\sin\alpha={\sin 2\alpha\over 2\cos\alpha}=\pm{\sin 2\alpha\over\sqrt{2(1+\cos 2\alpha)}}=\frac{{\mathrm{sgn}\,b\cdot a\over\sqrt{a^2+b^2}}}{\sqrt{2(1+\cos 2\alpha)}}=\frac{{\mathrm{sgn}\,b\cdot a\over\sqrt{a^2+b^2}}}{\sqrt{2(1\pm\sqrt{1-\sin^2 2\alpha)}}}