Calculating sin' und cos' without power series methods

1. The limit calculation

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

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

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

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

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

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

   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

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