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