Define \(\sin x\) as the solution to \(y''=-y\) with \(y(0)=0\) and \(y'(0)=1\text{,}\) and define \(\cos x\) as the solution to the same differential equation with \(y(0)=1\) and \(y'(0)=0\text{.}\)
Can show that \(\DS\dv{x} \sin x=\cos x\) by letting \(C(x)=\DS\dv{x} \sin x\) and noticing that \(C\) also satisfies the same differential equation and has the same initial conditions as \(\cos x\text{.}\) Similarly, can show that \(\DS\dv{x} \cos x=-\sin x\text{.}\)
Other properties such as the sum/difference identities and the fact that \(\sin^2 x + \cos^2 x = 1\) can also be shown by using the differential equation and initial conditions.
\begin{align*}
\dv{x}\tan x \amp = \sec^{2} x \amp \dv{x}\sec x \amp = \sec x \tan x\\
\dv{x}\cot x \amp = -\csc^{2} x \amp \dv{x}\csc x \amp = -\csc x \cot x
\end{align*}
Every one of these can be thought of backwards to become an antiderivative formula.