algebra precalculus - If $450^circ

The problem says:

If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$

I solved it in the following way:

$$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:

A) $0.6$

B) $\frac{4}{5}$

C) $-\frac{4}{5}$

D) $-0.6$

E) $0.96$

I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.

Answer

$$\frac {7^2}{24^2}=\frac {1+\cos 2\alpha}{1-\cos 2\alpha} \implies$$ $$\implies 7^2(1-\cos \alpha)=24^2(1+\cos \alpha)\implies$$ $$\implies 7^2- 7^2\cos 2\alpha = 24^2+ 24^2 \cos 2\alpha\implies$$ $$\implies 7^2-24^2= (7^2+24^2)\cos 2\alpha =25^2 \cos 2\alpha\implies$$ $$\implies -527=625\cos 2\alpha .$$

The missing negative sign on the LHS of the above line is your first error.

Your second error is writing $\cos \frac {\alpha}{2}=\sqrt {\frac {1+\cos \alpha}{2}}\;.$ We have $|\cos \frac {\alpha}{2}|=\sqrt { \frac {1+\cos \alpha}{2} }\;.\;$.... If $450^o<\alpha<340^o$ then $225^o<\frac {\alpha}{2}<270^o,$ implying $\cos \frac {\alpha}{2}<0.$

In general if $\cot x=\frac {a}{b}$ then the proportion of $\cos^2 x$ to $\sin^2 x$ is $a^2$ to $b^2$, so let $\cos^2 x=a^2y$ and $\sin^2 x=b^2y$. Since $1=\cos^2 x +\sin^2 x$, we have $y=a^2+b^2$, so $\cos^2 x =\frac {a^2}{a^2+b^2}$ and $\sin^2 x =\frac {b^2}{a^2+b^2}\;$ and therefore $\;|\cos x|=\frac {|a|}{\sqrt {a^2+b^2}}$ and $|\sin x|=\frac {|b|}{\sqrt {a^2+b^2}}.$

So if $\cot \alpha =\frac {-7}{24}$ and $\cos \alpha<0$ then $\cos \alpha =-\frac {7}{\sqrt {7^2+24^2}}=-\frac {7}{25}.$