Question

已知sin（α+3/4π）=5/13，cos（π/4-β）=3/5，且-π/4<α<π/4，π/4<β<3π/4，求cos[2(α-β)]的值。

高一数学题

Answer 1

∵-π/4<α<π/4∴π/2<α+3/4π<π
又sin（α+3/4π）=5/13
∴cos(α+3/4π)=-12/13
∵π/4<β<3π/4∴-π/2<π/4-β<0
∵cos（π/4-β）=3/5
∴sin(π/4-β)=-4/5
∴cos[(α+3π/4)+(π/4-β)]
=cos(α+3π/4)cos(π/4-β)-sin(α+3π/4)sin(π/4-β)
=-12/13*3/5-5/13*(-4/5)=-16/65
又cos[(α+3π/4)+(π/4-β)]=cos(π+α-β)=-cos(α-β)
∴cos(α-β)=16/65
∴cos[2(α-β)]=2cos²(α-β)-1=512/4225-1=-3713/4225

Answer 2

-π/4<α<π/4==>π/2<α+3/4π<π==>cos(α+3/4π)=-12/13
π/4<β<3π/4==>-π/2<π/4-β<0==>sin(π/4-β)=-4/5
令α+3/4π=x π/4-β=y
则sinx=5/13 cosx=-12/13
siny=-4/5 cosy=3/5
由cos[2(α-β)]=cos[2(α-β)+2π]=cos[2(α+3/4π)+2(π/4-β)]=cos(2x+2y)
=2cos^2(x+y)-1
cos(x+y)=cosxcosy-sinxsiny=(-12/13)*(3/5)-(5/13)*(-4/5)=-16/65
故cos[2(α-β)]=-3713/4225
这数值很怪，我怀凝是不是没有2倍关系

Answer 3

cos[2(α-β)]=cos[2(α+3/4π+（π/4-β)]
令α+3/4π=a,π/4-β=b
2a范围 pi到2pi
2b范围 -pi到0
两者正弦都小于0
原式=cos(2a)cos(2b)-sin(2a)sin(2b)
=(1-2(sin a)^2)(1-2(sin b)^2)-sqrt(1-cos(2a)^2)sqrt(1-cos(2b)^2)
=(119/169)*(7/25)-(120/169)*24/25
=（833-2880）/（169*25）
=-2047/4225