Why do both sine and cosine exist?
You seem to be asking why we name them both, rather than why they exist, since the very relationships you've written shows that if one exists, the other does, too.
Essentially, all mathematical notation and names, except for a very small subset, are for convenience/clarity/human communication. Math does not require that we name any function consistently across separate proofs, but it becomes much easier to communicate and think about things when we do.
I prefer the relationship:
$$\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)$$ since this is symmetric and obviously geometric when $0\leq x\leq \pi/2$, and because this is also the relationship between "tangent" and "cotangent" and "secant" and "cosecant." It indicates a duality in these functions.
It is certainly something of a paradox that adding more names often simplifies our understanding. In particular, if you defined only one, it would give you a sense that one of these functions was "primary." There is a hint of that error even in the names "sine" and "cosine," which vaguely implies that "sine" is primary, but it would be particularly strong if we only defined "sine" and never defined "cosine." We would have a harder time grasping the duality that happens in trig functions.
If you actually must start with one function, most mathematicians wouldn't start with $\cos x$ or $\sin x,$ they'd start with the complex-valued function: $\operatorname{cis}(x)=\cos(x)+i\sin(x).$ This has the property $\operatorname{cis}(x)\operatorname{cis}(y)=\operatorname{cis}(x+y),$ and can also be written as $e^{ix}.$ You can define $\cos x$ and $\sin x$ in terms of $\operatorname{cis} x:$
$$\cos x=\frac{1}{2}\left(\operatorname{cis}(x)+\operatorname{cis}(-x)\right)\\ \sin x=\frac{1}{2i}\left(\operatorname{cis}(x)-\operatorname{cis}(-x)\right)$$
Addition: I recently heard, in a lecture that covered some of the accomplishments of the early Muslim world, that they acquired knowledge of the sine function from India, and developed the other trigonometric functions. So if you want someone to blame, we can blame them. 🤓
$$ \sin^2 \alpha + \cos^2 \alpha =1, $$ $$ \sin^2 \alpha + \sin^2 \left( \alpha + \frac{\pi}{2} \right) =1. $$ Which one do you prefer?
Post scriptum: of course this is not a deep answer, but I think that sometimes mathematicians prefer elegance to logical "economy".
The question is ridiculous. (I'm afraid to say.)
It's equivalent to asking "Why are there words for both North and South?"
(Of course you could just use "negative North" at all times, if desired.)
In my opinion, one would only ask the question at hand, if, one has rather naively "just noticed" - let's put it that way - that sine and cosine are complementary.
Note too that, indeed, in English co-sine is simply "sine" ... with the appropriate prefix!! Just as you'd expect.
Again to make analogy, one could ask questions such as "why do we label both matter and antimatter!" or "Why label both up and down?"
It's clear, traditional, and expected in languages that there are matching terms for complementary qualities {rather than, let us say "minimalistically," using only the one and then the negative of it} ...
heaven and hell, paradis et enfer.
I think it's conceptually cleaner to have both $\cos$ and $\sin$ as distinct notations, for the following reason: in my opinion, it's a bit of a "coincidence" that each of the functions $\{\cos,\sin\}$ can be described as translations of the other.
My preferred definition of these two functions is the following: first, position yourself at $(1,0)$ on the unit circle. Then start walking anticlockwise at unit speed. It follows that if the time elapsed is $t$, your $x$-coordinate will be $\cos t$ and your $y$-coordinate will be $\sin t$.
But notice that, in general, this way of producing pairs of functions (namely: positioning yourself on a curve and walking at unit speed, and then projecting onto the $x$ and $y$ coordinate axes) won't usually result in a pair of functions such that each can be defined as a translation of the other. The ability to define $\cos$ and $\sin$ in terms of each other is specifically a quirk of circles centered at the origin. In some sense, it's kind of a coincidence.
By the way, this viewpoint explains why $\cos^2 t + \sin^2 t = 1$; it's because you're walking on the curve defined by $x^2+y^2= 1$. This also explains why $\cos(0) = 1$ and $\sin(0)= 0$; it's because we positioned ourselves at $(1,0)$ to begin with. It also explains why $(\cos' t)^2+(\sin't)^2= 1$; it's because you're walking at unit speed. And finally, this explains why $\sin'(0)>0;$. It's because we chose to start walking anticlockwise. I'm pretty sure these four conditions (listed below for your convenience) completely characterize the ordered pair ($\cos,\sin$) among all ordered pairs of differentiable functions $\mathbb{R} \rightarrow \mathbb{R}$.
- $\cos^2 t + \sin^2 t = 1$
- $\cos(0) = 1, \sin(0) = 0$
- $(\cos't)^2+(\sin't)^2 = 1$
- $\sin'(0) > 0$