How do I verify that $\sin (\theta)$ and $\cos (\theta)$ are functions? I am studying pre-calculus mathematics at the moment, and I need help in verifying if  $\sin (\theta)$ and $\cos (\theta)$ are functions? I want to demonstrate that for any angle $\theta$ that there is only one associated value of $\sin (\theta)$ and $\cos (\theta)$. How do I go about showing this?

I'm afraid that from a pre-calculus point of view, there is no consistent definition of the functional relationships that allows to discuss them as real functions (real valued in one real variable).
The big gap to be closed is the notion of an angle. In geometry, an angle is defined by a pair of rays from the same point. These most often occur at corners of triangles. All geometrically similar pairs of rays define the same angle. You can arrange such an equivalent pair of rays to have the first ray horizontal pointing to the right, then the second ray is uniquely characterized by a point on the unit circle (after fixing an origin and a unit length). The coordinates of that point then are the cosine and sine of that (equivalence class of) angle(s).
Then there are some angles that can be constructed and calculated geometrically by constructing regular n-gons and thus dividing the unit circle into n equal parts. n=3,4,6,8,12 are easy, n=5 is possible without too much effort. By bisection one can enclose any other angle in fractions of the full circle, starting from these known fractions, but this goes close to limit arguments, which already is calculus.
One can identify points on the unit circle with rotation matrices or complex numbers, which simplifies the angle arithmetic. Doubling an angle is then squaring the matrix or complex number, bisecting an angle corresponds to a square root. This gives functional equations that in their components are the trigonometric identities. Demanding continuity and differentiability at the zero angle will also give the trigonometric functions as unique solutions. However, this uniqueness is again enforced by calculus arguments.
