Express cos(α2)cos(\frac{\alpha}{2}) and sin(α2)sin(\frac{\alpha}{2}) in terms of sin(α)sin(\alpha)

[Aion]

Problem.

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

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

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

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

$$2sin(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\pm\sqrt{1-sin(\alpha)}$$$$2cos(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\mp\sqrt{1-sin(\alpha)}$$
Since

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

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

$$2k\pi+\frac{\pi}{4} < \frac{\alpha}{2} < 2k\pi + \frac{5\pi}{4}$$
And negative otherwise.

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


Case (1): For $\frac{-\pi}{2} \leq \alpha \leq \frac{-\pi}{4}$

$$sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\leq 0$$$$sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0$$.
Therefore

$$sin(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})$$$$cos(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})$$
Case (2): For $\frac{-\pi}{4} \leq \alpha \leq \frac{\pi}{4}$

$$sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0$$$$sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0$$.

Thus
$$sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})$$$$cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})$$
Case (3): For $\frac{\pi}{4} \leq \alpha \leq \frac{\pi}{2}$

$$sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0$$$$sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\geq 0$$.

Therefore
$$sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})$$ $$cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})$$

 

[mario99]

Problem.

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

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

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

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

$$2sin(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\pm\sqrt{1-sin(\alpha)}$$$$2cos(\frac{\alpha}{2})=\pm\sqrt{1+sin(\alpha)}\mp\sqrt{1-sin(\alpha)}$$
Since

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

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

$$2k\pi+\frac{\pi}{4} < \frac{\alpha}{2} < 2k\pi + \frac{5\pi}{4}$$
And negative otherwise.

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


Case (1): For $\frac{-\pi}{2} \leq \alpha \leq \frac{-\pi}{4}$

$$sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\leq 0$$$$sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0$$.
Therefore

$$sin(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})$$$$cos(\frac{\alpha}{2})=-\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})$$
Case (2): For $\frac{-\pi}{4} \leq \alpha \leq \frac{\pi}{4}$

$$sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0$$$$sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\leq 0$$.

Thus
$$sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})$$$$cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})$$
Case (3): For $\frac{\pi}{4} \leq \alpha \leq \frac{\pi}{2}$

$$sin(\frac{\alpha}{2})+cos(\frac{\alpha}{2})\geq 0$$$$sin(\frac{\alpha}{2})-cos(\frac{\alpha}{2})\geq 0$$.

Therefore
$$sin(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}+\sqrt{1-sin(\alpha)})$$ $$cos(\frac{\alpha}{2})=\frac{1}{2}(\sqrt{1+sin(\alpha)}-\sqrt{1-sin(\alpha)})$$

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

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}$

$\cos \alpha = \sqrt{1 - \sin^2 \alpha}$

Substitute this in the first expression:

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}}$

You can do the same for $\displaystyle \cos \frac{\alpha}{2}$

 

[Aion]

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

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}$

$\cos \alpha = \sqrt{1 - \sin^2 \alpha}$

Substitute this in the first expression:

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}}$

You can do the same for $\displaystyle \cos \frac{\alpha}{2}$

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:

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

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

 

[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:

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

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

This is the same as I did.

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}$

To be more precise about the interval, convert it to $\displaystyle \frac{\alpha}{2}$.

$\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}$

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

Or

$\displaystyle \frac{3\pi}{2} \leq \alpha < 2\pi$

And negative when

$\displaystyle 2\pi < \alpha \leq \frac{5\pi}{2}$

The same concept can be applied to $\displaystyle \cos \frac{\alpha}{2}$.

 

[Aion]

This is the same as I did.

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}$

To be more precise about the interval, convert it to $\displaystyle \frac{\alpha}{2}$.

$\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}$

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

Or

$\displaystyle \frac{3\pi}{2} \leq \alpha < 2\pi$

And negative when

$\displaystyle 2\pi < \alpha \leq \frac{5\pi}{2}$

The same concept can be applied to $\displaystyle \cos \frac{\alpha}{2}$.

This is the same as I did.

$\displaystyle \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \sqrt{1 - \sin^2 \alpha}}{2}} = \pm \sqrt{\frac{2 - 2\sqrt{1 - \sin^2 \alpha}}{4}} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}$

To be more precise about the interval, convert it to $\displaystyle \frac{\alpha}{2}$.

$\displaystyle \frac{3\pi}{4} \leq \frac{\alpha}{2} \leq \frac{5\pi}{4}$

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

Or

$\displaystyle \frac{3\pi}{2} \leq \alpha < 2\pi$

And negative when

$\displaystyle 2\pi < \alpha \leq \frac{5\pi}{2}$

The same concept can be applied to $\displaystyle \cos \frac{\alpha}{2}$.

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

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

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

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

$$\displaystyle \cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \sqrt{1 - \sin^2 \alpha}}{2}} = - \sqrt{\frac{2 + 2\sqrt{1 - \sin^2 \alpha}}{4}} = -\frac{1}{2}\sqrt{2 +2\sqrt{1 - \sin^2 \alpha}}$$

 

[Aion]

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

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

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

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

$$\displaystyle \cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \sqrt{1 - \sin^2 \alpha}}{2}} = - \sqrt{\frac{2 + 2\sqrt{1 - \sin^2 \alpha}}{4}} = -\frac{1}{2}\sqrt{2 +2\sqrt{1 - \sin^2 \alpha}}$$

Also

$$\sin \frac{\alpha}{2} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}=\pm\frac{1}{2}(\sqrt{(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})^2}=\pm\frac{1}{2}(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})$$

 

[mario99]

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

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

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

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

$$\displaystyle \cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \sqrt{1 - \sin^2 \alpha}}{2}} = - \sqrt{\frac{2 + 2\sqrt{1 - \sin^2 \alpha}}{4}} = -\frac{1}{2}\sqrt{2 +2\sqrt{1 - \sin^2 \alpha}}$$

It seems that you got the idea.

Also

$$\sin \frac{\alpha}{2} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}=\pm\frac{1}{2}(\sqrt{(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})^2}=\pm\frac{1}{2}(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})$$

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

$$\sin \frac{\alpha}{2} = \pm\frac{1}{2}\sqrt{2 - 2\sqrt{1 - \sin^2 \alpha}}=\pm\frac{1}{2}(\sqrt{(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})^2}=\pm\frac{1}{2}(\sqrt{1-sin(\alpha)}-\sqrt{1+sin(\alpha)})$$

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

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

 

[mario99]

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

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

I agree.



Bravo to catch where was your mistake!