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
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![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](/uploads/image/z/5126421-21-1.jpg?t=%E5%B7%B2%E7%9F%A5cosx%3D1%2F7%2Ccos%28x%2B%CE%B2%29%3D-11%2F14%2Cx%E5%B1%9E%E4%BA%8E%280%2C%CF%80%2F2%29%2Cx%2B%CE%B2%E5%B1%9E%E4%BA%8E%28%CF%80%2F2%2C%CF%80%29%2C%E6%B1%82%CE%B2%E7%9A%84%E5%80%BC)
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