分类：教育/科学-学习帮助
问题：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)=________
回答：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