Express [imath]cos(\frac{\alpha}{2})[/imath] and [imath]sin(\frac{\alpha}{2})[/imath] in terms of [imath]sin(\alpha)[/imath]

[Aion]

Problem.

Express [imath]cos(\frac{\alpha}{2})[/imath] and [imath]sin(\frac{\alpha}{2})[/imath] in terms of [imath]sin(\alpha)[/imath] if [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath].

Here's my attempt at solving this problem. Could someone please review my reasoning and let me know if it's correct?

[math]sin(\alpha)=2sin(\frac{\alpha}{2})cos(\frac{\alpha}{2})[/math][math]sin^2(\frac{\alpha}{2})+cos^2(\frac{\alpha}{2})=1[/math]
By addition and subtraction of the above identities we obtain

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})=\pm\sqrt{1-sin(\alpha)}[/math]
Adding and subtracting again, we have

[math]2sin(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\pm\sqrt{1-sin(\alpha)}[/math][math]2cos(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\mp\sqrt{1-sin(\alpha)}[/math]
Since

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})=\sqrt{2}(\frac{1}{\sqrt{2}}sin(\alpha)+\frac{1}{\sqrt{2}}cos(\alpha))=\sqrt{2}sin(\frac{\alpha}{2}+\frac{\pi}{4})[/math]
The expression is positive if

[math]2k\pi-\frac{\pi}{4} < \frac{\alpha}{2} < 2k\pi + \frac{3\pi}{4}[/math]
Similarly
[math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})=\sqrt{2}(\frac{1}{\sqrt{2}}sin(\alpha)-\frac{1}{\sqrt{2}}cos(\alpha))=\sqrt{2}sin(\frac{\alpha}{2}-\frac{\pi}{4})[/math]
And this positive if

[math]2k\pi+\frac{\pi}{4} < \frac{\alpha}{2} < 2k\pi + \frac{5\pi}{4}[/math]
And negative otherwise.

The condition for [imath]\alpha[/imath] as [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath] can be rewritten as [imath]\frac{-\pi}{2} \leq \alpha \leq \frac{\pi}{2}[/imath]. We can now draw the possible cases on a circle.


Case (1): For [imath]\frac{-\pi}{2} \leq \alpha \leq \frac{-\pi}{4}[/imath]

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\leq 0[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0[/math].
Therefore

[math]sin(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math][math]cos(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})[/math]
Case (2): For [imath]\frac{-\pi}{4} \leq \alpha \leq \frac{\pi}{4}[/imath]

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0[/math].

Thus
[math]sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})[/math][math]cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math]
Case (3): For [imath]\frac{\pi}{4} \leq \alpha \leq \frac{\pi}{2}[/imath]

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\geq 0[/math].

Therefore
[math]sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math] [math]cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})[/math]

 

[mario99]

Problem.

Express [imath]cos(\frac{\alpha}{2})[/imath] and [imath]sin(\frac{\alpha}{2})[/imath] in terms of [imath]sin(\alpha)[/imath] if [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath].

Here's my attempt at solving this problem. Could someone please review my reasoning and let me know if it's correct?

[math]sin(\alpha)=2sin(\frac{\alpha}{2})cos(\frac{\alpha}{2})[/math][math]sin^2(\frac{\alpha}{2})+cos^2(\frac{\alpha}{2})=1[/math]
By addition and subtraction of the above identities we obtain

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})=\pm\sqrt{1-sin(\alpha)}[/math]
Adding and subtracting again, we have

[math]2sin(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\pm\sqrt{1-sin(\alpha)}[/math][math]2cos(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\mp\sqrt{1-sin(\alpha)}[/math]
Since

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})=\sqrt{2}(\frac{1}{\sqrt{2}}sin(\alpha)+\frac{1}{\sqrt{2}}cos(\alpha))=\sqrt{2}sin(\frac{\alpha}{2}+\frac{\pi}{4})[/math]
The expression is positive if

