# How to prove $\frac{\cos\theta \cdot \theta}{\sin\theta} = \frac{\sin\theta}{\theta}$

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 $0 < \theta < \frac{\pi}{2}$.

• What is your definition of $\cos\theta$ and $\sin\theta$? – detnvvp Aug 19 '13 at 23:12
• $\cos\theta$ 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. – user88528 Aug 19 '13 at 23:16
• You can take a look here: math.stackexchange.com/questions/75130/… – detnvvp Aug 19 '13 at 23:20

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.
• I should have mentioned that the domain is 0 $<$ $\theta$ $<$ $\frac{\pi}{2}$. I added that information in the question. – user88528 Aug 19 '13 at 23:23
• @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. – Cameron Buie Aug 19 '13 at 23:26
• 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. – user88528 Aug 20 '13 at 1:15