已知(tanx)^2=2(tany)^2+1,求证(siny^)2=2(sinx^)2-1

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已知(tanx)^2=2(tany)^2+1,求证(siny^)2=2(sinx^)2-1
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已知(tanx)^2=2(tany)^2+1,求证(siny^)2=2(sinx^)2-1
已知(tanx)^2=2(tany)^2+1,求证(siny^)2=2(sinx^)2-1

已知(tanx)^2=2(tany)^2+1,求证(siny^)2=2(sinx^)2-1
证明最重要的是根据形式,选择合适的入手点.这里从右边结论出发,为什么呢?因为注意到(sinx)^2的形式可以化成tan(x)的平方形式.
(siny^)2=2(sinx^)2-1
(siny^)2/((siny)^2+(cosy)^2)=(2(sinx^)2-1)/((sinx)^2+(cosx)^2)
(tany^)2/(1+(tany)^2)=(2(tanx)^2-((tanx)^2+1))/(1+(tanx)^2)
(tany^)2/(1+(tany)^2)=((tanx)^2-1)/(1+(tanx)^2)
(tany^)2(1+(tanx)^2)=((tanx)^2-1)(1+(tany)^2)
(tany^)2+(tany^)2(tanx)^2=(tanx)^2(tany)^2-(tany)^2)+(tanx)^2-1
2(tany^)2+1=(tanx)^2
证毕.