How does $\sqrt{\frac{2(cos(x)-1)}{cos(x)}} \approx x$?

At first I thought it was an example of binomial approximation but I was unable to approximate is to just $x$. I could only get $$\left(\frac{2cos(x)}{cos(x)}-\frac{2}{cos(x)}\right)^{\frac{1}{2}}$$ $$\left(2-\frac{2}{cos(x)}\right)^{\frac{1}{2}} \approx 2-\frac{1}{cos(x)}$$

Am I using the binomial approximation theorem correctly? Or is there another way of approximating this value?

  • $\begingroup$ $\cos(x)= 1- x^2/2 + \cdots$ $\endgroup$ – gammatester Oct 29 '17 at 20:11
  • $\begingroup$ Is it in a neighbourhood of $0$? Then the radicand is negative. $\endgroup$ – Bernard Oct 29 '17 at 20:11
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    $\begingroup$ I think you have a sign problem. For $x$ near $0$ your expression is imaginary. Perhaps you meant $1-\cos x$ in the numerator? $\endgroup$ – lulu Oct 29 '17 at 20:14
  • $\begingroup$ @lulu unfortunatly no, it is $cos(x)-1$ $\endgroup$ – Niamh O'SS Oct 29 '17 at 20:17
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    $\begingroup$ No, really, what you wrote down is not correct. Try it for small positive values of $x$. With the sign change I proposed you get $|x|$ for small $x$. $\endgroup$ – lulu Oct 29 '17 at 20:25

Near $0$?

Well, first of all you have a sign error. For small $x\neq 0$ we have $0<\cos x<1$ so the expression inside the radical is clearly negative. Accordingly, I believe you are trying to show that $$\sqrt {2\,\frac {1-\cos x}{\cos x}}\approx |x|$$ for $x$ near $0$. At least, this has the advantage of being true.

To see it, note that $$\cos x \approx 1 -\frac {x^2}2\implies \frac 1{\cos x}\approx1+\frac {x^2}2$$ Here the second approximation follows from the first by means of first stage of the Geometric series $\frac 1{1-z}\approx 1+z$.

It follows that the modified expression is approximately $$\sqrt {2\times \left(1+\frac {x^2}2\right)\times \frac {x^2}2}\approx \sqrt {x^2+\frac {x^4}2}\approx |x|$$

  • $\begingroup$ The jump from$$\cos x\approx\dots$$to$$\frac1{\cos x}\approx\dots$$is probably not immediately clear. $\endgroup$ – Simply Beautiful Art Oct 29 '17 at 21:13
  • $\begingroup$ @SimplyBeautifulArt Thanks, I'll add a sentence. $\endgroup$ – lulu Oct 29 '17 at 21:17

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