F X 6x Cos X Where is X Continuous

2. Prove $\cos(x)$ is continuous

Prove $\cos(x)$ is continuous

Use that $|\sin u|\leq\min\{|u|,1\}$, then $2\left|\sin\left(\dfrac{x+a}{2}\right)\right|\cdot\left|\sin\left(\dfrac{x-a}{2}\right)\right|\leq 2\left|\dfrac{x-a}{2}\right|=|x-a|<\epsilon$ if we take $\delta=\epsilon$.

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Comments

• I'm stuck at a particular step and could use some help.

  By definition, a function is continuous at $x=a$ iff $\lim_{x \to a} f(x) = f(a)$.

  So I assume to prove $\cos(x)$ is continuous we must use the definition of a limit to show that:

  $$\lim_{x \to a} \cos(x) = \cos(a) \iff \forall \epsilon > 0, \exists \delta>0 : 0 < |x - a| < \delta \implies |\cos(x)-\cos(a)| < \epsilon$$

  From some triangle proofs it can be shown that:

  $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$

  $\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha) \sin(\beta)$

  Subtract the two equations:

  $$\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2\sin(\alpha) \sin(\beta)$$

  Let $x = \alpha + \beta$ and $a = \alpha - \beta$.

  Then $\alpha = \frac{x+a}{2}$ and $\beta = \frac{x-a}{2}$, implying:

  $$\cos(x) - \cos(a) = -2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)$$

  So

  $$\left|\cos(x)-\cos(a)\right| = 2\left|\sin\left(\frac{x+a}{2}\right)\right| \cdot \left|\sin\left(\frac{x-a}{2}\right)\right|$$

  I feel like I am close because I was able to change it so at least we have an $x-a$ term but now I'm not sure where to go from here.

  • You can ignore the first factor, $<1$, then you need the inequality $|\sin t|<|t|$.

  • @YvesDaoust Proving $|\sin t| < |t|$ is the hard part, unable to find a good way without circular arguments somewhere

  • see my comment in the accepted answer.

• I see that $|\sin u| \leq 1$ but where is the $\leq |u|$ piece coming from and how does any of that imply we can drop out the $\left| \sin\left(\frac{x+a}{2}\right)\right|$ term and then drop the $\sin$ from the other?

• Then $u=(x+a)/2$, $|\sin u|\leq 1$.

• That inequality you can argue case by case. First that $u\geq 0$, and consider $\varphi(u)=u-\sin u$ and use Mean Value Theorem or something like that.

• So $\left| \sin\left(\frac{x+a}{2}\right)\right| \leq 1$ means $2\left|\sin\left(\dfrac{x+a}{2}\right)\right|\cdot\left|\sin\left(\dfrac{x-a}{2}\right)\right| \leq 2\left|\sin\left(\dfrac{x-a}{2}\right)\right|$, now we need some way to drop out the sin by showing $|\sin(u)| \leq |u|$?

• Is there any other way to show |$\sin(u)| \leq |u|$ other than MVT?

• Another cheating way to finish your proof is that, we may not need such a good inequality that $|\sin u|\leq|u|$. Rather, if we consider $\varphi(u)=\sin u/u$ for $u\in[-1,1]$, $u\ne 0$, and $\varphi(0)=1$, then $\varphi$ is continuous, and hence it has maximum because $[-1,1]$ is a closed interval, so you get $|\sin u|/|u|\leq C$ for some constant $C>0$, and hence $|\sin u|\leq C|u|$, this is sufficient for the proof of your continuity question.

• I'm trying to show cos is continuous in order to help prove $\lim_{h \to 0} \frac{\sin(x)}{x} = 1$ as part of a geometric proof, is this circular with that?

• Then it is a very hard mission. Yes, in some sense it is circular. But the point is, how do you define $\sin$? I am wondering how to derive $|\sin u|\leq|u|$ without recourse to Mean Value Theorem. You know, some author defines $\sin$ or $\cos$ to be the entire function on the complex plane satisfying some functional equation, so the continuity is clear, and certain power series expansion would lead immediately to the fact that $\sin x/x\rightarrow 1$ as $x\rightarrow 0$.

• But to get the power series expansion don't you need to compute Taylor series which relies on finding $f^{(k)}(a)$ terms, i.e. derivatives of $\sin(x)$, which in turn ultimately involves terms of form $\lim_{h \to 0} \frac{\sin(x)}{x}$ when you need to prove those as well?

• So I have mentioned that some author defines $\sin$ to be like $x-x^{3}/3!+\cdots$, there is no need to do Taylor series anymore, because that is the definition. But that is not a pleasure definition of course.

• By the way do you really need the fact that $\cos$ being continuous to derive that $\sin x/x\rightarrow 1$ by geometric proof?

• @user525966: notice that the limit of $\sin u/u$ is depending on the angular unit and it cannot be obtained from trigonometric identities. In a way, it is the definition of the radian.

• The geometric proof I have involves squeeze theorem on $ \lim_{x\to 0} \cos(x) \leq \lim_{x\to 0} \sin(x)/x \leq \lim_{x\to 0} 1$ so I'd need to show each piece is continuous before using limit definition.

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