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

Question

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.

Answer 1

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

Answer 2

• Use the fact that the function $f(x) = \cos(x)$ is uniformly continuous. ( Lagrange Mean Value theorem.)

Answer 3

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".