Deriving the power series for cosine, using basic geometry Looking for the derivation of cosine lead to https://www.quora.com/How-do-I-calculate-cos-sine-etc-without-a-calculator and the MacLauren series.
$$\cos(x)=1−\frac{x^2}{2!}+\frac{x^4}{4!}−\frac{x^6}{6!}+\dotsc$$
Wondering if one could show how the cosine series function is derived, starting from basic geometry. Looking at that equation above, I'm not sure where the numbers and variables came from.
Note, I am hoping for a derivation starting with "A triangle has 3 sides", super simple, not from the Taylor series or idea of derivatives which already has a lot of context (but I would like to see derivatives and Taylor series in the process). I would like to see the connections from:


*

*basic geometry $\to$ stuff

*stuff $\to$ taylor series

*taylor series $\to$ stuff

*stuff $\to$ cosine power series

 A: I'll amend this eight(!)-year-old answer with more detail.

We begin with the fact that a triangle has three sides. :) In particular, a right triangle has one hypotenuse and two legs. If we take the hypotenuse to have length $1$, and one of the triangle's acute angles to have (radian) measure $\theta$, then the leg opposite $\theta$ has length $\sin\theta$, while the leg adjacent to $\theta$ has length $\cos\theta$. (That's the geometric definition of these values.)
In the diagrams below, $\overline{OP}$ is the hypotenuse of the right triangle in question, and we construct arc $\stackrel{\frown}{PP_0}$ of the unit circle about $O$. Note that, because the radius is $1$, we have $|\stackrel{\frown}{P_0P}| = \theta$.
Following a remarkable construction by Y. S. Chaikovsky (presented in this very readable American Mathematical Monthly article by Leo Gurin), we subdivide the $\stackrel{\frown}{PP_0}$ into $n$ equal parts, recursively building a collection of similar isosceles triangles in various stages. (Each stage has one fewer triangle than its predecessor.) The diagram shows the triangles for $n=4$ and $n=16$, as well as for the limiting case ("$n=\infty$").  

For each $n$, the bases of the first stage of triangles form a polygonal approximation of the circular arc $\stackrel{\frown}{P_0P}$; the bases of the second-stage triangles approximation the involute $P_1P$ of that arc; the bases of the third-stage triangles approximate the involute $P_2P$ of that involute; and so on. Moreover, the construction guarantees that the leg of the largest isosceles triangle at each stage has length equal to that of the polygonal path formed by the bases of the previous stage:
$$|\overline{P_{i-1}P_{i}}| = |\widehat{P_{i-1}P}| \tag{1}$$
At the first stage, each triangle has leg-length $1$ and base-length $s := 2\sin\frac{\theta}{2n}$. At the second stage, the smallest triangle has a previous base for a leg, so its base-length is $s^2$; in general, at stage $i$, the smallest triangle's base-length is $s^{i}$. Chaikovsky discovered a clever (but not difficult) combinatorial argument (omitted here) that the total length of all bases at a particular stage is an integer multiple of that smallest base, namely

$$|\widehat{P_{i-1}P}| = \binom{n}{i}\;s^i \quad\text{which we can write as}\quad \frac{1}{i!}\prod_{j=0}^{i-1}\left(2n\sin\frac{\theta}{2n}\cdot \frac{n-j}{n}\right) \tag{$\star$}$$

(a formula that conveniently works for $i=0$ as well, if we rename point $O$ to $P_{-1}$).
Now, as $n$ increases, the various polygonal paths better-approximate their corresponding smooth curves. This is guaranteed by the only sophisticated fact we need from elementary Calculus: $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \qquad\text{so that}\qquad \lim_{n\to \infty}2n\sin\frac{\theta}{2n} = \theta \tag{2}$$
Also, the fraction $(n-j)/n$ better-approximates $1$. Consequently, in the limit, the polygonal paths simplify to curves while the big product in $(\star)$ simplifies to $\theta^i$. Recalling $(1)$, we can write

