0<β<派/4
派/4<α<3派/4
且cos(派/4-α)=4/5
sin(3派/4+β)=5/13
求cosα,cos(α+β)

解
cos(π/4-α)=cos(α-π/4)=4/5
sin(π/4-α)=-3/5,
sin(α-π/4)=3/5 (因π/4<α<3π/4）

sin(3π/4+β)=sin[π-(π/4-β)]=sin(π/4-β)
即sin(π/4-β)=5/13
cos(3π/4+β)=-12/13,
cos(π/4-β)=12/13 (因0<β<π/4）

所以
cosα
=cos[(α-π/4)+π/4]
=cos(α-π/4)cos(π/4)-sin(α-π/4)sin(π/4)
=(4/5)(√2/2）-（3/5)(√2/2）
=√2/10

cos(α+β)
=-sin[π/2+(α+β)]
=-sin[(3π/4+β)-(π/4-α)]
=-sin(3π/4+β)cos(π/4-α)+cos(3π/4+β)sin(π/4-α)]
=-(5/13)(4/5)+(-12/13)(-3/5)
=16/65

另外3月30日的题应是这样：
sin(α+β)
=cos[π/2+(α+β)]
=cos[(3π/4+β)-(π/4-α)]
=cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/4-α)
=(-12/13)(3/5)+(5/13)(-4/5)
=-56/65