[math]2k\pi-\frac{\pi}{4} < \frac{\alpha}{2} < 2k\pi + \frac{3\pi}{4}[/math]
Similarly
[math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})=\sqrt{2}(\frac{1}{\sqrt{2}}sin(\alpha)-\frac{1}{\sqrt{2}}cos(\alpha))=\sqrt{2}sin(\frac{\alpha}{2}-\frac{\pi}{4})[/math]
And this positive if

[math]2k\pi+\frac{\pi}{4} < \frac{\alpha}{2} < 2k\pi + \frac{5\pi}{4}[/math]
And negative otherwise.

The condition for [imath]\alpha[/imath] as [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath] can be rewritten as [imath]\frac{-\pi}{2} \leq \alpha \leq \frac{\pi}{2}[/imath]. We can now draw the possible cases on a circle.


Case (1): For [imath]\frac{-\pi}{2} \leq \alpha \leq \frac{-\pi}{4}[/imath]

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\leq 0[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0[/math].
Therefore

[math]sin(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math][math]cos(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})[/math]
Case (2): For [imath]\frac{-\pi}{4} \leq \alpha \leq \frac{\pi}{4}[/imath]

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0[/math].

Thus
[math]sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})[/math][math]cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math]
Case (3): For [imath]\frac{\pi}{4} \leq \alpha \leq \frac{\pi}{2}[/imath]

[math]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0[/math][math]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\geq 0[/math].

Therefore
[math]sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math] [math]cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})[/math]

I did not see what you have done. But isn't it straightforward to use the half angle formula:

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}[/imath]

[imath]\cos \alpha = \sqrt{1 - \sin^2 \alpha}[/imath]

Substitute this in the first expression:

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}}[/imath]

You can do the same for [imath]\displaystyle \cos \frac{\alpha}{2}[/imath]

 

[Aion]

I did not see what you have done. But isn't it straightforward to use the half angle formula:

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}[/imath]

[imath]\cos \alpha = \sqrt{1 - \sin^2 \alpha}[/imath]

Substitute this in the first expression:

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}}[/imath]

You can do the same for [imath]\displaystyle \cos \frac{\alpha}{2}[/imath]

You're right that simplifies the problem a lot! However, the interval at which we should take each sign of the expression remains to be determined.

 

[Aion]

The book I'm using has a solution but it doesn't match mine. That's why I'm trying to figure out what I did wrong.

Here it is:

[math]cos(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{2+2\sqrt{1-sin^2(\alpha)}}=-\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math][math]sin(\frac{\alpha}{2})=\pm\frac{1}{2}(\sqrt{2-2\sqrt{1-sin^2(\alpha)}}[/math]
And plus sign be taken if [imath]\frac{3\pi}{2} \leq\alpha\leq 2\pi[/imath], the minus sign if [imath]2\pi\leq \alpha \leq\frac{5\pi}{2}[/imath].

For all [imath]\alpha[/imath] in the interval [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath], we can write [math]sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1-sin\alpha}-\sqrt{1+sin\alpha}).[/math]

 

[mario99]

The book I'm using has a solution but it doesn't match mine. That's why I'm trying to figure out what I did wrong.

Here it is:

[math]cos(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{2+2\sqrt{1-sin^2(\alpha)}}=-\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})[/math][math]sin(\frac{\alpha}{2})=\pm\frac{1}{2}(\sqrt{2-2\sqrt{1-sin^2(\alpha)}}[/math]
And plus sign be taken if [imath]\frac{3\pi}{2} \leq\alpha\leq 2\pi[/imath], the minus sign if [imath]2\pi\leq \alpha \leq\frac{5\pi}{2}[/imath].

For all [imath]\alpha[/imath] in the interval [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath], we can write [math]sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1-sin\alpha}-\sqrt{1+sin\alpha}).[/math]

This is the same as I did.

