The small angle approximations are given by:

\[\sin x \approx x\]

\[\tan x \approx x\]

\[\cos x \approx 1-\frac{x^2}{2}\]

when x is measured in radians.

## Proof for \(\sin x\)

Consider the following sector:

The area of the sector is: \(\frac{1}{2} \times 1^2 \times x=\frac{1}{2} x\)

The area of the triangle is: \(\frac{1}{2}\times 1^2 \times \sin x=\frac{1}{2} \sin x\)

As \(x \to 0\), the area of the sector is approximately equal to the area of the triangle:

\[\therefore \frac{1}{2} \sin x \approx \frac{1}{2} x\]

\[\therefore \sin x \approx x\]

## Proof for \(\cos x\)

From the double angle formula for \(\cos\), we know that:

\[\cos 2x = \cos^2 x -\sin^2 x\]

\[\cos 2x = (1-\sin^2 x)-\sin^2 x\]

\[\therefore \cos 2x = 1-2\sin^2 x\]

If we subtitute \(\frac{x}{2}\) for \(x\), we get:

\[\cos x = 1-2\sin^2 \frac{x}{2}\]

Therefore as \(x \to 0\) :

\[\cos x =1-2\sin^2 \frac{x}{2} \to 1-2\left(\frac{x}{2}\right)^2=1-2\left(\frac{x^2}{4}\right)=1-\frac{x^2}{2}\]

\[\therefore \cos x \to 1-\frac{x^2}{2}\]

\[Q.E.D.\]

## Proof for \(\tan x\)

As \(x \to 0\) :

\[\tan x = \frac{\sin x}{\cos x} \to \frac{x}{1-\frac{x^2}{2}}\]

As \(x \to 0\) , \(\frac{x^2}{2}\) becomes negligible. Therefore:

\[\tan x \to \frac{x}{1}\]

\[\tan x \to x\]

\[Q.E.D.\]

## Using Maclaurin Series

If you are familiar with Maclaurin series, you can use these to show that the small angle approximations are true (and even improve upon them). However, these will use the derivatives of \(\sin\) and \(\cos\), the proofs of which rely on small angle approximations (and can be found here). This is therefore not a replacement for the above derivations (since we can't use Maclaurin series without having derived small angle approximations), simply an improvement upon them.

The Maclaurin series of a function, \(f(x)\), is given by:

\[f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!} x^2 +\frac{f'''(0)}{3!} x^3 + \cdots + \frac{f^{(r)}(0)}{r!} x^r +\cdots=\sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!} x^r \]

Therefore the Maclaurin series for \(\sin x\) is given by:

\[\sin x=\sin(0)+\cos(0)x+\frac{-\sin(0)}{2!} x^2 +\frac{-\cos(0)}{3!} x^3 +\frac{\sin(0)}{4!}+ \cdots \]

\[\sin x=0+ x-\frac{0}{2}x^2-\frac{x^3}{6}+\frac{0}{24}+ \cdots\]

\[\sin x= x-\frac{x^3}{6}+ \cdots\]

Notice that every second term will have \(\sin(0)\) and so cancels, whilst the terms with \(\cos(0)\) alternate signs. Therefore we get:

\[\sin x= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \cdots=\sum_{r=0}^{\infty} \frac{(-1)^r }{(2r+1)!} x^{2r+1}\]

Repeating this process for \(\cos x\) gives:

\[\cos x = \sum_{r=0}^{\infty} \frac{(-1)^{r}}{(2r)!} x^{2r}\]

We can do this for \(\tan x\) as well, although expressing it as a sum is difficult. You get:

\[\tan x = \sum_{r=1}^{\infty} \frac{B_{2r}(-4)^r (1-4^r)}{(2n)!} x^{2r-1}\]

where \(B_k\) is the k^{th} Bernoulli number.