高中数学导学导练求正弦函数问题
解法一:
【1】由0<β<π/4,sin(β+3π/4)=5/13 可得 cos(β+3π/4)=-12/13
【2】由π/4<α<3π/4,cos(π/4-α)=3/5 可得 sin(π/4-α)=-4/5
【所以】sin(α+β)=sin[(β+3π/4)-(π/4-α)-π/2]=-cos[(β+3π/4)-(π/4-α)]
=-[cos(β+3π/4)cos(π/4-α)+sin(β+3π/4)sin(π/4-α)]
=-[(-12/13)(3/5)+(5/13)(-4/5)]=56/65。
解法二
【1】0<β<π/4,sin(β+3π/4)=5/13 ==> cos(β...全部
解法一:
【1】由0<β<π/4,sin(β+3π/4)=5/13 可得 cos(β+3π/4)=-12/13
【2】由π/4<α<3π/4,cos(π/4-α)=3/5 可得 sin(π/4-α)=-4/5
【所以】sin(α+β)=sin[(β+3π/4)-(π/4-α)-π/2]=-cos[(β+3π/4)-(π/4-α)]
=-[cos(β+3π/4)cos(π/4-α)+sin(β+3π/4)sin(π/4-α)]
=-[(-12/13)(3/5)+(5/13)(-4/5)]=56/65。
解法二
【1】0<β<π/4,sin(β+3π/4)=5/13 ==> cos(β+3π/4)=-12/13
即(√2/2)(cosβ-sinβ)=5/13……①;(√2/2)(cosβ+sinβ)=12/13……②
由①、②可得cosβ=17√2/26,sinβ=7√2/26。
【2】π/4<α<3π/4,cos(π/4-α)=3/5 ==> sin(π/4-α)=-4/5
(√2/2)(cosα+sinα)=3/5……③;(√2/2)(cosα-sinα)=-4/5……⑥
由③、④可得cosα=-√2/10,sinα=7√2/10。
【结论】sin(α+β)=sinαcosβ+cosαsinβ=[7√2/10][17√2/26]+[-√2/10][7√2/26]=56/65。收起