- 高一数学填空题3已知sin(α+β)=1,则sin(2α+β)+
- 已知sin(α+β)=1,则sin(2α+β)+sin(2α+3β)=________
- sin(α+β)=1, cos(α+β)=0;
sin(2α+β)+sin(2α+3β)
= sin(a+(α+β)) + sin((2α+2β)+β)
= sinα*cos(α+β) + cosα*sin(α+β)
+ sin(2α+2β)*cosβ + cos(2α+2β)*sinβ
=cosα+2sin(α+β)*cos(α+β)*cosβ+(cos^2(α+β)-sin^2(α+β))*sinβ
=cosα-sinβ;
α+β=2kπ+π/2;
α=2kπ+π/2-β
cosα = sinβ;
sin(2α+β)+sin(2α+3β)
=cosα-sinβ
=0