3
$\begingroup$

I am working on a problem that is looking to prove $\lim_{x\to0}\frac{\sin\theta}{\theta} = 1$. At the particular point I am working on, I have to prove $1 < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta}$ can be written as $\cos\theta < \frac{\sin\theta}{\theta} < 1$. I assume that you need to multiply everything by $\cos\theta$, but I am not sure how to prove that $\frac{\cos\theta \cdot \theta}{\sin\theta} = \frac{\sin\theta}{\theta}$. The domain is $0 < \theta < \frac{\pi}{2}$.

$\endgroup$
3
  • $\begingroup$ What is your definition of $\cos\theta$ and $\sin\theta$? $\endgroup$ – detnvvp Aug 19 '13 at 23:12
  • $\begingroup$ $\cos\theta$ is the $adj$$/$$hyp$ in a triangle; $\sin\theta$ is the $opp$$/$$hyp$. In the unit circle, this would mean that sine is just the opposite side of the angle, and cosine is the adjacent because the hypotenuse is 1. Both functions have a range between -1 and 1. $\endgroup$ – user88528 Aug 19 '13 at 23:16
  • $\begingroup$ You can take a look here: math.stackexchange.com/questions/75130/… $\endgroup$ – detnvvp Aug 19 '13 at 23:20
3
$\begingroup$

You cannot prove this, because it is not true in general. Instead, take the reciprocal of everything, and reverse the direction of the inequalities, so that $$1<\frac{\theta}{\sin\theta}<\frac1{\cos\theta}$$ becomes $$1>\frac{\sin\theta}{\theta}>\cos\theta.$$ You can do this because all quantities involved are positive (for positive $\theta$ sufficiently close to $0$). You should be able to show that if $0<x<y,$ then $\frac1x>\frac1y.$ That's the result we're using, here.

$\endgroup$
4
  • 2
    $\begingroup$ Or rather, if he can prove it, then we've got a very serious problem :) $\endgroup$ – Thomas Andrews Aug 19 '13 at 23:19
  • $\begingroup$ I should have mentioned that the domain is 0 $<$ $\theta$ $<$ $\frac{\pi}{2}$. I added that information in the question. $\endgroup$ – user88528 Aug 19 '13 at 23:23
  • $\begingroup$ @user2398046: In fact, for every $0<\theta<\frac\pi2,$ we have $$\frac{\theta\cdot\cos\theta}{\sin\theta}\neq\frac{\sin\theta}\theta.$$ The interval $0<\theta<\frac\pi2$ is precisely the "positive $\theta$ sufficiently close to $0$" to which I referred in my answer, actually. $\endgroup$ – Cameron Buie Aug 19 '13 at 23:26
  • $\begingroup$ I wasn't saying your answer was wrong. I was just saying it would have helped people answering to know for sure that $\frac{\theta}{\sin\theta}$ is positive in the problem. $\endgroup$ – user88528 Aug 20 '13 at 1:15
3
$\begingroup$

You need to take the reciprocal of each term. In this case it is permitted to do that and also invert the sense of the inequality because all terms are positive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy