tan(arctanX+arctanY)=(X+Y)/(1-XY)
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tan(arctanX+arctanY)=(X+Y)/(1-XY)
想了很久,没转过来......
还有 sin(2arctanX)=2X/(1+X^2)
想了很久,没转过来......
还有 sin(2arctanX)=2X/(1+X^2)
![tan(arctanX+arctanY)=(X+Y)/(1-XY)](/uploads/image/z/9834080-32-0.jpg?t=tan%28arctanX%2BarctanY%29%3D%28X%2BY%29%2F%281-XY%29)
❶证明:tan(arctanX+arctanY)=(X+Y)/(1-XY)
证明:tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tanarctanx)(tanarctany)]
=(x+y)/(1-xy)=右边.
【如果看不明白,可令arctanx=α,arctany=β,那么tanα=tanarctanx=x,tanβ=tanarctany=y
❷证明sin(2arctanx)=2x/(1+x²)
证明:2sin(arctanx)cos(arctanx)=2(x/√(1+x²)][√(1/1+x²)]=2x/(1+x²)=右边.
这是因为α=arctanx,-π/2
证明:tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tanarctanx)(tanarctany)]
=(x+y)/(1-xy)=右边.
【如果看不明白,可令arctanx=α,arctany=β,那么tanα=tanarctanx=x,tanβ=tanarctany=y
❷证明sin(2arctanx)=2x/(1+x²)
证明:2sin(arctanx)cos(arctanx)=2(x/√(1+x²)][√(1/1+x²)]=2x/(1+x²)=右边.
这是因为α=arctanx,-π/2
tan(arctanX+arctanY)=(X+Y)/(1-XY)
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