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 $(A^{-1})' \left (\dfrac{x}{2} \right)$.

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

Textbook's Proof enter image description here


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


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