$$|\overline{P_{i-1}P_{i}}| = \frac{1}{i!}\theta^i \tag{$\star\star$}$$

So what?
Well, observe that, in the limiting diagram, the path $OP_1P_2P_3P_4\cdots$ forms a spiral that appears to (and actually happens to) converge on point $P$. The segments of that path are either perfectly horizontal or perfectly vertical: With each horizontal step, the path alternately over- and under-shoots $P$'s horizontal offset from $O$, while each vertical step does likewise for the vertical offset. But those offsets are precisely $\cos\theta$ and $\sin\theta$! Therefore,

$$\begin{align}
\cos\theta = |\overline{OP_0}| - |\overline{P_1P_2}| + |\overline{P_3P_4}| - \cdots &= \sum_{i\;\text{even}}(-1)^{i/2}\;|\overline{P_{i-1}P_{i}}| \;\;\;\;= \sum_{i\;\text{even}} (-1)^{i/2}\;\frac{1}{i!}\theta^i \\[4pt]
\sin\theta = |\overline{P_0P_1}| - |\overline{P_2P_3}| + |\overline{P_4P_5}| - \cdots &= \sum_{i\;\text{odd}}(-1)^{(i-1)/2}\;|\overline{P_{i-1}P_{i}}| = \sum_{i\;\text{odd}} (-1)^{(i-1)/2}\;\frac{1}{i!}\theta^i
\end{align}$$ 

