sint+2cost+2dt的积分

何呢戴4416高中数学三角函数公式
翁贩鬼13422812494 ______ Sqrt(a^2+b^2)sin(x+arctan(a/b))=Sqrt(a^2+b^2)*[sinxcosarctan(a/b)+cosxsinarctan(a/b)]=acosx+bsinx 实质就是和角公式的逆推,这个逆推公式在高中三角函数的题目中还比较常用,建议熟记.

何呢戴44161/(sinx+2)的不定积分如何求 -
翁贩鬼13422812494 ______ 1/(sinx+2)=(1/2)/(0.5*sinx+1)dx=1/(sin(x/2)cos(x/2)+1)d(x/2) 令t=x/2 原式=(1/sint*cost+1)dt 分子分母都除以(cost)^2=(1/(cost)^2)/{[1/(cost)^2]+tant})dt=1/{[1/(cost)^2]+tant}d(tant)=[1/(1+(tantt)^2)+tant]d(tant) 令u=tant={1/(1+u^2)+u}du=1/[(0.5+u)^...

何呢戴4416sin(x+t) - sinx=2cos(x+t/2)sint/2 如何证明? -
翁贩鬼13422812494 ______ 证明:sin(x+t)-sinx=sinxcost+cosxsint-sinx=sinx(cost-1)+cosxsint=-2sinx(sint/2)^2+2cosxcost/2sint/2=2sint/2(-sinxsint/2+cosxcost/2)=2sint/2cos(x+t/2)

何呢戴4416复数|z|=√2,那么|z - 1+i|最大值为 -
翁贩鬼13422812494 ______ 设z=a+bi |z|=√(a²+b²)=√2 a²+b²=2 设a=√2sinx, b=√2cosx,|z-1+i|=√[(√2sinx-1)²+(√2cost+1)²]=√(2sin²x-2√2sinx+1+2cos²x+2√2cosx+1)=√[4-2√2(sinx-cosx)]=√[4-2√2.√2(√2/2sinx-√2/2cosx)]=√[4-4sin(x-π/4)] 当sin(x-π/4)=-1时 |z-1+i|有最大值√(4+4)=2√2

何呢戴4416若sin2分之x - cos2分之x=3分之1求sinx的值 -
翁贩鬼13422812494 ______ sin(π-x)-cos(π-x)=√2/3 sinx+cosx=√2/3(利用诱导公式,奇变偶不变,符号看限象) (sinx+cosx)^2=(√2/3)^2=4/9=sinx^2+sinx^2+cosx^2+cosx^2 又sinx^2+cosx^2=1 所以2sinxcosx=-5/9 π/20,cosx0 sinx-cosx=√(sinx-cosx)^2=√(sinx+cosx)^2-...

何呢戴4416求函数y=sinx+2cosx+2的值域 -
翁贩鬼13422812494 ______[答案] y=√5(1/√5sinx+2/√5cosx)+2 =√5(sinx+α)+2 所以函数y=sinx+2cosx+2的值域是[2-√5,2+√5]

何呢戴4416y=sin2x+cos2x化简成y=Asin(wx+t)形式,并求最小正周期 -
翁贩鬼13422812494 ______ 原式=根号2(根号2/2 sin2x+根号2/2 cos2x) =根号2sin(π/4+2x) 所以T=2π/w=2π/2=π

何呢戴4416定积分sinx/(sinx+cosx)从0到π//2为为什么等于定积分cosx/(sinx+cosx)从0到π/2 -
翁贩鬼13422812494 ______[答案] 设t=π/2-x sinxdx/(sinx+cosx)=-cosdt/(sint+cost) sinxdx/(sinx+cosx)从0积到π/2等于-cosdt/(sint+cost)从π/2积到0等于cosdt/(sint+cost)从0积到π/2

何呢戴4416sin(x+T) - sinx= - T怎么变成2sinT/2*cos(x+T/2)= - T -
翁贩鬼13422812494 ______[答案] sin(x+T)-sinx=sin(x+T/2+T/2)-sin(x+T/2-T/2)=sin(x+T/2)*cosT/2+cos(x+T/2)*sinT/2-[sin(x+T/2)*cosT/2-cos(x+T/2)*sinT/2]=2sinT/2*cos(x+T/2)所以2sinT/2*cos(x+T/2)=sin(x+T)-sinx=-T

何呢戴4416如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等? -
翁贩鬼13422812494 ______[答案] 只需令x=pi/2-t,则当x=0,t=pi/2,当x=pi/2,t=0,dx=-dt,那么 ∫(0,pi/2)cosx/(sinx+cosx)dx =-∫(pi/2,0)sint/(sint+cost)dt =∫(0,pi/2)sinx/(sinx+cosx)dx 所以 ∫(0,pi/2)cosx/(sinx+cosx)dx=∫(0,pi/2)sinx/(sinx+cosx)dx =(1/2)[∫(0,pi/2)cosx/(sinx+cosx)dx+∫(0,pi/2)sinx/...