# If $\{x_n\}$ is a cauchy sequence then show that $\{\cos x_n\}$ is also a cauchy sequence.

If $\{x_n\}$ is a Cauchy sequence then show that $\{\cos x_n\}$ is also a Cauchy sequence.

Let $y_n=\cos x_n$ then for $m>n$, $|y_m-y_n|\leq |\cos x_m-\cos x_n|\leq |\cos x_m|+|\cos x_n|\leq 1+1=2$

Please correct my proof using the definition of Cauchy sequence.

• "correct my proof" I can't find any. – drhab May 10 '16 at 11:21
• I don't think your proof is correct. It's better to use $$\cos x_m-\cos x_n=2 \sin \left(\frac{x_n-x_m}{2} \right) \sin \left(\frac{x_n+x_m}{2} \right)$$ which becomes arbitrarily small with $x_n-x_m$ arbitrarily small – Yuriy S May 10 '16 at 11:22
• What is $n$ in the statement "for $m>n$"? – Jack M May 10 '16 at 11:22
• Your proof is completely wrong, sorry. You must estimate $\cos x_m - \cos x_n$ in terms of $x_m-x_n$. A trivial estimate does not suffice. – Siminore May 10 '16 at 11:22
• So... for example, to show that $z_n=1+\frac1n$ is Cauchy, you would write $|z_n-z_m|=\left|1+\frac1n-\left(1+\frac1m\right)\right|\leqslant\left|1+\frac1n\right|+\left|1+\frac1m\right|$? Really? – Did May 10 '16 at 11:31

I think I'll elaborate on my comment. Use the well-known trigonometric indentity:

$$\cos x_m-\cos x_n=2 \sin \left(\frac{x_n-x_m}{2} \right) \sin \left(\frac{x_n+x_m}{2} \right)$$

According to the usual definition of a Cauchy sequence, if we choose $\epsilon>0$ then there exists some $N$ such that for any $n,m>N$ it follows that $|x_m-x_n|<\epsilon$.

Using this definition and the above relation we write:

$$|\cos x_m-\cos x_n|=2| \sin \left(\frac{x_n-x_m}{2} \right) \sin \left(\frac{x_n+x_m}{2} \right)| \leq 2 \frac{|x_m-x_n|}{2} \cdot 1=|x_m-x_n|$$

Obviously from $\{x_k\}$ being Cauchy follows $\{\cos x_k\}$ being Cauchy as well

• How to get $| \sin \left(\frac{x_n-x_m}{2} \right) | \leq \frac{|x_m-x_n|}{2}$ ? – user1942348 May 10 '16 at 13:13
• @user1942348 from the definition of $\sin x$. Either from the series or from geometrical definition. – Yuriy S May 10 '16 at 13:18
• $|\sin x|\leq |x|$ – Kushal Bhuyan May 10 '16 at 13:45
• @YuriyS The series to show that $|\sin x|\leqslant|x|$? How? – Did May 10 '16 at 21:21
• Use the fact that the function $f(x) = \cos(x)$ is uniformly continuous. ( Lagrange Mean Value theorem.)
• The OP is specifically looking for a correction of their proof, not advice on how to write a different proof. – Jack M May 10 '16 at 11:21
• @JackM Em. I personally believe that seeing this hint, he can rectify the answer himself. And essentially all proofs of the question are going to based on the same idea. – crskhr May 10 '16 at 11:22

You gave it away. Start with $$|y_m-y_n|=|\cos x_m-\cos x_n|\leq\ ?\ |x_m-x_n|$$ instead, whereby you have to replace the question mark by something meaningful (and correct). A hint: Use the MVT, you will obtain $$|y_m-y_n|\leq|x_m-x_n|\ .$$ The rest is "pure logic".

• Thank you very much. I also want to get the result of such style. I am unable to detect the ? terms . I would be grateful if you suggest it for me. – user1942348 May 10 '16 at 13:10
• Do you mean $|y_m-y_n|=|\cos x_m-\cos{x_n}|\leq\ \sin{ \theta}\ |x_m-x_n|\leq |x_m-x_n|$ as $-1\leq \sin{\theta}\leq 1$ where $x_m<\theta<x_n$? – user1942348 May 10 '16 at 14:46
• Sir, Am I correct? – user1942348 May 10 '16 at 15:00