Question

高一数学几道题

已知sin（αβ）cosα—cos(β-α)sinα=3/5,β是第三象限角，求sin(β+5π/4)的值
已知sin（α-β）cosα—cos(β-α)sinα=3/5,β是第三象限角，求sin(β+5π/4)的值

Answer 1

解：由sin（α-β）cosα—cos(β-α)sinα=3/5
得sin（α-β）cosα—cos(α-β)sinα=3/5
即sin[（α-β）-α]=3/5，
sinβ= -3/5，β是第三象限角，∴cosβ= -√（1-sin²β）= -4/5
sin(β+5π/4)= -sin(β+π/4)= -[sinβ×cos(π/4)+cosβ×sin(π/4)]=（4√2）/5

Answer 2

sin（α-β）cosα—cos(β-α)sinα=sin（α-β）cosα—cos(α-β)sinα=sin[(α-β)-α]= - sinβ=3/5
cosβ=-4/5
sin(β+5π/4)=sin[π+(β+π/4)] =-sin(β+π/4) =-[sinβcos π/4+cosβsin π/4]
=-[-3/5*√2/2+(- 4/5*√2/2)]=4√2/5