We need to be a little careful with this graph. \(\cos \left( x \right)\) has a period of \(2\pi \), but we’re not dealing with \(\cos \left( x \right)\) here. We are dealing with \(\cos \left( {2x} \right)\). In this case notice that if we plug in \(x = \pi \) we will get

\[\cos \left( {2\left( \pi \right)} \right) = \cos \left( {2\pi } \right) = \cos \left( 0 \right) = 1\]

In this case the function starts to repeat itself after \(\pi \) instead of \(2\pi \)! So, this function has a period of \(\pi \). So, we can expect the graph to repeat itself 8 times in the range \( - 4\pi \le x \le 4\pi \). Here is that graph.

Sure enough, there are twice as many cycles in this graph.

In general, we can get the period of \(\cos \left( {\omega \,x} \right)\) using the following.

\[{\rm{Period}} = \frac{{2\pi }}{\omega }\]

If \(\omega > 1\) we can expect a period smaller than \(2\pi \) and so the graph will oscillate faster. Likewise, if \(\omega < 1\) we can expect a period larger than \(2\pi \) and so the graph will oscillate slower.

Note that the period does not affect how large cosine will get. We still have

\[ - 1 \le \cos \left( {2x} \right) \le 1\]