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Given $\cot{\theta} = \frac{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}{\sqrt{1+\sin(x)}-\sqrt{1-\sin(x)}}$ I have to find its differential coefficient w.r.t $x$ i.e. $\dfrac{d \theta }{dx}$.

Now I can find it in the following two ways:

(1). When I write $\sqrt{1-\sin(x)}=\cos(x/2)-\sin(x/2)$, I get $\cot(\theta) = \frac{\cos(x/2) + \sin(x/2) + \cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2) - \cos(x/2) + \sin(x/2)} = \cot(x/2)$ $\theta=x/2 \implies \frac{d \theta}{dx} = \frac12$

(2). When I write $\sqrt{1-\sin(x)}=\sin(x/2)-\cos(x/2)$, I get $\cot(\theta) = \frac{\cos(x/2) + \sin(x/2) - \cos(x/2) + \sin(x/2)}{\cos(x/2) + \sin(x/2) + \cos(x/2) - \sin(x/2)} = \tan(x/2)$ $\theta=\pi/2 - x/2 \implies \frac{d \theta}{dx} = -\frac12$ Which of the above two answers is correct and why? Please help me know it. Thanks.

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    thank you very much Marvis for your help and also for giving the above link2012-05-20

2 Answers 2

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Neither. When you deal in half-angle formulas, you have to be careful about sign.

Notice that $\sqrt{1-\sin x}$ is always positive, while $\sin (x/2)-\cos(x/2)$ and $\cos(x/2)-\sin(x/2)$ may be positive or negative depending on $x$. The correct statement is $\sqrt{1-\sin x}=|\cos(x/2)-\sin(x/2)|$; from there, you'll need to do some case analysis to compute the derivative you want.

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    Thanks for your answer Micah2012-05-20
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Micah has a good answer as to why you have an inconsistency. To make your task easier, consider rationalizing your denominator:

$\begin{align} \frac{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}{\sqrt{1+\sin(x)}-\sqrt{1-\sin(x)}}\cdot\frac{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}&=\frac{2+2\sqrt{1-\sin^2(x)}}{2\sin(x)}\\ \cot(\theta)&=\frac{1+|\cos(x)|}{\sin(x)} \end{align}$

Now the chain rule, quotient rule, and the derivative of $|x|$ (which is $|x|/x$) will give you $\frac{d\theta}{dx}$ in terms of $x$ and $\csc^2(\theta)$ without dealing with any more square roots or half-angle formulas. If you like, $\csc^2(\theta)$ can be subbed out for $1+\cot^2(\theta)$, which can be written in terms of $x$ alone.

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    Thanks to you too alex.jordan2012-05-20