# Proof Correction: $\cos ' (x) = -2 \sin (x)$

Preliminary Information

Let $$A(y) = \cfrac{y \sqrt{1 - y^2}}{2} + \int_{y}^{1} 1\sqrt{1 - t^2} dt \ \ \ \text{on [-1, 1]}$$

Moreover $$A'(y) = \dfrac{-1}{2\sqrt{1 - y^2}}$$

Define $$\cos x$$ as $$A(\cos x) = \cfrac{x}{2}$$ and $$\sin x = \sqrt{1 - \cos ^{2} x}$$

Problem: Find $$\cos' (x)$$.

$$A(\cos x) = \dfrac{x}{2}$$; since $$A$$ is decreasing it is one-one and $$A^{-1}$$ a function. Consider that $$A^{-1}\left( \dfrac{x}{2} \right) = \cos x$$. Let us find then $$(A^{-1})' \left (\dfrac{x}{2} \right)$$.

$$$$\begin{split} (A^{-1})' \left (\dfrac{x}{2} \right) &= \dfrac{1}{A'(A^{-1} \left (\dfrac{x}{2} \right)} \\ &= -2 \sin x \end{split}$$$$

Textbook's Proof

You forgot to apply the chain rule and multiply by $$(x/2)'$$ when differentiating $$(A^{-1})(x/2)$$.