Can I approximate sine and cosine without derivatives? Assuming I don't know derivatives (and Taylor series) can I manage to approximate sine and cosine of a generic given (rational) angle in radians?
 A: Yes. Taylor series are just one way to construct a polynomial approximation of a function. There are many other ways; for example, you could use a Lagrange polynomial. Constructing this polynomial doesn't require any knowledge of derivatives.
As an aside ... Taylor series approximations are not very good in many situations. They are very good near one single point, and get steadily worse as you move away from this point. When constructing an approximation of sine or cosine functions, you probably want something that's uniformly good throughout some entire interval of interest. A Lagrange polynomial using Chebyshev nodes is easy to compute and reasonably good (though not optimal).
A: You could use 
$$\cos(\frac{\pi}2) = 0$$ 
$$\sin(\frac{\pi}{2}) = 1$$
and the half-angle formulas (for $0 \le \alpha \le \frac{\pi}2$) 
$$\cos(\frac{\alpha}{2}) = \sqrt{\frac12(1 + \cos{\alpha})}$$
$$\sin(\frac{\alpha}{2}) = \sqrt{\frac12 (1 - \cos \alpha)}$$ 
and the angle-sum formulas
$$ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$
$$ \sin(\alpha + \beta) = \cos \alpha \sin \beta + \cos \beta \sin \alpha$$
to approximate any sin and cos value to any arbitrary accuracy.  Approximate your angle as $\dfrac{k \pi}{2^j}$, then break that into sum of a bunch of dyadic fractions ($\frac{\pi}2$, $\frac{\pi}4$, $\frac{\pi}8$, etc.), calculate all the $\sin$ and $\cos$ values using the half-angle formulas, then determine the $\sin$ and $\cos$ for your sum of dyadics using the angle sum formulas.
A: Here is a way that can be used to approximate cosine.
This method can be proved using only geometry and algebra.
Lets find cos($\pi$/7). First choose a cosine multiple angle identity. $cos(3x) = 4cos^3(x) - 3cos(x)$
Let a = the cosine identity number cos(ax). a = 3
Choose the number of iterations to be used. How about i=3. Calculate $$1 - \frac{\theta^2}{2a^{2i}}$$
$$1 - \frac{(\pi/7)^2}{2*3^{2*3}} = 0.999861852$$
Then plug that into the identity for i times.
$f(x) = 4x^3 - 3x$
Note that f(x) is the third chebyshev polynomial of the first order.
$f(f(f(0.999861852))) = 0.900966971$
$cos(\pi/7) = 0.900966971$
My calculator says $cos(\pi/7) = 0.900968868$
The difference between them = $1.8967813 * 10^{-6}$
You can do a similar thing with sine.
Find a multiple angle identity for sine that is only in terms of sine just as the cosine identity is only in terms of cosine.
One of the identities that doesn't require square roots is:
$sin(3x) = -4sin^3(x) + 3sin(x)$
$f(x) = -4x^3 + 3x$
Instead of calculating $$1 - \frac{θ^2}{2a^{2i}}$$
Calculate $$\frac{θ}{a^i}*\sqrt{1 - \frac{θ^2}{4a^{2i}}}$$
Then follow the same steps used for cosine.
Or could just calculate cosine and then $\sqrt{1 - cos^2(x)}$ If you do that, you have to keep the quadrant of the angle in mind.
As the number of iterations goes to infinity the algorithm will converge to cos(x). The error will go down as the number of iterations goes up in theory. However, because of decimal round off error in calculators and such, higher numbers of iterations can cause more error. If you analyze the error patters, starting from 1 and going up, you will see that the error goes down for a time, until it reaches its lowest point, then it starts to go back up.
Proof:
Start with a right triangle with hypotenuse = 1. Place this triangle on the unit circle in the first quadrant. Let a equal the x side of the triangle. Let b equal the y side of the triangle. We are going to approximate the arc length of the unit circle from the point (1, 0) to (a, b). We can do this by finding the midpoint of the arc many times and then multiplying the distance between (1, 0) and the midpoint by 2 to the power of the number of splits. First lets look at the distance formula for (x, y) and (1, 0).
