若a∈(0,π/2),β∈(π/2,π),cosβ=-1/3,sin(a+β)=7/9,求cosa的值

Question

若a∈(0,π/2),β∈(π/2,π),cosβ=-1/3,sin(a+β)=7/9,求cosa的值

Answer 1

∵cosβ=-1/3,β∈(π/2,π),
∴sinβ=2√2/3
∵a+β∈(π/2,3π/2),
sin(a+β)=7/9>0
∴a+β∈(π/2,π),cos(a+β)=-4√2/9
∴cosa=cos[(a+β)-β]
=cos(a+β)cosβ-sin(a+β)sinβ
=(-4√2/9)(-1/3)-(7/9)(2√2/3)
=(-10√2)/27