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$$y=\sqrt\frac{1+\cosθ}{1-\cosθ}$$ my professor said that the answer is $$y'=\frac{1}{\cosθ-1}$$ she said use half angle formula but I just end up with $\frac{(-2\sinθ)\sqrt{(1-\cosθ)(1+\cosθ)}}{2(1-\cosθ)^2(1+\cosθ)}$ I used the quotient rule. I also know that ${(1-\cosθ)^2}$ can be $\sinθ^2$ but i try to apply my identities but it's still wrong.

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    $\begingroup$ $1-\cos^2 \theta=\sin \theta$ and NOT $(1-\cos \theta)^2=\sin ^2 \theta$. Please see that. $\endgroup$
    – creative
    Sep 19, 2014 at 9:42
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    $\begingroup$ @Abstraction You made a typo ;-) $\endgroup$ Sep 19, 2014 at 10:14
  • $\begingroup$ @Mickey Is there any restriction on the domain of $\theta$? $\endgroup$ Sep 24, 2014 at 14:16
  • $\begingroup$ Your professor is wrong. $\endgroup$
    – egreg
    Sep 24, 2014 at 14:46

6 Answers 6

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$$y=\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}=\sqrt{\frac{2\cos^2\theta/2}{2\sin^2\theta/2}}=|\cot\theta/2|$$

Also,

$$\frac{\mathrm{d}}{\mathrm{d}\theta}\cot(\theta/2)=-\frac12\frac1{\sin^2\theta/2}$$

Hence, when $\cos\theta/2\neq0$ and $\sin\theta/2\neq 0$, that is when $\theta\notin\pi\Bbb Z$,

$$\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\frac12\frac{\mathrm{sign}(\cot \theta/2)}{\sin^2\theta/2}$$

And since $2\sin^2\theta/2=1-\cos\theta$,

$$\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\frac{\mathrm{sign}(\cot \theta/2)}{1-\cos\theta}$$

And yet one more simplification, when $\sin\theta/2\neq 0$,

$$\mathrm{sign}(\cot \theta/2)=\mathrm{sign}(\cos (\theta/2)\sin (\theta/2))=\mathrm{sign}(\sin\theta)$$

So, still for $\theta\notin\pi\Bbb Z$

$$\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\frac{\mathrm{sign}(\sin\theta)}{1-\cos\theta}$$

Since it may not be absolutely clear, here is a plot of $y$, to show how $y$ differs from $\cot\theta/2$. Notice the angle when the curve reaches $0$: $y$ is thus not differentiable here.

enter image description here

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We have

$\dfrac{(-2\sinθ)\sqrt{(1-\cosθ)(1+\cosθ)}}{2(1-\cosθ)^2(1+\cosθ)}=-\dfrac{2\sin\theta|\sin\theta|}{2(1-\cos\theta)(1-\cos^2\theta)}$

=-sign$(\sin\theta)\dfrac1{1-\cos\theta}$

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  • $\begingroup$ can you expand the solution please, Did you use the quotient rule? $\endgroup$
    – Mickey
    Sep 19, 2014 at 9:46
  • $\begingroup$ @Mickey, I've started where you left of $\endgroup$ Sep 19, 2014 at 9:46
  • $\begingroup$ @Mickey, $$(1-\cosθ)(1+\cosθ)=\sin^2\theta$$ $$\implies\sqrt{(1-\cosθ)(1+\cosθ)}=|\sin\theta|$$ $\endgroup$ Sep 19, 2014 at 9:48
  • $\begingroup$ I got the numerator but how did you get rid of the denominator $(1+cosθ)$ ? $\endgroup$
    – Mickey
    Sep 19, 2014 at 9:51
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    $\begingroup$ @Mickey In numerator, it's NOT $\sin^2\theta$, it's $\sin\theta|\sin\theta|=\mathrm{sign}(\sin \theta)\sin^2\theta$. Absolutely not the same thing. $\endgroup$ Sep 19, 2014 at 10:21
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$\dfrac{1+\cos\theta}{1-\cos\theta}=\dfrac{(1+\cos\theta)(1-\cos\theta)}{(1-\cos\theta)^2}=\left(\dfrac{\sin\theta}{1-\cos\theta}\right)^2$

