# The tangent function, and evaluating $$\pi$$

## Tangent

### Tan

The $$\tan(\theta )$$ function is defined as:

$$\tan(\theta ):=\dfrac{\sin(\theta )}{\cos(\theta )}$$

#### Behaviour around $$0$$

$$\sin(0)=0$$

$$\cos(0)=1$$

$$\tan(0):=\dfrac{\sin(0)}{\cos(0)}$$

$$\tan(0)=\dfrac{0}{1}$$

$$\tan(0)=0$$

#### Behaviour around $$\cos(\theta )=0$$

$$\tan(\theta )=\dfrac{\sin(\theta )}{\cos(\theta )}$$

So $$\tan (\theta )$$ is undefined where $$\cos(\theta )=0$$.

This happens where:

$$\theta=\dfrac{\tau }{4}+\dfrac{1}{2}n\tau$$

$$\theta=\dfrac{1}{4}\tau (1+2n)$$

Where $$n\in \mathbb{Z}$$.

#### Derivatives

$$\tan(\theta )=\dfrac{\sin(\theta )}{\cos(\theta )}$$

$$\dfrac{\delta }{\delta \theta } \tan(\theta )=\dfrac{\delta }{\delta \theta }\dfrac{\sin(\theta )}{\cos(\theta )}$$

$$\dfrac{\delta }{\delta \theta } \tan(\theta )=\dfrac{\cos(\theta )}{\cos(\theta )}+ \dfrac{\sin^2(\theta )}{\cos^n(\theta )}$$

$$\dfrac{\delta }{\delta \theta } \tan(\theta )=1+\tan^2(\theta )$$

Note this is always positive. This means:

$$\lim_{\cos(\theta )\rightarrow 0^+}=-\infty$$

$$\lim_{\cos(\theta )\rightarrow 0^-}=\infty$$

### Inverse functions

#### Inverse trigonometric functions

$$\sin (\arcsin (\theta )):=\theta$$

$$\cos (\arccos (\theta )):=\theta$$

$$\tan (\arctan (\theta )):=\theta$$

### Integrals

#### Cosine and sine

$$\arccos (\theta)$$, $$\arcsin (\theta )$$ and difficulty of inversing

In order to determine $$\tau$$ we need inverse functions for $$\cos (\theta )$$ or $$\sin (\theta )$$.

These are the $$\arccos (\theta )$$ and $$\arcsin (\theta )$$ functions respectively.

However this is not easily calculated. Instead we look for another function.

#### Calculating $$\arctan (\theta )$$

So we want a function to inverse this. This is the $$\arctan (\theta )$$ function.

If $$y=\tan (\theta )$$, then:

$$\theta =\arctan (y)$$

We know the derivative for $$\tan (\theta )$$ is:

$$\dfrac{\delta }{\delta \theta }\tan (\theta )=1+\tan^2(\theta )$$

$$\dfrac{\delta y}{\delta \theta }=1+y^2$$

So

$$\dfrac{\delta \theta }{\delta y}=\dfrac{1}{1+y^2}$$

$$\dfrac{\delta }{\delta y}\arctan (y)=\dfrac{1}{1+y^2}$$

So the value for $$\arctan (k)$$ is:

$$\arctan (k)=\arctan (a)+\int_a^k\dfrac{\delta }{\delta y}\arctan (y) \delta y$$

$$\arctan (k)=\arctan (a)+\int_a^k\dfrac{1}{1+y^2} \delta y$$

What do we know about this function? We know it can map to multiple values of $$\theta$$ because the underlying $$\sin (\theta )$$ and $$\cos (\theta )$$ functions also loop.

We know that one of the results for $$\arctan (0)$$ is $$0$$.

### Calculating $$\tau$$

As we note above, $$\sin (\theta )=\cos (\theta )$$ at $$\theta =\tau *\dfrac{1}{8}$$

This is also where $$\tan (\theta )=1$$.

$$\arctan (k)=\arctan (a)+\int_a^k\dfrac{1}{1+y^2} \delta y$$

We start from $$a=0$$.

$$\arctan (k)=\arctan (0)+\int_0^k\dfrac{1}{1+y^2} \delta y$$

We know that one of the results for $$\arctan (0)$$ is $$0$$.

$$\arctan (k)=\int_0^k\dfrac{1}{1+y^2} \delta y$$

We want $$k=1$$

$$\arctan (1)=\int_0^1\dfrac{1}{1+y^2} \delta y$$

$$\dfrac{\tau }{8}=\int_0^1\dfrac{1}{1+y^2} \delta y$$

$$\tau =8\int_0^1\dfrac{1}{1+y^2} \delta y$$

We know that the $$\cos (\theta )$$ and $$\sin (\theta )$$ functions cycle with period $$\tau$$.

Therefore $$cos (n.\tau )=\cos (0)$$