You might try to regularize the sum, $$\sum_{n=1}^\infty \frac{ \cos(\alpha n) \sin(n+1) }{n}= -\frac{1}{4} i e^{-i} \left(\ln \left(1-e^{i (\alpha-1)}\right)+\ln \left(1-e^{-i (\alpha+1)}\right)-e^{2 i} \left(\ln \left(1-e^{-i (\alpha-1)}\right)+\ln \left(1-e^{i (\alpha+1)}\right)\right)\right)$$ $$=\text{Re}\, \biggl(\tfrac{1}{2} i e^i \ln \left[2 e^i (\cos 1-\cos \alpha)\right]\biggr),\;\;\alpha\neq\pm 1\;\;\text{mod}\;(2\pi).$$

You might try to regularize the sum, $$S(\alpha)=\sum_{n=1}^\infty \frac{ \cos(\alpha n) \sin(n+1) }{n}= -\tfrac{1}{4} i \left[e^{-i} \ln \left(1-e^{i (\alpha-1)}\right)+e^{-i} \ln \left(1-e^{-i (\alpha+1)}\right)-e^{ i} \ln \left(1-e^{-i (\alpha-1)}\right)-e^{ i}\ln \left(1-e^{i (\alpha+1)}\right)\right]$$ $$=\text{Re}\, \biggl(\tfrac{1}{2} i e^i \ln \left[2 e^i (\cos 1-\cos \alpha)\right]\biggr),\;\;\alpha\neq\pm 1\;\;\text{mod}\;(2\pi).$$

Here is a plot of $S(\alpha)$.

You might try to regularize the sum, $$\sum_{n=1}^\infty \frac{ \cos(\alpha n) \sin(n+1) }{n}= -\frac{1}{4} i e^{-i} \left(\ln \left(1-e^{i (\alpha-1)}\right)+\ln \left(1-e^{-i (\alpha+1)}\right)-e^{2 i} \left(\ln \left(1-e^{-i (\alpha-1)}\right)+\ln \left(1-e^{i (\alpha+1)}\right)\right)\right).$$ This is finite for $\alpha\neq 1$.

You might try to regularize the sum, $$\sum_{n=1}^\infty \frac{ \cos(\alpha n) \sin(n+1) }{n}= -\frac{1}{4} i e^{-i} \left(\ln \left(1-e^{i (\alpha-1)}\right)+\ln \left(1-e^{-i (\alpha+1)}\right)-e^{2 i} \left(\ln \left(1-e^{-i (\alpha-1)}\right)+\ln \left(1-e^{i (\alpha+1)}\right)\right)\right)$$ $$=\text{Re}\, \biggl(\tfrac{1}{2} i e^i \ln \left[2 e^i (\cos 1-\cos \alpha)\right]\biggr),\;\;\alpha\neq\pm 1\;\;\text{mod}\;(2\pi).$$