 Periodic functions > The cosine function
123456The cosine function

## Theory

Take a look at the applet: Cosine function

Above you see the graph of $f\left(x\right)=cos\left(x\right)$ with $x$ in radians on $\left[0,2\pi \right]$ . The solutions of $cos\left(x\right)=c$ are shown ( $c$ is a constant).

The solution of $cos\left(x\right)=c$ within $\left[-\frac{1}{2}\pi ,\frac{1}{2}\pi \right]$ is arccosine of c: $x=arccos\left(c\right)$ .
There (often) is another solution within a range of one period.
Due to the symmetry of the graph that other solution is $x=-arccos\left(c\right)$ .

Because of the period of $2\pi$ all solutions of $cos\left(x\right)=c$ are given by:
$x=arccos\left(x\right)+k\cdot 2\pi \vee x=-arccos\left(x\right)+k\cdot 2\pi$ where $k$ k is any integer.

The equation $cos\left(x\right)=c$ only has solutions if $-1\le c\le 1$ .

The graph of the cosine function strongly resembles the graph of the sine function. Therefore there are several relations between the two.

Take a look at the applet: Unit circle

The graph of $f\left(x\right)=cos\left(x\right)$ with $x$ in radians, the standard cosine graph strongly resembles the standard sine graph and the period also is $2\pi$ . It has only been shifted to the left by 1/2Π.

This means that $cos\left(x\right)=sin\left(x+\frac{1}{2}\pi \right)$ .

Furthermore, it is possible to draw the sine and cosine of the same angle in one unit circle. When you do this, you see that the coördinates op point $P$ can be written as: ${x}_{P}=cos\left(\alpha \right)$ and ${y}_{P}=sin\left(\alpha \right)$ .
Using Pythagoras' theorem you can check that:
${\left(sin\left(\alpha \right)\right)}^{2}+{\left(cos\left(\alpha \right)\right)}^{2}=1$
To decrease the number of brackets you can use:
${sin}^{2}\left(\alpha \right)+{cos}^{2}\left(\alpha \right)=1$ There are some values that are easy to use:

• $cos\left(0\right)=1$

• $cos\left(\frac{1}{6}\pi \right)=\frac{1}{2}\sqrt{3}$

• $cos\left(\frac{1}{4}\pi \right)=\frac{1}{2}\sqrt{2}$

• $cos\left(\frac{1}{3}\pi \right)=\frac{1}{2}$

• $cos\left(\frac{1}{2}\pi \right)=0$

and vice versa:

• $arccos\left(0\right)=\frac{1}{2}\pi$

• $arccos\left(\frac{1}{2}\right)=\frac{1}{3}\pi$

• $arccos\left(\frac{1}{2}\sqrt{2}\right)=\frac{1}{4}\pi$

• $arccos\left(\frac{1}{2}\sqrt{3}\right)=\frac{1}{6}\pi$

• $arccos\left(1\right)=1$

Use these values, when exact answers are required.