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?
 A: Usually one begins studying the $\sin$ function in connection with right triangles. Here the notion of angle is not questioned, so I shall stick with this intuitive concept. It follows that we have $\sin \theta$ defined for $0\leq\theta\leq{\pi\over2}$.
As a next step one considers polar angles in the $(x,y)$-plane, measured counterclockwise from the positive $x$-axis, and one identifies these angles with the length of the corresponding arc on the unit circle $S^1$. When $0\leq\theta\leq{\pi\over2}$ then $\sin\theta$ is  the $y$-coordinate of the point where the second leg of the angle in question intersects $S^1$. It is then  natural to define $\sin \theta$ for all $\theta\in[0,2\pi]$ in this way, and a further step leads to the definition of $\sin\theta$ for all $\theta\in{\mathbb R}$: For arbitrary $\theta>0$ spool a thread of length $\theta$ counterclockwise around $S^1$ with initial point at $(1,0)$. The $y$-coordinate of the endpoint is then defined to be $\sin\theta$. It is then intuitively obvious that the function $\sin$ is periodic.
This sounds all very simple and natural. The difficulties begin when we try to make the above thought  process mathematically precise. In the first place we need a precise notion of angle: axiomatic euclidean geometry allows comparison and addition of angles, but does not provide an identification of angles with real numbers, which is inherent in the concept of the $\sin$ function.
A: The proof is based simply on similar triangles. If a right-angled triangle has an angle $\theta$ then the other two angles are $90^{\circ}$ and $(90-\theta)^{\circ}$. If two triangles have the same angles then they are similar.
My picture shows two similar triangles: $\triangle OAB$ and $\triangle OA'B'$. 
Since $\theta = \angle AOB$ then, by definition 
$$\sin\theta = \frac{\|AB\|}{\|OB\|}$$
Since $\theta = \angle A'OB'$ then, by definition
$$\sin\theta = \frac{\|A'B'\|}{\|OB'\|}$$
We can show that $\sin \theta$ has a single, unique value if we can show that the two ratios agree.
Let $T$ be the linear transformation given by an enlargement, centre $O$, with scale factor $\lambda$, such that $T(A) = A'$ and $T(B) = B'$. We have $\|A'B'\|=\lambda\|AB\|$ and $\|OB'\| = \lambda\|OB\|$, hence
$$\frac{\|A'B'\|}{\|OB'\|} = \frac{\lambda\|AB\|}{\lambda\|OB\|}=\frac{\|AB\|}{\|OB\|}$$
This shows that the ratio of the opposite side to the hypotenuse is the same for any two similar, right-angled triangles with angle $\theta$. That means that $\sin\theta$ is uniquely, well-defined.
(N.B. Similarity allows rotation and reflection as well as enlargement. However, rotations and reflections preserve lengths and so preserve ratios of lengths.)

A: 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.
A: It follows from the definition of $\sin(\theta)$ as a ratio of length of $\theta$'s opposite side to hypotenuse (Wikipedia).
As long as hypotenuse length is not equal to zero (otherwise the triangle degenerates into line or point), the division is a function which yields only one associated value.
A: More precisely, in Analysis, sin(x) is defined as the limit of a specific series.
Before, we know that the below series converges for every $x\in\mathbb{R}$:
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
And then we define :
$$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
By this definition, $\sin(x)$ has all properties of the elementary definition (ratio of sides in a right-angled triangle).
A: If you want to verify this intuitively for yourself, then this answer could prove to be helpful. But remember, this is not a proof.
First try to visualize what $\sin\theta$ looks like. If you need some help, this is what it looks like.

Now remember that if $\sin\theta$ is a function, then for a certain $\theta$, there can be only one value of $\sin\theta$.
Now pick a random point on the $x-$axis. This is going to be your value for $\theta$. Now a draw a line perpendicular to the $x-$axis through that point. Now is it possible for that line to intersect the sine curve at more than one point? Try it out for different values of $x$. No, it is completely impossible for that line to intersect that curve at more than one point.
This means that for a certain $\theta$, there is only one associated value of $\sin\theta$.
$\sin\theta$ and $\cos\theta$ are pretty much the same. Notice that $\cos\theta=\sin(\theta +\frac{\pi}{2})$. So if you move the sine curve by $\frac{\pi}{2}$ to the right, you're going to get the curve for $\cos\theta$. And $\sin\theta$ and $\cos\theta$ look exactly the same. So you can use the same argument.
Finally, I repeat that this is not a proof. But I hope this helps you see it intuitively.
The figure is taken from here.
