Question

高中数学导学导练求正弦函数问题

只有图片一张

最好解析一下

Answer 1

∵π/4<a<3π/4
∴-3π/4<-a<-π/4
-π/2<π/4-a<0
∴sin(π/4-a)=-√[1-(3/5)^2]=-4/5
∵0<b<π/4
∴3π/4<b+3π/4<π
∴cos(b+3π/4)=-√[1-(5/13)^2]=-12/13
sin(a+b)=-cos[(b+3π/4)-(π/4-a)]=-[-(12/13)(3/5)+(5/13)(-4/5)]=56/65

Answer 2

∵π/4<a<3π/4
∴-3π/4<-a<-π/4
-π/2<π/4-a<0
∴sin(π/4-a)=-√[1-(3/5)^2]=-4/5
∵0<b<π/4
∴3π/4<b+3π/4<π
∴cos(b+3π/4)=-√[1-(5/13)^2]=-12/13
sin(a+b)=-cos[(b+3π/4)-(π/4-a)]=-[-(12/13)(3/5)+(5/13)(-4/5)]=56/65