Question

数学分析法问题

若tan(α+β)=2tanα，求证3sinβ=sin(2α+β)

Answer 1

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