Тригонометрия - да се намерят стойностите - несъответствие

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Здравейте, решавах една задача по два начина, а получавам различен отговор

[tex]\[\begin{align}
& tg\frac{\alpha }{2}=-\frac{3}{5} \\
& \alpha \in (180{}^\circ ;270{}^\circ )- treti. kv\\
& A)\cos \left( \frac{\pi }{4}-\frac{\alpha }{2} \right)=? \\
& B)\sin \alpha =? \\
& \\
& \frac{\alpha }{2}\in (90{}^\circ ;135{}^\circ )-vtori . kv \\
& B) \\
& \text{sin}\alpha \text{=}\frac{2tg\frac{\alpha }{2}}{1+t{{g}^{2}}\frac{\alpha }{2}}=\frac{-\frac{6}{5}}{1+\frac{9}{25}}=-\frac{6}{5}:\frac{25}{34}=-\frac{15}{17} \\
& A) \\
& \left| \begin{align}
& tg\frac{\alpha }{2}=-\frac{3}{5}=\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}}-->sin {\frac{\alpha }{2 }}=- \frac35 cos {\frac{\alpha }{2 }} \\
& {{\sin }^{2}}\frac{\alpha }{2}+{{\cos }^{2}}\frac{\alpha }{2}=1 \\
\end{align} \right. \\
& \frac{9}{25}{{\cos }^{2}}\frac{\alpha }{2}+{{\cos }^{2}}\frac{\alpha }{2}=1|.25 \\
& 9{{\cos }^{2}}\frac{\alpha }{2}+25{{\cos }^{2}}\frac{\alpha }{2}=25 \\
& {{\cos }^{2}}\frac{\alpha }{2}=\frac{25}{34}--> cos { \frac{ \alpha}{2}}= - \frac {5 \sqrt {34}}{34} \\
& \\
& \cos \left( \frac{\pi }{4}-\frac{\alpha }{2} \right)=\cos \frac{\pi }{4}\cos \frac{\alpha }{2}+\sin \frac{\pi }{4}\sin \frac{\alpha }{2}= \\
& =-\frac{\sqrt{2}}{2}.\frac{5\sqrt{34}}{34}+\frac{\sqrt{2}}{2}.\frac{3\sqrt{34}}{34}=\frac{-4\sqrt{17}}{68}= -\frac {\sqrt {17}}{17} \\
& \text{ } \\
& \text{si}{{\text{n}}^{2}}\alpha \text{+co}{{\text{s}}^{2}}\alpha \text{=1} \\
& \text{co}{{\text{s}}^{2}}\alpha =1-\frac{225}{289}=\frac{64}{289}--> cos \alpha = - \frac {8}{17} \\
& \text{cos}\frac{\alpha }{2}=-\sqrt{\frac{1+\cos \alpha }{2}}=-\sqrt{\frac{1-\frac{8}{17}}{2}}=-\sqrt{\frac{9}{34}}=-\frac{3\sqrt{34}}{34} \\
& \text{sin}\frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{2}}=-\sqrt{\frac{1+\frac{8}{17}}{2}}=\sqrt{\frac{25}{34}}=\frac{5\sqrt{34}}{34} \\
& \cos \left( \frac{\pi }{4}-\frac{\alpha }{2} \right)=\cos \frac{\pi }{4}\cos \frac{\alpha }{2}+\sin \frac{\pi }{4}\sin \frac{\alpha }{2}= \\
& =-\frac{3\sqrt{34}}{34}.\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}.\frac{5\sqrt{34}}{34}=\frac{-6\sqrt{17}+10\sqrt{17}}{34.2}= \frac {\sqrt {17}}{17} \\
\end{align}\][/tex]

Защо се получава разлика? Къде греша