若cos(a+b)=4/5,cos(a-b)=1/3,且a+b∈(7π/4,2π),a-b∈(3π/4,π),则cos2a=

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若cos(a+b)=4/5,cos(a-b)=1/3,且a+b∈(7π/4,2π),a-b∈(3π/4,π),则cos2a=
若cos(a+b)=4/5,cos(a-b)=1/3,且a+b∈(7π/4,2π),a-b∈(3π/4,π),则cos2a=

若cos(a+b)=4/5,cos(a-b)=1/3,且a+b∈(7π/4,2π),a-b∈(3π/4,π),则cos2a=
a+b∈(7π/4,2π)
a-b∈(3π/4,π)
cos(a+b)=4/5,cos(a-b)=1/3
∴sin(a+b)=-3/5,sin(a-b)=2√2/3
cos2a
=cos(a+b+a-b)
=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=4/5*1/3-(-3/5)*2√2/3
=(4+6√2)/15

cos2a=cos(a+b+a-b)
=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=4/5*1/3+3/5*2√2/3
=(6√2+4)/15