Question

已知cos(π/4-a)=3/5,sin(3π/4+b)=5/13,其中<a<3π/4,0<b<π/4,求sin(a+b)的值

请告诉我答案及解题过程!!!谢谢!!!

Answer 1

cos(π/4-a)=3/5
0<a<3π/4 -3π/4<-a<0
-π/2<π/4-a<π/4
因为当0<π/4-a<π/4
cos(π/4-a)=3/5<√2/2=cos(π/4)
π/4-a>π/4与0<π/4-a<π/4矛盾
所以-π/2<π/4-a<0
sin(π/4-a)=-√(1-(cos(π/4-a))^2)=-3/5

sin(3π/4+b)=5/13
0<b<π/4 3π/4<3π/4+b<π
cos(3π/4+b)=√(1-(sin(3π/4+b))^2)=12/13

sin(a+b)=-cos(a+b+π/2)=-cos((3π/4+b)-(π/4-a))
=-[cos(3π/4+b)cos(π/4-a)+sin(3π/4+b)sin(π/4-a)]
=-[12/13*3/5+5/13*(-3/5)]
=-16/65