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.