$cosα=cos(α+β)-cosβ=-2sin\frac{α+2β}{2}sin\frac{α}{2}$ ,considering the arbitrariness of $β$ and the periodicity of $sin$ ,so $sin\frac{α+2β}{2}$ is a independent variable， \begin{aligned}set \large{sin\frac{α+2β}{2}=m,(m∈[-1,1]) \\ \implies cos^{2}α=4m^{2}sin^{2}\frac{α}{2} \\ \implies \frac{2cos^{2}α}{1-cosα}=4m^{2} ∈[0,4] \\ \implies 2cos^{2}α≤4-4cosα \\ cos^{2}α+2cosα-2≤0 \\ \implies -1-\sqrt{3}≤cosα≤-1+\sqrt{3}} \end{aligned}

as a result of which,the maximum value is $\boxed{\sqrt{3}-1}$