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}[/imath]

To be more precise about the interval, convert it to [imath]\displaystyle \frac{\alpha}{2}[/imath].

[imath]\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath]

So, [imath]\displaystyle \sin \frac{\alpha}{2}[/imath] will be positive in this interval when [imath]\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} < \pi[/imath]

Or

[imath]\displaystyle \frac{3\pi}{2} \leq \alpha < 2\pi[/imath]

And negative when

[imath]\displaystyle 2\pi < \alpha \leq \frac{5\pi}{2}[/imath]

The same concept can be applied to [imath]\displaystyle \cos \frac{\alpha}{2}[/imath].

 

[Aion]

This is the same as I did.

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}[/imath]

To be more precise about the interval, convert it to [imath]\displaystyle \frac{\alpha}{2}[/imath].

[imath]\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath]

So, [imath]\displaystyle \sin \frac{\alpha}{2}[/imath] will be positive in this interval when [imath]\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} < \pi[/imath]

Or

[imath]\displaystyle \frac{3\pi}{2} \leq \alpha < 2\pi[/imath]

And negative when

[imath]\displaystyle 2\pi < \alpha \leq \frac{5\pi}{2}[/imath]

The same concept can be applied to [imath]\displaystyle \cos \frac{\alpha}{2}[/imath].

This is the same as I did.

[imath]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}[/imath]

To be more precise about the interval, convert it to [imath]\displaystyle \frac{\alpha}{2}[/imath].

[imath]\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath]

So, [imath]\displaystyle \sin \frac{\alpha}{2}[/imath] will be positive in this interval when [imath]\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} < \pi[/imath]

Or

[imath]\displaystyle \frac{3\pi}{2} \leq \alpha < 2\pi[/imath]

And negative when

[imath]\displaystyle 2\pi < \alpha \leq \frac{5\pi}{2}[/imath]

The same concept can be applied to [imath]\displaystyle \cos \frac{\alpha}{2}[/imath].

Thanks for the help, Mario. I overcomplicated things unnecessarily. By converting the interval for [imath]\frac{\alpha}{2}[/imath] everything makes sense .

Since we are working in the interval [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath], [imath]cos(\alpha) \geq 0[/imath], and so [imath]cos(\alpha)=\sqrt{1-sin^2(\alpha)}[/imath].

Hence [math]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}[/math]
And [imath]sin(\frac{\alpha}{2})[/imath] is positive if [imath]\frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \pi[/imath] and negative if [imath]\pi \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath].

Similarly, since [imath]cos(\frac{\alpha}{2})[/imath] is negative in the interval [imath]\frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath]

[math]\displaystyle \cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \sqrt{1 - \sin^2 \alpha}}{2}} = - \sqrt{\frac{2 + 2\sqrt{1 - \sin^2 \alpha}}{4}} = -\frac{1}{2}\sqrt{2 +2\sqrt{1 - \sin^2 \alpha}}[/math]

 

[Aion]

Thanks for the help, Mario. I overcomplicated things unnecessarily. By converting the interval for [imath]\frac{\alpha}{2}[/imath] everything makes sense .

Since we are working in the interval [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath], [imath]cos(\alpha) \geq 0[/imath], and so [imath]cos(\alpha)=\sqrt{1-sin^2(\alpha)}[/imath].

Hence [math]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}[/math]
And [imath]sin(\frac{\alpha}{2})[/imath] is positive if [imath]\frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \pi[/imath] and negative if [imath]\pi \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath].

Similarly, since [imath]cos(\frac{\alpha}{2})[/imath] is negative in the interval [imath]\frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath]

[math]\displaystyle \cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \sqrt{1 - \sin^2 \alpha}}{2}} = - \sqrt{\frac{2 + 2\sqrt{1 - \sin^2 \alpha}}{4}} = -\frac{1}{2}\sqrt{2 +2\sqrt{1 - \sin^2 \alpha}}[/math]

Also

[math]\sin \frac{\alpha}{2} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}=\pm\frac{1}{2}(\sqrt{(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})^2}=\pm\frac{1}{2}(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})[/math]

 

[mario99]

Thanks for the help, Mario. I overcomplicated things unnecessarily. By converting the interval for [imath]\frac{\alpha}{2}[/imath] everything makes sense .

