If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong?

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.

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

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 $\cos \alpha =-\frac {7}{\sqrt {7^2+24^2}}=-\frac {7}{25}.$

It's probably easier to use $$\cos^2\alpha = \frac{1}{\sec^2\alpha} = \frac{1}{1+\tan^2\alpha} = \frac{1}{1+\frac{1}{\cot^2\alpha}}$$ to find $\cos\alpha$, this gives $\cos\alpha=\pm\frac7{25}$. The given range for $\alpha$ tells which of these applies.

I'd suggest you check your calculation of $\cos\alpha$.

Hint:

$$\cot\alpha=\frac{1-\tan^2\frac\alpha2}{2\tan\frac\alpha2}=\frac{1-t^2}{2t}$$ or

$$t^2+2\cot\alpha\ t-1=0,$$

or

$$t=-\cot\alpha\pm\sqrt{1-\cot^2\alpha}=-\frac34,\frac43.$$

Then $$\cos\frac\alpha2=\pm\frac1{\sqrt{1+t^2}}=\pm\frac1{\sqrt{2-2\cot\alpha\ t}}=\pm\frac45,\pm\frac35.$$

Then given range of $\alpha$ tells you that

$$-\frac1{\sqrt2}<\cos\frac\alpha2<0.$$

$$\cot\alpha = -\frac{7}{24} \implies \tan\alpha = -\frac{24}{7}$$ Since $1 + \tan^2\alpha = \sec^2\alpha$, $$\sec^2\alpha = 1 + \left(-\frac{24}{7}\right)^2 = 1 + \frac{576}{49} = \frac{625}{49} \implies |\sec\alpha| = \frac{25}{7}$$ Observe that $450^\circ < \alpha < 540^\circ \implies \cos\alpha < 0$. Hence, $$\sec\alpha = -\frac{25}{7} \implies \cos\alpha = -\frac{7}{25}$$ Since $450^\circ < \alpha < 540^\circ$, $225^\circ < \frac{\alpha}{2} < 270^\circ$. Thus, $\cos(\frac{\alpha}{2}) < 0$. Therefore, $$\cos\left(\frac{\alpha}{2}\right) = -\sqrt{\frac{1 + \cos\alpha}{2}} = -\sqrt{\frac{1 - \frac{7}{25}}{2}} = -\sqrt{\frac{\frac{18}{25}}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$

$$\cot x =\frac{-7}{24}=\frac{\cos x}{\sin x}$$ $$\frac{49}{576}=\frac{\cos^2 x}{1-\cos^2 x}$$ It gives $$\cos x=\frac{-7}{25}$$ And $$\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\frac{3}{5}=0.6$$ Hence (A) is correct

EDIT:

After concerning with experts , I concluded that $\cos \frac{x}{2}$ is negative and hence (D) is correct

• This is wrong, $\cos\frac\alpha2<0$.
– user65203
Aug 14 '17 at 8:52
• @YvesDaoust $90°\le x \le 180°$ implies $45°\le \frac{x}{2} \le 90°$ imples $\cos \frac{x}{2}$ is positive ... what is wrong with this ?? Aug 14 '17 at 8:58
• $D)$ is correct Aug 14 '17 at 9:02
• But isn't $450°=360°+90°\equiv 90°$ and similarly $540°\equiv 180°$ which gives $\cos (\frac{x}{2})>0$ @N.F.Taussig?? Aug 14 '17 at 9:03
• No. Since $450^\circ < \alpha < 540^\circ$, we may conclude that $\alpha$ is a second-quadrant angle. However, that does not tell us that $\frac{\alpha}{2}$ is a first-quadrant angle. In fact, $225^\circ < \frac{\alpha}{2} < 270^\circ$, so $\frac{\alpha}{2}$ is a third-quadrant angle. Aug 14 '17 at 9:10