已知cosx=1/7,cos(x+β)=-11/14,x属于(0,π/2),x+β属于(π/2,π),求β的值

已知cosx=1/7,cos(x+β)=-11/14,x属于(0,π/2),x+β属于(π/2,π),求β的值
已知cosx=1/7,cos(x+β)=-11/14,x属于(0,π/2),x+β属于(π/2,π),求β的值

已知cosx=1/7,cos(x+β)=-11/14,x属于(0,π/2),x+β属于(π/2,π),求β的值
sinx=4根号3/7
sin(x+β)=5根号3/14
那么cosβ=cos[(x+β)-x]
=cos(x+β)cosx+sin(x+β)sinx
=-11/14*1/7+5根号3/14*4根号3/7
=1/2
β=π/3