Question

已知A属于(0,pai\2),B属于(0,pai) sinA+cosB=-8\65，sin（A+B）=-33\65 求sinB+cosA 要过程！！！

Answer 1

sinA+cosB=-8\65
(sinA+cosB)^2=（-8\65）^2
(sinA)^2+(cosB)^2+2sinAcosB=(8/65)^2 (1)

sin（A+B）=-33\65
sinA*cosB+sinB*cosA=-33/65

sinB+cosA =K
(sinB+cosA)^2 =K^2
(sinB)^2+(cosA)^2+2sinBcosA=K^2 (2)

(1)+(2):
[(sinA)^2+(cosB)^2+2sinAcosB]+[(sinB)^2+(cosA)^2+2sinBcosA]=(8/65)^2+K^2
[(sinA)^2+(cosB)^2+(sinB)^2+(cosA)^2]+2(sinA*cosB+sinB*cosA)=(8/65)^2+K^2
2+2*(-33/65)=(8/65)^2+K^2
K^2=64/65-64/65^2
K^2=64/65(1-1/65)
K^2=64/65*64/65
A属于(0,pai\2),B属于(0,pai)
sinB>0 cosA>0
sinB+cosA =K>0
K^2=(64/65)^2
K=64/65
sinB+cosA =K=64/65