Question

已知sin(α+3/4π)=5/13，cos（π/4-β）=3/5，且-π/4<α<π/4，π/4<β<3/4π,求cos2（α-β）的值

已知sin(α+3/4π)=5/13，cos（π/4-β）=3/5，且-π/4<α<π/4，π/4<β<3/4π,求cos2（α-β）的值
要过程 谢谢

Answer 1

∵-π/4<α<π/4，π/4<β<3/4π
∴π/2<α+3/4π<π，0>π/4-β>-π/2
∴sin(α+3/4π)= 5/闹碧旁13 => cos(α+3/4π) = -12/13
cos(π/4-β）= 3/5 => sin(π/4-β) = -4/5
根据公式cos(a+b)=cosacosb-sinasinb
cos(α-β)=-cos(π+α-β)=-cos[(α+3/4π)+(π/4-β)]
=sinsin-coscos 为了慧穗方便具体值省略
=(5/13) (-4/5) -(12/液橡13) (3/5)
=-4/13-99/65=-119/65
然后求cos^2(α-β) = 1161/4225

Answer 2

-π/4<α<π/4 π/2<α+3/4π<π ∴ cos(α+3/4π)<0
π/4<β<3/4π -3π/4<-β<-1/信乎丛4π -π/2<π/4-β<0 ∴sin(π/4-β)<0
∴cos(α+3/滑樱4π)=-√[1-sin²(α+3/4π)]=-12/13
sin(π/4-β)=-√顷吵[1-cos²(π/4-β)]=-4/5
∴sin[(α+3/4π)+(π/4-β)]=sin(α-β+π)=-sin(α-β)
=sin(α+3/4π)cos(π/4-β)+cos(α+3/4π)sin(π/4-β)
=5/13×3/5+(-12/13)×(-4/5)=63/65
∴sin(α-β)=-63/65
∴cos2(α-β)=1-2sin²(α-β)=1-2(-63/65)²=-3713/4225