So,$\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}}=$sign of $(\sin\theta)\cdot\dfrac{\sin\theta}{1-\cos\theta}$

Now apply quotient Rule

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  • $\begingroup$ @Mickey, Here is another method $\endgroup$ Sep 19, 2014 at 10:00
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The half angle formulas are often useful: $$\cos^2(x)=\dfrac{1+\cos(2x)}{2} \text{ and } \sin^2(x)=\dfrac{1-\cos(2x)}{2}$$

$$y=\sqrt{\dfrac{(1+\cos\theta)/2}{(1-\cos\theta)/2}}=\sqrt{\dfrac{\cos^2(\theta/2)}{\sin^2(\theta/2)}}=\sqrt{\cot^2(\theta/2)}=\left|\cot(\theta/2)\right|$$

The derivative of $\cot(x)$ is $(\cot(x))'=-(1+\cot^2(x))=\dfrac{-1}{\sin^2x}$

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  • $\begingroup$ I think $y'= -csc(x)^2$ of $cot(x)$ ? so it's $-1/sin2x$ right? anyway got the answer from lab thank you too! gonna try this out with the other problems $\endgroup$
    – Mickey
    Sep 19, 2014 at 10:00
  • $\begingroup$ Yes, you're right. Sorry for the mistake. $\endgroup$ Sep 19, 2014 at 10:06
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$$ y^2=\frac {1+\cos\theta}{1-\cos\theta}.$$

Differentiating gives

$$\frac {d}{d\theta} y^2=\frac {d}{d\theta} \frac {1+\cos\theta}{1-\cos\theta} = -\frac{2\sin\theta}{(1-\cos\theta)^2}. $$

Using the chain rule gives for the left hand side

$$2y\frac {dy}{d\theta}. $$ Hence, $$\frac {dy}{d\theta}=-\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{(1-\cos\theta)^2},$$ a plot of which agrees with that of Jean-Claude Arbaut's answer.

Assuming this to be the case then retrospectively it would seem that

$$\text{sgn}(\sin\theta) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}.$$

Interesting.


Out of curiosity, let $\theta= \arcsin x$. Then

$$\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}=\frac{x}{\sqrt{1-\sqrt{1-x^2}} \sqrt{\sqrt{1-x^2}+1}}=\frac{x}{\sqrt{x^2}}=\frac{x}{\mid x\mid}=\text{sgn}(x) .$$

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Using half angle formula $$\cos\theta =2\cos^2(\theta/2)-1$$ and $$\cos\theta =1-2\sin^2(\theta/2).$$

So,

\begin{equation*} \begin{split} y & = \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\\ & = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}}\\ & = \frac{\cos(\theta/2)}{\sin(\theta/2)} \quad \text{ for } 0^{\circ}\le \frac{\theta}{2}\le 90^\circ \end{split} \end{equation*} Now, by using quotient rule $y'=\frac{-1/2}{\sin^2(\theta /2)}$, using half angle formula again we have $$y'=\frac{-1}{1-\cos\theta}.$$


Your professor's answer is correct provided that the denominator is $1-\cos\theta$ not $\cos\theta -1.$

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  • $\begingroup$ No, it's wrong. $\sqrt{x^2}=|x|$, not $x$. $\endgroup$ Sep 19, 2014 at 10:37
  • $\begingroup$ @Jean-ClaudeArbaut what can you say about $(-x)^2=x^2$? What we have under that square root is positive, whatever the value of $\cos(\theta/2)$. $\endgroup$ Sep 19, 2014 at 10:47
  • $\begingroup$ You write the square root equals $$\frac{\cos\theta/2}{\sin\theta/2}$$ Which is of course not always positive. How can a square root be negative, exactly? I repeat, $$\sqrt{x^2}=|x|$$ And $|x|\neq x$ when $x<0$. $\endgroup$ Sep 19, 2014 at 10:50
  • $\begingroup$ @Jean-ClaudeArbaut can you explain the down vote? $\endgroup$ Sep 19, 2014 at 10:52
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    $\begingroup$ I just explained. You are writing that $y=\cot \theta/2$, though $y$ is always positive or zero, by definition. $\endgroup$ Sep 19, 2014 at 10:53

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