Let the arcsine of x, written asin(x), be the integral of 1/sqrt(1-t2), as t runs from 0 to x. The function is not defined outside of x = ±1. By symmetry, asin(-x) = -asin(x). Note that asin(1) is an indefinite integral; let's hope it converges.
Actually we can show this integral is finite. Replace 1-t2 with 1-t, which is smaller, hence the fraction (integrand) is larger. Now we are integrating 1 over sqrt(t), from 0 to 1. The integral is 2sqrt(t), and the area is 2, hence the arcsine of 1 is no more than 2.
Let π symbolize twice the arcsine of 1. In other words, asin(1) = π/2, and asin(-1) = -π/2.
Let the sine of x, written sin(x), be the inverse function, mapping the range, [-π/2,π/2], back to [-1,1].
Write asin(sin(x)) = x and differentiate. By the chain rule, the derivitive of sin(x) is sqrt(1-sin2(x)). Let's call this derivative the cosine of x, or cos(x). Take the derivative of sqrt(1-sin2(x)), cancel 2cos(x) above and below, and get -sin(x). Thus the second derivative of sin(x) is -sin(x), and the second derivative of cos(x) is -cos(x). Successive derivatives cycle through sin(x), cos(x), -sin(x), -cos(x), forever.
We defined cosine as the derivative of sine, and showed this was the square root of 1 minus sine squared. In other words, sin(t)2 + cos(t)2 = 1. Taken together, sine and cosine always define a point on the unit circle. We usually let sine be the y coordinate, while cosine is the x coordinate. As t runs from -π/2 to π/2, cosine and sine trace the right half of the unit circle from 0,-1 counterclockwise up to 0,1. The sine increases monotonically with time, and the cosine starts at 0, moves out to 1, and drops back to 0. Also, if you are familiar with vector calculus, the speed of this partical is the square root of the sum of the squares of the component derivatives, or (-sin(t))2 + cos(t)2, or 1. The particle moves along at constant speed as it traces the semi-circle. As mentioned earlier, one can define sine and cosine in this manner, measuring the coordinates as we move around the circle at constant speed, then compute all the derivatives and integrals. Don't be surprised if your text book does it this way, but I like to start with integrals and work from there.
No matter how we got here, by definition or by inference, sin(t) and cos(t) trace the right half of the unit circle as t runs from -π/2 to π/2. We can extend this function forward and backward in time, moving the particle around the unit circle. The function is periodic, with period 2π. If we add 2π to t, the sine and cosine will be the same, having traveled once around the unit circle. Since the derivative, or cosine, is 0 at the top and bottom of the circle, the extended function is differentiable where the pieces are joined together, hence it is everywhere differentiable. Using a similar argument, the cosine function is everywhere differentiable. Thus both functions are infinitely differentiable everywhere, cycling through sin(t), cos(t), -sin(t), -cos(t).
And what is the value of π? Since our particle moves at speed one, it is the distance traveled as t runs from -π/2 to +π/2, which is the distance around the right half of the circle. In other words, the distance around the unit circle is 2π. It turns out, π is a transcendental number, that can only be approximated. Most people are happy with 22/7, or 3.14. If you want to go to Mars, 3.1415926535 should suffice. If you are interested in the structure of π, as an irrational number, it has been calculated out to 4 billion digits.