(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)

显示全部楼层 · 2008-8-9 14:10:57

化简~
谢谢

千问 · 2008-8-9 14:10:57

(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)=(sinα)^2 + (sinβ)^2 + 2sinαsinβ (cosαcosβ - sinαsinβ)= (sinα)^2 + (sinβ)^2 + 2sinαcosαsinβcosβ - 2(sinα)^2(sinβ)^2= (sinα)^2 * [1 - (sinβ)^2] + (sinβ)^2 *[ 1 -(sinα)^2] + 2sinαcosαsinβcosβ= (sinα)^2 * (cosβ)^2 + (sinβ)^2 *(cosα)^2 + 2sinαcosαsinβcosβ= (sinαcosβ + cosαsinβ)^2= [sin(

千问 · 2008-8-9 14:10:57

(sinα)^2+(sinβ)^2+2sinαsinβcos(α+β)=(sinα)^2+(sinβ)^2+2sinαsinβ(cosαcosβ-sinαsinβ)=(sinα)^2+(sinβ)^2+2sinαsinβcosαcosβ-2(sinα)^2*(sinβ)^2=(sinα)^2(1-sinβ*sinβ)+(sinβ)^2(1-sin