$\sqrt{(x - 1)^2 + y^2} = \sqrt{x^2 - 2x + 1 + y^2}$
Since it is a point on the arc of a circle, (x, y) satisfies $x^2 + y^2 = 1$
$\sqrt{(x - 1)^2 + y^2} = \sqrt{1 - 2x + 1}$
$\sqrt{(x - 1)^2 + y^2} = \sqrt{2 - 2x}$
So we have found a way to find the distance between the midpoint and (1, 0) with only using the x value of the midpoint. Now let's find how to get the x value of the midpoint using only the a value of the triangle. This is actually just the cosine half angle formula, but it can be proved algebraically using what we have here. To prove it algebraically just set the distance between the midpoint and (1, 0) equal to the distance from the midpoint to (a, b). Then use $x^2 + y^2 = 1$ and $a^2 + b^2 = 1$ and solve the system of equations for x in terms of a.
You get: $$x = \sqrt{\frac{1 + a}{2}}$$
We can plug that formula into itself to split the arc many times.
That gives us the formula for arccosine.
$$arccos(x) = 2^i * \sqrt{2 - 2\sqrt{\frac{1 + \sqrt{
\frac{1 + ^{...}a}{2}
}}{2}}}$$You can use this to get a formula for all of the other trig functions.
For cosine just solve for a in the equation.
The goes on forever but you can see the pattern in the inverse easily.
cos(x) becomes:
$$2( 2( 2( 1 - \frac{θ^2}{2*2^{2i}} )^2 - 1 )^2 - 1 )^2 - 1$$
Note that this is the formula for cos(2x) plugged into itself.
$cos(2x) = 2cos^2(x) - 1$ which we use as $2x^2 - 1$
After thinking about why this works and stuff, I came up with the generalization above.
A: I will only use things that are known in basic trigonometry, i.e., the fundamental trigonometric identity ($\sin^2(x)+\cos^2(x)=1$), the sine/cosine of the sum or difference of two angles, the usual values at $0$, $\pi/4$ and $\pi/2$ and the formulas of the half angle that can be deduce for the previous formulas.
First of all, imagine that we want to calc the sine or cosine of the angle $\alpha$. After a translation by an entire multiple of $\pi$, something of the form $k\pi$ with $k\in\mathbb{Z}$, I can assume that $\alpha$ is in $[-\pi/2,\pi/2]$ due to the the "antiperiodicity" of sine and cosine of period $\pi$, due to
$$\cos(x+k\pi)=(-1)^k\cos(x)\text{ and }\sin(x+k\pi)=(-1)^k\sin(x)\text{.}$$
And here, after using the symmetry of the cosine or antisymmetry of the sine, i.e. $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$; we can, furthermore, suppose that $\alpha\in[0,\pi/2]$.
Secondly, we know that for $x,y\in[0,\pi/2]$, if $x\leq y$, then
$$\cos(x)\geq \cos(y)\text{ and }\sin(x)\leq \sin(y)\text{.}$$
For this, remember that
$$\cos(y)-\cos(x)=-2\sin\left(\frac{y-x}{2}\right)\sin\left(\frac{x+y}{2}\right)\leq 0$$
and
$$\sin(y)-\sin(x)=2\sin\left(\frac{y-x}{2}\right)\cos\left(\frac{x+y}{2}\right)\geq 0\text{.}$$
We have only use the formula for the sine/cosine of the sum/difference of two angles and the fact that sine and cosine for angles in $[0,\pi/2]$ are non-negative.
Third of all, remember that for $\alpha\in[0\pi/2]$,
$$\cos\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1+cos(2\alpha)}{2}}\text{ and }\sin\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1-cos(2\alpha)}{2}}\text{.}$$
Therefore, we can recursively calculate
$$c_n=\cos\left(\frac{\pi}{2^n}\right)\text{ and }s_n=\sin\left(\frac{\pi}{2^n}\right)\text{,}$$
using the formulas
$$c_{n+1}=\sqrt{\frac{1+c_n}{2}}\text{ and }s_{n+1}=\sqrt{\frac{1-c_n}{2}}\text{.}$$
Finally, given $\alpha\in [0,\pi/2]$, express it as
$$\alpha=\sum_{n=1}^\infty a_n\frac{\pi}{2^n}$$
with $a_n\in\{0,1\}$, by using the binary expansion of $\alpha/\pi$, and then you will have for all $m$ that
$$\cos\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}+\frac{\pi}{2^{m}}\right) \leq\cos(\alpha)\leq \cos\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}\right)$$
and
$$\sin\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}\right)\leq \sin(\alpha)\leq \sin\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}+\frac{\pi}{2^{m}}\right)\text{.}$$
You can easily check that the LHS and RHS in the both expression can be calculated as polynomials in the $c_n$ and $s_n$ by applying the formula for the sine/cosine of the sum of two angles.
