# Proof

(related to Proposition: Continuity of Cosine and Sine)

Let $$(x_n)_{n\in\mathbb N}$$ be a convergent real sequence with $\lim_{n\to\infty}x_n=a$ for some real number $$a\in\mathbb R$$. We want to show that cosine $$\cos:\mathbb R\mapsto \mathbb R$$ and sine $$\sin:\mathbb R\mapsto \mathbb R$$ are continuous real functions, which means by definition that $\begin{array}{rcl}\lim_{n\to\infty}\cos(x_n)&=&\cos(a)\\\lim_{n\to\infty}\sin(x_n)&=&\sin(a).\end{array}$

We use the fact that real numbers are embedded in the complex numbers. Therefore, we can consider all real sequence members $$x_n$$ to be special complex numbers with $$x_n=\Re( x_n)$$. Then $$(x_n)_{n\in\mathbb N}$$ is a convergent complex sequence. Due to the rule of multiplying a complex sequence by a complex number (here, we multiply all $$x_n$$ by the imaginary unit $$i$$, we get the result

$\lim_{n\to\infty}ix_n=ia.$

Because the complex exponential function is continuous, we get in particular on the unit circle the next result

$\lim_{n\to\infty}\exp(ix_n)=\exp(ia).$

We also know that a complex sequence is convergent if and only if its real and imaginary parts are convergent. It follows from the Euler's formula that

$\lim_{n\to\infty}\exp(ix_n)=\lim_{n\to\infty}\cos(x_n)+i\lim_{n\to\infty}\sin(x_n)=\cos(a)+i\sin(a).$

By comparing the real and imaginary parts of the last equation, we get the desired result:

$\begin{array}{rcl}\lim_{n\to\infty}\cos(x_n)&=&\cos(a)\\\lim_{n\to\infty}\sin(x_n)&=&\sin(a).\end{array}$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983