大师作文网若cos(π/4+x)=3/5 17π/12
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若cos(π/4+x)=3/5 17π/12
17π/12＜x＜7π/4,得5π/3＜x+π/4＜2π
cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75
再问： 不对吧 是cos²x 不是sin²x 而且是cos(π/4+x）不是cos(x-π/4)你再看看 别玩复制粘贴 OK？
再答： cos(π/4+x) =cosπ/4cosx-sinπ/4sinx =√2/2(cosx-sinx) =3/5 cosx-sinx=3√2/5 …………（1） （cosx-sinx）^2=(cosx)^2+(sinx)^2-2sinxcosx=1-2sinxcosx=18/25 2sinxcosx=1-18/25=7/25 所以sin2x=2sinxcosx=7/25 由（1）得 sinx=cosx-3√2/5 两边平方 sin²x=1-cos²x=cos²x-6√2/5*cosx+18/25 2cos²x-6√2/5*cosx-7/25=0 50cos²x-30√2cosx-7=0 cosx=(3√2±4√2)/10 cosx=-√2/10或7√2/10 17π/12
cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75
再问： 不对吧 是cos²x 不是sin²x 而且是cos(π/4+x）不是cos(x-π/4)你再看看 别玩复制粘贴 OK？
再答： cos(π/4+x) =cosπ/4cosx-sinπ/4sinx =√2/2(cosx-sinx) =3/5 cosx-sinx=3√2/5 …………（1） （cosx-sinx）^2=(cosx)^2+(sinx)^2-2sinxcosx=1-2sinxcosx=18/25 2sinxcosx=1-18/25=7/25 所以sin2x=2sinxcosx=7/25 由（1）得 sinx=cosx-3√2/5 两边平方 sin²x=1-cos²x=cos²x-6√2/5*cosx+18/25 2cos²x-6√2/5*cosx-7/25=0 50cos²x-30√2cosx-7=0 cosx=(3√2±4√2)/10 cosx=-√2/10或7√2/10 17π/12
cos(π/4+x)=3/5,17π/12
若cos（π/4+x)=3/5,求值
若cos（π/4+x）=3/5,17/12π
若cos(π/4+x)=3/5 17π/12
一、已知cos(π/4+x)=3/5,17π/12
已知cos(π/4+x）=3/5,17π/12
已知cos(x+π/4)=3/5且17π/12
已知cos(π/4 +x)=3/5,17π/12
已知cos(π/4+x)=3/5,17π/12
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π
已知cos(π/4+x)=3/5 ,0
cos(派/4+x)=3/5,17派/12