Since we are working in the interval [imath]\frac{3\pi}{2} \leq \alpha \leq \frac{5\pi}{2}[/imath], [imath]cos(\alpha) \geq 0[/imath], and so [imath]cos(\alpha)=\sqrt{1-sin^2(\alpha)}[/imath].

Hence [math]\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}[/math]
And [imath]sin(\frac{\alpha}{2})[/imath] is positive if [imath]\frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \pi[/imath] and negative if [imath]\pi \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath].

Similarly, since [imath]cos(\frac{\alpha}{2})[/imath] is negative in the interval [imath]\frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}[/imath]

[math]\displaystyle \cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \sqrt{1 - \sin^2 \alpha}}{2}} = - \sqrt{\frac{2 + 2\sqrt{1 - \sin^2 \alpha}}{4}} = -\frac{1}{2}\sqrt{2 +2\sqrt{1 - \sin^2 \alpha}}[/math]

It seems that you got the idea.

Also

[math]\sin \frac{\alpha}{2} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}=\pm\frac{1}{2}(\sqrt{(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})^2}=\pm\frac{1}{2}(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})[/math]

I did not check this one, but since you worked in this problem a lot, I trust in your calculations to be correct.

 

[Aion]

Also

[math]\sin \frac{\alpha}{2} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}=\pm\frac{1}{2}(\sqrt{(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})^2}=\pm\frac{1}{2}(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})[/math]

No that's quite right, all we can conclude is that its equal to the absolute value. [math]\sin \frac{\alpha}{2} = \pm\frac{1}{2}\left(\sqrt{\left(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)}\right)^2}\right)=\pm\frac{1}{2}\left| \sqrt{1-sin(\alpha})-\sqrt{1+sin(\alpha)}\right|[/math]
However by the above reasoning in my original post for any [imath]\frac{\alpha}{2} \in (\frac{3\pi}{4},\frac{5\pi}{4})[/imath] we know that

[imath]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\leq 0[/imath] and [imath]sin(\frac{\alpha}{2})-sin(\frac{\alpha}{2})\geq0[/imath], which implies that [imath]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})=-\sqrt{1+\sin(\alpha)}[/imath] and [imath]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})=\sqrt{1-\sin(\alpha)},[/imath] hence [imath]2sin(\frac{\alpha}{2})=\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)}[/imath] for all [imath]\alpha \in (\frac{3\pi}{2},\frac{5\pi}{2})[/imath].

 

[mario99]

No that's quite right, all we can conclude is that its equal to the absolute value. [math]\sin \frac{\alpha}{2} = \pm\frac{1}{2}\left(\sqrt{\left(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)}\right)^2}\right)=\pm\frac{1}{2}\left| \sqrt{1-sin(\alpha})-\sqrt{1+sin(\alpha)}\right|[/math]
However by the above reasoning in my original post for any [imath]\frac{\alpha}{2} \in (\frac{3\pi}{4},\frac{5\pi}{4})[/imath] we know that

[imath]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\leq 0[/imath] and [imath]sin(\frac{\alpha}{2})-sin(\frac{\alpha}{2})\geq0[/imath], which implies that [imath]sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})=-\sqrt{1+\sin(\alpha)}[/imath] and [imath]sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})=\sqrt{1-\sin(\alpha)},[/imath] hence [imath]2sin(\frac{\alpha}{2})=\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)}[/imath] for all [imath]\alpha \in (\frac{3\pi}{2},\frac{5\pi}{2})[/imath].

I agree.



Bravo to catch where was your mistake!