三角函数 证明题~
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三角函数 证明题~
三角函数 证明题~
三角函数 证明题~
(1+sinx+cosx)cosx
=cosx+sinxcosx+cos²x
(1-sinx+cosx)(1+sinx)
=1+sinx-sinx-sin²x+cosx+sinxcosx
=1-sin²x+cosx+sinxcosx
=cos²x+cosx+sinxcosx
(1-sinx+cosx)(1+sinx)=(1+sinx+cosx)cosx
(1+sinx+cosx)/(1-sinx+cosx)=(1+sinx)/cosx
设y=x/2
左边=(2cos^2y+2sinycosy)/(2cos^2y-2sinycosy)
=(cosy+siny)/(cosy-siny)
右边=(sin^2y+2sinycosy+cos^2y)/(cos^2y-sin^2y)
=[(siny+cosy)^2]/[(cosy+siny)(cosy-siny)]
=(siny+cosy)/(cosy-siny)
所以左边=右边
原题得证
∵1=[sin(x/2)]^2 + [cos(x/2)]^2 ， sinx=2×sin(x/2)×cos(x/2)
∴1+sinx=[sin(x/2) + cos(x/2)]^2 ， 1-sinx=[cos(x/2) - sin(x/2)]^2
∵cosx=[cos(x/2)]^2 - [sin(x/2)]^2 =[cos(x/2) + sin(x/2...
全部展开
∵1=[sin(x/2)]^2 + [cos(x/2)]^2 ， sinx=2×sin(x/2)×cos(x/2)
∴1+sinx=[sin(x/2) + cos(x/2)]^2 ， 1-sinx=[cos(x/2) - sin(x/2)]^2
∵cosx=[cos(x/2)]^2 - [sin(x/2)]^2 =[cos(x/2) + sin(x/2)]×[cos(x/2) - sin(x/2)]
∴等式左边分子=[sin(x/2) + cos(x/2)]×{[sin(x/2) + cos(x/2)]+[cos(x/2) - sin(x/2)]}
=[sin(x/2) + cos(x/2)]×[2cos(x/2)]
分母=[cos(x/2) - sin(x/2)]×{[cos(x/2) - sin(x/2)]+[cos(x/2) + sin(x/2)]}
=[cos(x/2) - sin(x/2)]×[2cos(x/2)]
上下约分=[sin(x/2) + cos(x/2)]/[cos(x/2) - sin(x/2)]
再通分=[sin(x/2) + cos(x/2)]×[cos(x/2) + sin(x/2)] / [cos(x/2) - sin(x/2)]×[cos(x/2) + sin(x/2)]
=[sin(x/2)]^2 + 2×sin(x/2)×cos(x/2) + [cos(x/2)]^2 / [cos(x/2)]^2 - [sin(x/2)]^2
=(1+sinx)/cosx
=右边
收起
追忆~