求证:cosx+sinx=根号2cos(x-π/4)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/26 22:55:04
求证:cosx+sinx=根号2cos(x-π/4)求证:cosx+sinx=根号2cos(x-π/4)求证:cosx+sinx=根号2cos(x-π/4)cosx+sinx=√2[((√2/2)co
求证:cosx+sinx=根号2cos(x-π/4)
求证:cosx+sinx=根号2cos(x-π/4)
求证:cosx+sinx=根号2cos(x-π/4)
cosx+sinx
=√2[((√2/2)cosx+(√2/2)sinx]
=√2[cosxcos(π/4)+sinxsin(π/4)]
=√2cos(x-π/4)
得证
展开即得
两边同时平方 然后右侧化消次就行了
cosx-sinx=根号2sinx,求证tanx=cosx-sinx/cosx+sinx谢谢了,
求证:cosx+sinx=根号2cos(x-π/4)
求证2sinxcosx/(sinx+cosx-1)(sinx-cos+1)=sinx/1-cosx2sinxcosx/(sinx+cosx-1)(sinx-cos+1)=sinx/(1-cosx)
求证cosx+sinx=√2cos(x-π/4)
求证[sinx(1+sinx)+cosx(1+cosx)][sinx(1-sinx)+cos(1-cosx)]=sin2x
根号2cos(x-π/2)怎么=sinx+cosx
证明:sinx-cosx=-根号2cos(x+派/4)
cosx+根号3sinx=2cos(x-π/3)
求证sinx-cosx=根号2sin(x-π/4)
求证cosX/(1+sinx)-sinx/(1+cosx)=2(cosx-sinx)/(1+sinx+cosx)
求证:(1+sinx+cosx)/(1+sinx-cosx)-(1+sinx-cosx)/(1+sinx+cosx)=2/tanx
求证:2(sinx-cosx)/(1+sinx+cosx)=sinx/(1+cosx)-cosx/(1+sinx)
求证:cosx/1+sinx-sinx/1+cosx=2(cosx-sinx)/1+sinx+cosx
求证.[(1+sinx+cosx+2sinx cosx)/(1+sinx+cosx)]=sinx+cosx
tanx=根号2,则(2cos^2x/2-sinx-1)/(sinx+cosx)=
证明2(cosx-sinx)/(1+sinx+cosx)=cosx/(1+sinx)-sinx/(1+cos)
(1)已知:(4sinx-2cosx)/(5cosx+3sinx)=6/11 求sinx乘以cosx的值(2)证明(cosx/1+sinx) - (sinx/1+cosx)=2(cosx-sinx)/1+sinx+cosx(3)求证:sin²x乘以tanx + cos²x乘以cotx+2sinx乘以cosx=tanx+cotx要用 sin²X+cos²X=1
已知sinαcosβ=cos^βsinx/2cosx+sin^2αcosx/2sinx,求证:tgx=sinα/cosβ