Let $X$ be uniformly distributed on $(a,b)$. I want to find the cdf of
$$ \sin^2(X) + \cos^2(X) $$ My feeling is that since $\sin^2(X) + \cos^2(X) = 1$, the cdf will be: $$F(1 \le x)= \begin{cases} 1, \text{if } 1 \le x \\ 0, \text{otherwise} \end{cases} $$ The only thing I am not sure about is whether $\sin^2(X) + \cos^2(X) = 1$, because $X$ is a random variable, not a real number. However, this identity works even if I replace $x$ with a real-valued function $g(x)$, namely $$\sin^2(g(x)) + \cos^2(g(x)) = 1$$ Random variables are real-valued functions so I don't see why this may fail.
I would very much appreciate if someone with more expertise than me can let me know if I am heading the right direction or if I am completely off.
A formal argument is: Let $(\Omega,\mathcal{F},P)$ a probabilistic space and $A:\Omega\to\mathbb{R}$ a random variable.
Set $g:\mathbb{R}\to\mathbb{R}$ such that Borel measurable. Then $g\circ A:\Omega\to\mathbb{R}$ is random variable.
Now, let $U$ a Borel set in $\mathbb{R}$. Suppose $U\subseteq(Im(g))^c$ (that is, $U$ doesn't have elements of $Im(g)$). Then $(g\circ A)^{-1}(U)=\emptyset$.
Thus $P((g\circ A)^{-1}(U))=0$.
Now, for your question:
Let $x\in\mathbb{R}$. Take $g:\mathbb{R}\to\mathbb{R}$ given by $g(x)=\sin^2 x+\cos^2x(=1)$. Then $g\circ X$ is a random variable in the probabilistic space where $X$ is defined. Moreover, $Im(g)=1$. Suppose $x<1$. The borel set $U=(-\infty,x]$ is contained in $Im(g)^c$
By the observations above, we have, $P((g\circ X)^{-1}(U))=0$. But $P((g\circ U)^{-1}(U))=P(g\circ X\in U)=P(\cos^2X+\sin^2X\le x)$. Thus $0=P(\cos^2X+\sin^2X\le x)$ for all $x<1$.
A similar argument shows that $P(\cos^2X+\sin^2X\le x)=1$ iff $x\ge 1$.
Now you can rest assured that what you've done is right.
Finally, please note that the unique condition on $X$ is $g\circ X$ to be well defined.