# Calculating sin' und cos' without power series methods

1. The limit calculation

$\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x-\mathrm{sin}0}{x-0}=\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1$

in [6.8.6] proves in fact the differentiability of sin at 0 with ${\mathrm{sin}}^{\prime }\left(0\right)=1$.

2. For $x\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ we use Pythagoras' theorem (see [4.3.*]) to get $\mathrm{cos}x=\sqrt{1-{\mathrm{sin}}^{2}\left(x\right)}$, which means that cos and $\sqrt{1-{\mathrm{sin}}^{2}}$ coincide locally at 0. As $1-{\mathrm{sin}}^{2}$ is differentiable at 0 and $\sqrt{\mathrm{X}}$ at 1, the chain rule ([7.6.11]) proves cos to be differentiable at 0 with

${\mathrm{cos}}^{\prime }\left(0\right)=\frac{-2\mathrm{sin}0\cdot {\mathrm{sin}}^{\prime }\left(0\right)}{2\sqrt{1-{\mathrm{sin}}^{2}\left(0\right)}}=0$.

3. Using the addition formulas for sine and cosine (see [4.3.*]) we get for all $x,a\in ℝ$:

$\begin{array}{l}\mathrm{sin}x=\mathrm{sin}\left(x-a+a\right)=\mathrm{sin}\left(x-a\right)\cdot \mathrm{cos}a+\mathrm{cos}\left(x-a\right)\cdot \mathrm{sin}a\hfill \\ \mathrm{cos}x=\mathrm{cos}\left(x-a+a\right)=\mathrm{cos}\left(x-a\right)\cdot \mathrm{cos}a-\mathrm{sin}\left(x-a\right)\cdot \mathrm{sin}a\hfill \end{array}$

With the results of 1. and 2. we may apply the chain rule again. Thus, with a factor rule ([7.6.6]) argument, we find that sin and cos are differentiable at a with the following derivation numbers

$\begin{array}{l}{\mathrm{sin}}^{\prime }\left(a\right)={\mathrm{sin}}^{\prime }\left(0\right)\cdot 1\cdot \mathrm{cos}a+{\mathrm{cos}}^{\prime }\left(0\right)\cdot 1\cdot \mathrm{sin}a=\mathrm{cos}a\hfill \\ {\mathrm{cos}}^{\prime }\left(a\right)={\mathrm{cos}}^{\prime }\left(0\right)\cdot 1\cdot \mathrm{cos}a-{\mathrm{sin}}^{\prime }\left(0\right)\cdot 1\cdot \mathrm{sin}a=-\mathrm{sin}a\hfill \end{array}$