I'm having trouble proceeding from $$\frac{\sin(\theta)}{1+\cos(\theta)}$$ to $$\tan\left(\frac{\theta}{2}\right)$$


Consider the function $f$ defined for all $(x,y)$ such that $y \neq 0$, with the rule $$f(x,y) = \frac{y}{\sqrt{x^2+y^2}+x}$$ Show that $$f(r\cos(\theta),r\sin(\theta)) = \tan\left(\frac{\theta}{2}\right)$$

So far I've done: $$f(r\cos(\theta),r\sin(\theta)) = \frac{r\sin(\theta)}{\sqrt{r^2\cos^2(\theta) + r^2\sin^2(\theta)}+r\cos(\theta)} = \frac{r\sin(\theta)}{\sqrt{r^2}+r\cos(\theta)}\\=\frac{\sin(\theta)}{1+\cos(\theta)}$$

Using $$\cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right)-1 \implies 2\cos^2\left(\frac{\theta}{2}\right)=\cos(\theta)+1$$ We get $$f(r\cos(\theta),r\sin(\theta)) = \frac{\sin(\theta)}{2\cos^2\left(\frac{\theta}{2}\right)}$$ But I can't see how to proceed from here to the required result. Thanks in advance for any help!


Hint: The numerator can be written as $$ \sin\theta = \sin \left(2 \cdot \frac{\theta}{2}\right) = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}. $$

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    $\begingroup$ beautiful, thanks man! so quick too $\endgroup$ – Patrick Jankowski Oct 30 '18 at 8:27
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    $\begingroup$ Thank you for that, have a nice day! $\endgroup$ – MisterRiemann Oct 30 '18 at 8:29
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    $\begingroup$ In what sense is this answer a hint? $\endgroup$ – Najib Idrissi Oct 30 '18 at 17:13
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    $\begingroup$ @NajibIdrissi In the sense that I did not solve the entire problem, I only showed how the OP can proceed to solve it by giving a hint. $\endgroup$ – MisterRiemann Oct 30 '18 at 17:14
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    $\begingroup$ Well that's true. How else was I supposed to help then? $\endgroup$ – MisterRiemann Oct 30 '18 at 17:51

It is a lot easier to use some trig. identities: $\sin (\theta) =2 \sin (\theta /2) \cos (\theta /2)$ and $1+\cos \theta =2\cos ^{2} (\theta /2)$. You will immediately get the result from these two formulas.


It's possible to understand many trigonometric identities with the unit circle.

Once you understand them, it's also much easier to remember them.

You can use two unit circles (one for $a$, one for $b$) to build a rhombus:

enter image description here

The sides of this rhombus have length 1. From this diagram, you can see that:

$$\tan\left(\frac{a+b}{2}\right) = \frac{\sin\left(\frac{a+b}{2}\right)}{\cos\left(\frac{a+b}{2}\right)}=\frac{\sin\left(a\right) + \sin\left(b\right)}{\cos\left(a\right)+\cos\left(b\right)}$$

In particular, with $a = 0$ and $b = \theta$, you have:

$$\tan\left(\frac{\theta}{2}\right) = \frac{0 + \sin\left(\theta\right)}{1 + \cos\left(\theta\right)}$$

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    $\begingroup$ These function names (sin, cos, tan) should not be in italics. $\endgroup$ – Andreas Rejbrand Oct 30 '18 at 18:33
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    $\begingroup$ @AndreasRejbrand: I'm still learning Mathjax. Corrected, thanks! $\endgroup$ – Eric Duminil Oct 30 '18 at 18:38
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    $\begingroup$ This is a very helpful and useful diagram and I'll make use of it from now on when I'm stuck in long triple trig integrals haha, thanks for this! $\endgroup$ – Patrick Jankowski Oct 31 '18 at 10:00

Let $L=\frac{\sin \theta}{1+\cos \theta}$ and $R=\tan(\theta/2)$.


\begin{align} \frac{\textrm{d}L}{\textrm{d}\theta} &= \frac{\cos\theta(1+\cos\theta)-\sin\theta (-\sin\theta)}{(1+\cos \theta)^2}\\ &=\frac{1}{1+\cos\theta} \end{align} Again \begin{align} \frac{\textrm{d}^2L}{\textrm{d}\theta^2} &= -(1+\cos\theta)^{-2}(-\sin\theta)\\ &=\frac{\sin\theta}{1+\cos\theta}\frac{1}{1+\cos\theta}\\ &=L\frac{\textrm{d}L}{\textrm{d}\theta} \end{align}


\begin{align} \frac{\textrm{d}R}{\textrm{d}\theta} &= \frac{\sec^2(\theta/2)}{2}\\ &= \frac{1+\tan^2(\theta/2)}{2}\\ 2\frac{\textrm{d}R}{\textrm{d}\theta} &= 1+R^2 \end{align}


\begin{align} 2\frac{\textrm{d}^2R}{\textrm{d}\theta^2} &= 2R\frac{\textrm{d}R}{\textrm{d}\theta}\\ \frac{\textrm{d}^2R}{\textrm{d}\theta^2} &= R\frac{\textrm{d}R}{\textrm{d}\theta} \end{align}


\begin{align} L&=R\\ \frac{\sin \theta}{1+\cos \theta}&=\tan(\theta/2) \end{align}


Assume that the questioner and his friend were asked to solve a differential equation $y''=yy'$ with $y(0)=0$ and $y'(0)=1/2$.

Making the story short, he and his friend found the solution, $y=\frac{\sin\theta}{1+\cos\theta}$ and $y=\tan(\theta/2)$, respectively.

He now can proceed from $\frac{\sin\theta}{1+\cos\theta}$ to $\tan(\theta/2)$ without doubt.

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    $\begingroup$ Well not the answer to my question but useful regardless, I'll accept it $\endgroup$ – Patrick Jankowski Oct 31 '18 at 10:03

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