若tan(α+β)=2tanα,求证3sinβ=sin(2α+β)
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![若tan(α+β)=2tanα,求证3sinβ=sin(2α+β)](/uploads/image/z/12295431-63-1.jpg?t=%E8%8B%A5tan%28%CE%B1%2B%CE%B2%29%3D2tan%CE%B1%2C%E6%B1%82%E8%AF%813sin%CE%B2%3Dsin%282%CE%B1%2B%CE%B2%29)
若tan(α+β)=2tanα,求证3sinβ=sin(2α+β)
若tan(α+β)=2tanα,求证3sinβ=sin(2α+β)
若tan(α+β)=2tanα,求证3sinβ=sin(2α+β)
因为tan(α+β)=2tanα=sin(α+β)/cos(α+β)=2sinα/cosα
所以sin(α+β)*cosα = 2cos(α+β)sinα
3sinβ = 3sin(α+β-α) = 3sin(α+β)cosα - 3cos(α+β)sinα
= 6cos(α+β)sinα - 3cos(α+β)sinα
= 3cos(α+β)sinα
sin(2α+β) = sin(α+β + α) = sin(α+β)cosα + cos(α+β)sinα
= 2cos(α+β)sinα + cos(α+β)sinα
= 3cos(α+β)sinα
所以
3sinβ=sin(2α+β)
证毕