Question

sinx+siny=1/2，cosx+cosy=1/3求sin（x+y）

Answer 1

解：∵ sinx+siny=1/2，cosx+cosy=1/3
∴ (sinx+siny)^2=1/4..........(1)
(cosx+cosy)^2=1/9..........(2)
(sinx+siny)(cosx+cosy)=1/6..........(3)
∵由(1)式+(2)式，再化简得 cosxcosy+sinxsiny=-59/72
∴cos(x-y)=-59/72 (应用差角公式)..........(4)
∵由(3)式，得 (sinxcosx+sinycosy)+(sinxcosy+cosxsiny)=1/6
==>(sin(2x)+sin(2y))/2+sin(x+y)=1/6 (应用倍角公式与和角公式)
==>sin(x+y)cos(x-y)+sin(x+y)=1/6 (应用和角公式)
==>sin(x+y)(cos(x-y)+1)=1/6
∴sin(x+y)=(1/6)/(cos(x-y)+1)
=(1/6)/(-59/72+1) (把(4)式代入)
=12/13。