MA101.13 Trigonometrical functions

Series definitions for the sin and cosine functions are:

\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...
\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...

These converge \forall x \in \mathbb{R}.

If we differentiate these term by term we can see that:

\frac{d(\sin{x})}{dx} = \cos{x}
\frac{d(\cos{x})}{dx} = -\sin{x}

Many other properties can be deduced from these power series.

The graph of y = \sin{x} is as shown:

We can see that sin is not an injection (domain of \mathbb{R}) and so there is no inverse. However the function f:[-\frac{\pi}{2},\frac{\pi}{2}]\to[-1,1], or f:x\to \sin{x} [called the cut down sine], has the graph:

and this is an injection, and has an inverse function f^{-1} with domain $latex[-1,1]$ and range [-\frac{\pi}{2},\frac{\pi}{2}]. f^{-1} is the unique real number (angle) between [-\frac{\pi}{2},\frac{\pi}{2}] whose sine is x. The f^{-1} is symbolised by \sin^{-1}{x} or \arcsin{x}.

The graph of \arcsin{x} is a reflection of y=\sin{x} (cut down) in the line y=x.

Knowing y=\arcsin{x} is true, you may deduce x=\sin{x} is true.

Example: \frac{\pi}{4}=arcsin(\frac{1}{\sqrt{2}}) \Rightarrow sin)\frac{\pi}{4} = \frac{1}{\sqrt{2}}.

Knowing x = sin(y) is true, you may not deduce y = arcsin(x).

Example: \frac{1}{\sqrt{2}}=sin(\frac{3\pi}{4}) \nRightarrow \frac{3\pi}{4} = arcsin(\frac{1}{\sqrt{2}}).

Theorem

\frac{d(arcsin(x))}{dx} = \frac{1}{\sqrt{1-x^2}}

Proof

\begin{aligned} \mbox{Let: } y &= arcsin(x) \\ \mbox{then } x &= sin(y) \\ \frac{dx}{dy} &= cos(y) \\ \frac{dy}{dx} &= \frac{1}{cos(y)} \\ &= \frac{1}{\sqrt{1-sin^2(y)}} \\ &= \frac{1}{\sqrt{1-x^2}} \end{aligned}

We take the positive square root because for [-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}] the \cos{y} \geq 0


Similarly we define the ‘so called’ cut down cosine – this is cosine but with the domain [0,\pi], and the cut down tangent with domain (-\frac{\pi}{2}, \frac{\pi}{2}). The inverses of these functions are arccos and arctan.

MA101.12 The exponential functions

We define

\exp{x} = 1 +x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+... which in fact converges \forall x \in \mathbb{R}.

\begin{aligned} \exp{x} \cdot  \exp{y} &= (1 +x+\frac{x^2}{2!}+...)(1 +y+\frac{y^2}{2!}+...) \\&= 1 +x+y+\frac{x^2}{2!}+xy+\frac{y^2}{2!}+...+\frac{x^n}{n!}+\frac{x^{n-1}y}{n-1!}+\frac{x^{n-2}y^2}{(n-2)!2!}+...+\frac{y^2}{2!}+... \\&= 1 +(x+y)+\frac{(x+y)^2}{2!}+...+\frac{(x+y)^n}{n!}+... \\&= \exp{(x+y)} \end{aligned}

We have assumed that we can multiply infinite series as if they were finite algebraic expressions. This is not always true, but it is for convergent power series.

The relationship \exp{x}\cdot\exp{y} = \exp{(x+y)} leads us to write \exp{x}=e^x so that e^x\cdot e^y = e^{x+y}.

Note, \exp{1} = e^1 = e = 1+1+\frac{1}{2!}+\frac{1}{3!}+...

On the assumption that a power series can be differentiated term by term \frac{d e^x}{dx} = 0 + 1 +\frac{x^2}{2!}+...=e^x.

Properties of exp

The graph of y=e^x is as shown, (with series definition this is not easy to see), but there is some evidence from the property e^{x_1}\cdot e^{x_2}=e^{x_1+x_2}

Observe from the graph the properties:

  1. \displaystyle \lim_{x\to\infty}e^x=\infty.
  2. \displaystyle \lim_{x\to-\infty}e^x=0.
  3. \forall x, e^x >0.
  4. e^x is continuous everywhere.
  5. D(e^x)=\mathbb{R}.
  6. e^x is an injection (or one to one).
    ie. x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2).

This means there will be an inverse function. We call it the log function. We have \log{y} equal to the unique x such that y=e^x.

y=e^x and x=\ln{y} mean of course the same thing. The graph of y=\ln{x} will be a reflection of y=e^x in the line y=x.

Note that the domain of ln = range of exp = \mathbb{R}^+.

Properties of ln
  1. If y=\ln{x} then x=e^y, so \frac{dx}{dy} = e^y.
    Therefore, \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^y} = \frac{1}{x}.
  2. \ln{x} is an injection.
    Proof:
    Let \ln{x}=\ln{y}.
    Put \ln{x}=p and \ln{y}=p.
    So, e^p=x and e^p=y.
    Whence x=y.
  3. \forall x, x=e^{\ln{x}}.
    Proof:
    Let y=e^{ln{x}}.
    Then \ln{y} = \ln{x}.
    Whence x=y by 2.
    So, x=e^{\ln{x}}.
  4. \ln{xy}=\ln{x}+\ln{y}.
    Proof:
    \begin{aligned} \ln{xy}&=\ln{e^{\ln{x}}\cdot e^{\ln{y}}} \\&= \ln{e^{\ln{x}+\ln{y}}} \\ xy &= e^{ln{x}+\ln{y}} \\ \mbox{Hence}\; \ln{x}+\ln{y} &= \ln{xy} \end{aligned}.
  5. \forall a \in \mathbb{R}^+
    (i) if  m in an integer then a^m = m\ln{a}.
    (ii) if \frac{m}{n} is in \mathbb{Q}, then equally we have \ln{a\frac{m}{n}}=\frac{m}{n}\ln{a}. (a^{\frac{m}{n}} \mbox{ means } \sqrt[n]{a^m}.
Definition

For a \in \mathbb{R}^+ and x \in \mathbb{R} we define a^x = e^{x\ln{a}}.

For example 3^{\sqrt{2}}=e^{\sqrt{2}\ln{3}}.

Note that this definition fits in with the usual one when x is rational.

For fixed a \in \mathbb{R}^+, a^x is a function.

Note: \frac{d(a^x)}{dx} = \frac{d(e^{\ln{a^x}}}{dx} = \frac{d(e^{x\ln{a}})}{dx} = \ln{a(e^{x\ln{a}})} = a^x\ln{a}.