sin2x等于-1/2
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sin2x+1=2tanx/(1+tan^2x)+1=4/5+1=9/5
假设sin2x*[(根号1-tanx)+(根号1+tanx)]=2sin2x则(根号1-tanx)+(根号1+tanx)=2两边平方得1-tanx+1+tanx+2根号(1-tan^2x)=42根号(
(1)根据图信息可知sina=4/5cosa=3/5∴(sina^2+sin2a)/(cosa^2+cos2a)=(sina^2+2sinacosa)/(cosa^2+cosa^2-sina^2)=(
第一个式子等於(1+cos2x)/2sin2x=1-cosx是错的应该是sin2x=2sinxcosx
证明:(cosxsinx)^2=sinx^2cosx^22sinx·cosx=12sinx·cosx=1sin2x
sinx=2cosx,sinx^2+(1/2sinx)^2=1,得sinx^2=4/5,sin2x=2sinxcosx=sinx^2=4/5希望采纳
x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3
tanx=3则:sinx/cosx=3sinx=3cosxsin2x/cos²x=2sinxcosx/cos²x=6cos²x/cos²x=6
题目条件不充分啊cos2x+sin2x=1cos2x=cos^2x-sin^2xsin2x=2sinxcosxcos2x+sin2x=cos^2x-sin^2x+2sinxcosx
∵cos2x=2cos²x-1,∴cos²x=1/2(cos2x+1)∴原式=1/(1/2cos2x+sin2x+1/2)=1/[根号5/2sin(2x+ψ)+1/2](tanψ=
=(2sin2x)/(1-cos2x)²=(4sinxcosx)/(2sin²x)²=cosx/sin³x
sinx/cosx=tanx=2sinx=2cosxsin²x=4cos²x因为sin²x+cos²x=1所以cos²x=1/5sin2x=2sinx
sinx/cosx=tanx所以sin2x/cos2x=tan2x
sin2x+cos2x=√2(√2/2sin2x+√2/2*cos2x)=√2(cospai/4sin2x+sinpai/4cos2x)=√2sin(2x+pai&#4
等式左边=2sinxcosx/[sinx+(cosx-1)][sinx-(cosx-1)]=2sinxcosx/[sin^2x-(cosx-1)^2]=2sinxcosx/(sin^2x-cos^2x
1/2sin4x
等式左边=2sinxcosx/[sinx+(cosx-1)][sinx-(cosx-1)]=2sinxcosx/[sin^2x-(cosx-1)^2]=2sinxcosx/(sin^2x-cos^2x
sin²2x+sin2x+cos2x=1sn2x+cos2x=1-sin²2x=cos²2xsin2x=cos²2x-cos2x----------------
sinx^2+cosx^2=1,这就相当于一个公式,中间的变数x当然可以换成任何值了!
∫cos2xdx上限是π/4下限是π/6=1/2*sin2x上限是π/4下限是π/6=1/2sinπ/2-1/2sinπ/3=1/2-根号3/4我是数学百事通,数学问题想不通,快上数学百事通!