3sinβ=sin(2α+β),α≠2kπ+π/2 ,α+β≠kπ+π/2 (k∈z)求证tan(α+β)=2tanα
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3sinβ=sin(2α+β),α≠2kπ+π/2 ,α+β≠kπ+π/2 (k∈z)求证tan(α+β)=2tanα
3sinβ=sin(2α+β),α≠2kπ+π/2 ,α+β≠kπ+π/2 (k∈z)求证tan(α+β)=2tanα
3sinβ=sin(2α+β),α≠2kπ+π/2 ,α+β≠kπ+π/2 (k∈z)求证tan(α+β)=2tanα
以a,b代替α,β sin(2a+b)=sin[(a+b)+a]=sin(a+b)cosa+cos(a+b)sina sinb=sin[(a+b)-a]=sin(a+b)cosa-cos(a+b)sina 所以 3sin(a+b)cosa-3cos(a+b)sina=sin(a+b)cosa+cos(a+b)sina 2sin(a+b)cosa=4cos(a+b)sina sin(a+b)cosa=2cos(a+b)sina tan (a+b)=2 tan a 得证.
化简sin(α+β)+sin(α-β)+2sinαsin(3π/2-β)=
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