大师作文网三角函数题……求高手指出错误……
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三角函数题……求高手指出错误……
题目是：已知cos(π/4+x）=3/5,17π/12＜x＜7π/4,求sin2x+2sin²x/1-tanx的值.
我的做法如下：
sin2x+2sin²x/1-tanx
=sin2x+2sin²x-1+1/1-tanx
=sin2x-sin2x+1/1-tanx
=1/1-tanx
=cosx/cosx-sinx
又∵17π/12＜x＜7π/4,∴5π/3＜x+π/4＜2π
∴sinx＜0,sin（x+π/4)＜0
又∵cosx=cos[(π/4+x)-π/4]
=cos(π/4+x)cosπ/4+sin(π/4+x)sinπ/4
=3/5*√2/2+(-4/5)*√2/2
=-√2/10
∴sinx=-7√2/10
∴原式=(-√2/10)/(-√2/10+7√2/10)=1/6
答案是-28/75,我已经知道怎么用另一种方法做出正确答案了……但我觉得我这个方法没错啊……
望高手解答此问……灌水的就可以滚了~
题目是：已知cos(π/4+x）=3/5,17π/12＜x＜7π/4,求sin2x+2sin²x/1-tanx的值.
我的做法如下：
sin2x+2sin²x/1-tanx
=sin2x+2sin²x-1+1/1-tanx
=sin2x-sin2x+1/1-tanx
=1/1-tanx
=cosx/cosx-sinx
又∵17π/12＜x＜7π/4,∴5π/3＜x+π/4＜2π
∴sinx＜0,sin（x+π/4)＜0
又∵cosx=cos[(π/4+x)-π/4]
=cos(π/4+x)cosπ/4+sin(π/4+x)sinπ/4
=3/5*√2/2+(-4/5)*√2/2
=-√2/10
∴sinx=-7√2/10
∴原式=(-√2/10)/(-√2/10+7√2/10)=1/6
答案是-28/75,我已经知道怎么用另一种方法做出正确答案了……但我觉得我这个方法没错啊……
望高手解答此问……灌水的就可以滚了~
sin2x+2sin²x/1-tanx
=sin2x+2sin²x-1+1/1-tanx
=(sin2x-cos2x+1)/(1-tanx)
=(2sinx*cosx-cosx^2+sinx^2+1)/(1-tanx)
=[(sinx+cosx)^2-(cosx+sinx)(cosx-sinx)]cosx/(cosx-sinx)
=2sinx*cosx(sinx+cosx)/(cosx-sinx)
又∵17π/12＜x＜7π/4,∴5π/3＜x+π/4＜2π
∴sinx＜0,sin（x+π/4)＜0
又∵cosx=cos[(π/4+x)-π/4]
=cos(π/4+x)cosπ/4+sin(π/4+x)sinπ/4
=3/5*√2/2+(-4/5)*√2/2
=-√2/10
∴sinx=-7√2/10
∴原式=2*(-√2/10)*(-7√2/10)(-8√2/10)/(6√2/10)
=-28/75
=sin2x+2sin²x-1+1/1-tanx
=(sin2x-cos2x+1)/(1-tanx)
=(2sinx*cosx-cosx^2+sinx^2+1)/(1-tanx)
=[(sinx+cosx)^2-(cosx+sinx)(cosx-sinx)]cosx/(cosx-sinx)
=2sinx*cosx(sinx+cosx)/(cosx-sinx)
又∵17π/12＜x＜7π/4,∴5π/3＜x+π/4＜2π
∴sinx＜0,sin（x+π/4)＜0
又∵cosx=cos[(π/4+x)-π/4]
=cos(π/4+x)cosπ/4+sin(π/4+x)sinπ/4
=3/5*√2/2+(-4/5)*√2/2
=-√2/10
∴sinx=-7√2/10
∴原式=2*(-√2/10)*(-7√2/10)(-8√2/10)/(6√2/10)
=-28/75