Suppose $f_q(x) = \sin^q x$. Next, we compute the inverse of $f_q(x)$: $$ \begin{aligned} y &= \sin^q x \\ y^{1/q} &= \sin x \\ \arcsin{y^{1/q}} &= x \\ \end{aligned} $$ After exchanging $x$ and $y$ in the most recent equation, we obtain: $$ y = f_q^{-1}(x) = \arcsin(x^{1/q}). $$ Now, if we derivate $f_q^{-1}(x)$, we obtain: $$ \begin{aligned} \frac{d}{dx} f_q^{-1}(x) &= \frac{1}{\sqrt{1 - x^{2/q}}} \frac{d}{dx} x^{1/q} \\ &= \frac{1}{\sqrt{1 - x^{2/q}}} \frac{x^{1/q}}{qx} \\ &= \frac{1}{qx^{1 - 1/q}\sqrt{1 - x^{2/q}}}. \end{aligned} $$ Now, we need the derivative of $\sin^q x$: $$ \begin{aligned} f'_q(x) &= \frac{d}{dx} \sin^q x \\ &= q \sin^{q - 1} x \cos x. \end{aligned}. $$ Since we know from the calculus books that $$ \frac{d}{dx} f_q^{-1}(x) = \frac{1}{f'_q(f_q^{-1}(x))}, $$ we must have $$ \begin{aligned} \frac{1}{\underbrace{qx^{1 - 1/q}}_{\alpha}\sqrt{1 - x^{2/q}}} &= \frac{1}{q\sin^{q-1}(\arcsin(x^{1/q})) \cos(\arcsin(x^{1/q}))} \\ &= \frac{1}{\underbrace{qx^{1 - 1/q}}_{\alpha}\cos(\arcsin(x^{1/q}))}, \end{aligned} $$ which leads to $$ I_q(x) = \cos(\arcsin(x^{1/q})) = \sqrt{1 - x^{2/q}}. $$ Exchanging $1/q$ with $q$ in $I_q(x)$ leads us to the desired identity $$ \cos(\arcsin(x^q)) = \sqrt{1 - x^{2q}}. $$
Now, my question is: will this work with any $q \in \mathbb{R}$?