That is, with some simple geometry, a dash of combinatorics, and the slightest touch of Calculus, we arrive at the power series representations for sine and cosine. As my other answer notes, a minor variation in the construction of the involutes (albeit with significantly-trickier combinatorics) leads to the series for tangent and secant. (I still don't have a counterpart for cotangent and cosecant, which remains the topic of my first Trigonography Challenge.) $\square$
A: $$cos(x)=1−\frac{x^2}{2!}+\frac{x^4}{4!}−\frac{x^6}{6!}+\dotsc$$
is the Taylor series for $\cos x$ about $a=0.$
Taylor polynomials are polynomials which match with the function up to n-tn derivative. That is $$ f(a)=p(a), f'(a)=p'(a),...f^{(n)}(a) =  p^{(n)}(a)$$
If we graph the function and the Taylor polynomial, on the same screen, we see they get closer and closer as you add more terms to the polynomial.
A: MacLaurin series of a function $f:\mathbb{R\to R}$ analytic at $x=0$ is just defined as $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$ with $f^{(n)}(0)$ being the nth derivative of $f$ in $x=0$.
In your scenario $$f^{(0)}(0)=f^{(2)}(0)=\dots=f^{(2n)}(0)=(-1)^n$$ whereas $$f^{(1)}(0)=f^{(3)}(0)=\dots=f^{(2n+1)}(0)=0$$ That's how you get this expansion.
EDIT: We do use some geometry when we consider the derivatives. You need to calculate (for instance) $$f''(x)=\lim_{t\to 0}\frac{f'(t)-f'(0)}{t}=\lim\frac{\cos t}{t}.$$ limits of the form $\lim\frac{\sin t}{t}$ or $\lim\frac{\cos t}{t}$ are often solved by basic geometry as you suggest. For instance see the video https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-derivtive-rules-opt-vids/v/sinx-over-x-as-x-approaches-0
A: I'll start with "a bit of a long shot that might work" to motivate a power series in the first place; I'll then proceed to a fairly elegant argument that it legitimises.
Starting from the small-angle approximation $\sin x\approx x$ (which is geometrically justifiable by considering a triangle inside a very small sector of a circle), $\cos^2 x+\sin^2 x = 1$ and $\cos 0=1$ implies $\cos x\approx 1-\frac{x^2}{2}$. Defining $\cos x,\,\sin x$ in the usual way in terms of a circle, $\cos x$ is an even function so $\cos x\approx 1-\frac{x^2}{2}+cx^4$ for some $c$. Then $\sin^2 x\approx x^2-(2c+\frac{1}{4})x^4$, and since $\sin x$ is an odd function we have $\sin x\approx x+kx^3$ for some $k$, viz. $\sin^2 x\approx x^2+2kx^4$.
This tells us $2k=-2c-\frac{1}{4}$, although I'm not sure how we show $c=\frac{1}{24}$ or equivalently $k=-\frac{1}{6}$. But in theory, it may be possible to do just that, and to further recurse through other coefficients, and if we obtain a recursion formula to show by induction it gives the expected coefficients. (If we can do that, the fact that each power series has an infinite radius of convergence will make them both exact for all $x$ despite our starting from a small-angle approximation.)
OK, now for the main course. (It uses complex numbers, but if you would prefer an argument that doesn't you can with some care rewrite this around that.) Define $\mathrm{cis}x:=\cos x+i\sin x$ with $i^2=-1$, so the compound-angle formulae (which you can derive geometrically) proves $\mathrm{cis}(x+y)=\mathrm{cis}x\mathrm{cis}y$. A power-series $\mathrm{cis}x=\sum_{n\ge 0}\frac{a_n}{n!}x^n$ will begin $1+ix-\frac{x^2}{2}$ by the above, so $a_1=i$, a fact that will be important in a moment. The $x^k y^l$ coefficient in $\mathrm{cis}(x+y)$ gives $\frac{a_k}{k!}\frac{a_l}{l!}=\frac{a_{k+1}}{(k+l)!}\binom{k+l}{k}=\frac{a_{k+l}}{k!l!}$ and $a_{k+l}=a_k a_l$. Thus $a_n = i^n$, and the desired series follow from the real and imaginary parts.
I realise that most of the above argument is algebraic, but it does start from geometry, never using any calculus.
A: Hmm, starting from basic geometry going into Taylor series is a bit of a stretch but...
Firstly, the cosine of an angle, in a right angled triangle, is the adjacent over the hypotenuse. 
In a unit circle (with radius one) on the Cartesian plane, if we pick a point on the circumference, the cosine will be the adjacent side, as the radius (which is the hypotenuse of a right angled triangle), will be equals to one. The Unit Circle and Trigonometry
The cosine of an angle is a function of the angle. That means that the cosine of the angle varies depending on the angle itself.
A function is something where you input a value, and an output comes out, based on the value. The value varies, and is called a variable.
And we can graph a function on a place, by seeing which values of the function correspond to the value of the variable. This is how we get our cosine graph. A nice little animation to help you understand, you can find out more about this by searching "unit circle cosine graph".
The Taylor series states that for a function, $$f(x),$$
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$
which expands to
$$f(a)+\frac{f’(a)}{1!}(x-a)+\frac{f’’(a)}{2!}+...$$
And so with the maclaurin series where we let a=0, we will get
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x-0)^n$$
Which expands to
$$f(0)+\frac{f’(0)}{1!}(x-0)+\frac{f’’(0)}{2!}(x-0)^2+...$$
And when we have a cosine function, we will get
$$cos(0)+\frac{sin(0)}{1}(x)+\frac{-cos(0)}{2}(x)^2+...$$
And then you finally get your cosine power series
$$1+0-\frac{x^2}{2}+...$$
I only did three terms but if you wanted to you could expand it out and see that the summation at 0 becomes the power series
A: The Spivak solution: starting from the geometrical intuition, use the integral calculus.
Spivak starts with the function
$$
A(x) = \frac{x\sqrt{1 - x^2}}2 + \int_x^1\sqrt{1 - t^2}\,dt,
\qquad x\in[-1,1]
$$
(What is $A$ intuitively? Hint $A' = \cdots$)
As $A(-1) = \pi/2$, $A(1) = 0$ and $A$ is strictly decreasing, for $x\in[0,\pi]$ we can define $\cos x$ as the only number in $[-1,1]$ s. t.
$$A(\cos x) = \frac{x}2$$
and
$$\sin x = \sqrt{1 - (\cos x)^2}.$$
Using the inverse function theorem, can be proved easily that in $(0,\pi)$
$$
\cos' = -\sin,\qquad\sin' = \cos
$$
Now, both functions can be extended to $\Bbb R$ by periodicity and the property of the derivatives keeps true. Using it, the Taylor series in $0$ are obvious.
