos(2sinx+x²)·(2sinx+x²) ' =lim【x→0】cos(2sinx+x²)·(2cosx+2x) =cos(2*0+0)·(2cos0+0) =cos0*2cos0 =1*2 =2≠1 所以当x趋近于零时,无穷小量sin(2sinx+x^2)与x不等价
甘先婷3026已知sin(X+π/4)=1/3,X∈(π/2,π),求sinX -
水钞胜13199303784 ______ sin(X+π/4)=1/3,X∈(π/2,π),说明X+π/4∈(π/2,π),sinx>0,cosxsin(X+π/4)=1/3=sinxcos(π/4)+cosxsin(π/4)=sinx+cosx 又sin^2x+cos^2x=1 sin^2x+(1/3-sinx)^2=12sin^2x-2/3sinx+1/9=118sin^2x-6sinx-8=09sin^2x-3sinx-4=0 sinx=(3±√153)/18 所以sinx=(3+√153)/18>0
甘先婷3026(√(1+tanx) - √(1+sinx))/(x√(1+sin^2x) - x)x趋近于0,求极限,刚才说错了,sin^2x就是(sinx)的平方 -
水钞胜13199303784 ______[答案] (√(1+tanx)-√(1+sinx))/(x√(1+sin^2x)-x) 分子分母同时有理化 =(tanx-sinx)(x√(1+sin^2x)+x)/[(x²*sin^2x)(√(1+tanx)+√(1+sinx))] =(tanx-sinx)(√(1+sin^2x)+1)/[(xsin^2x)(√(1+tanx)+√(1+sinx))] 代入lim√(1+sin^2x)+1=2,lim√(1+tanx)+√(1+...
甘先婷3026求下列函数的最大值、周期、单调区间 -
水钞胜13199303784 ______ (1)y=2sin^2x+2sinxcosx=1-cos2x+sin2x=sin(2x-pi/4)+1最大值2,周期pi2kpi-pi/2<=2x-pi/4<=2kpi+pi/2解得kpi-pi/8<=x<=kpi+3pi/8单调递增...
甘先婷3026已知sinX+cosX=1/3,0<X<π,求sin 2X 与cos 2X值
水钞胜13199303784 ______ sinX+cosX=1/3sin^2x+cos^2x=1(sinx+cosx)^2=1+2sinxcosx=1/92sinx*cosx=sin2x=-8/9<0TT<2X<2TT,TT/2<X<TT,(sinx>0)sinx>cosx(sinx+cosx>0)TT/2<x<3TT/4TT<2x<TT*3/2COS2X<0COS2X=-17^0.5/9
甘先婷3026f(x)=(sinx+sin3x)/(cosx+cos3x)怎么化简为2sin2xcosx/2cos2xcosx
水钞胜13199303784 ______ sin2x=sinxcosx+cosxsinx=2sinxcosx , cos2x=2(cosx)^2-1 =1-2(sinx)^2 1) sin3x=sin(2x+x)=sin2xcosx+cos2xsinx=(sinxcosx+cosxsinx)cosx+cos2xsinx=2sinx(cosx)^2+sinxcos2x sinx+sin3x=sinx[1+2(cosx)^2+cos2x]=sinx[4(cosx)^2]=2*2sinxcosx*...
甘先婷3026x与y均为锐角,且sin(x+y)=2sinx,则x与y的大小关系是?
水钞胜13199303784 ______ sin(x+y)=2Sinx<=1 则X<=30 . 假设x=y 则sin(x+y)=Sin(2x)<2sinx 矛盾 假设 y<x 由于单调递增 Sin(x+y)<Sin(2x)<2Sinx 矛盾 只能是y>x 其中x=30 y=60 就是一个解
甘先婷3026已知函数f(x)=sinx+sin(x+π2),x∈R.(1)求f(x)的最小正周期;(2)求f(x)的最大值和最小值;(3)若f(α)=34,求sin 2α的值. -
水钞胜13199303784 ______[答案] (1)∵f(x)=sinx+sin( π 2+x)=sinx+cosx= 2sin(x+ π 4)∴函数f(x)=sin x+sin(x+ π 2)的最小正周期是2π. (2)∵x∈R,-1≤sinx≤1 (2)f(x)=sinx+sin( π 2+x)=sinx+cosx= 2sin(x+ π 4) ∴f(x)的最大值为 2,最小值为- 2…(8分) (3)∵f(α)=sinα+sin(α+ π 2)=sinα+...
甘先婷30262sinx(sinx+cosx) = - 根号2sin(2x - n/4)吗?哪里错 -
水钞胜13199303784 ______ 2sinx(sinx+cosx)=2sin^2 x+2sinxcosx =1-cos2x+sin2x =1+sin2x-cos2x =1+根号2sin(2x-π/4)