已知sinx+sinb=√2,cosx+cosb=√2/3,求tan(x+b)的值

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已知sinx+sinb=√2,cosx+cosb=√2/3,求tan(x+b)的值

已知sinx+sinb=√2,cosx+cosb=√2/3,求tan(x+b)的值
已知sinx+sinb=√2,cosx+cosb=√2/3,求tan(x+b)的值

已知sinx+sinb=√2,cosx+cosb=√2/3,求tan(x+b)的值
sinx+sinb = 2 sin[(x+b)/2] cos[(x-b)/2] = √2
cosx+cosb = 2 cos[(x+b)/2] cos[(x-b)/2] =√2/3
两式相除,得到
sin[(x+b)/2] / cos[(x+b)/2]=3
tan[(x+b)/2] =3
令 (x+b)/2=θ,则 x+b=2θ,tanθ=3
tan(x+b) = tan2θ = 2tanθ/[1-(tanθ)^2] = 2x3/(1-3^2) = -3/4