In Example 2 we showed that the period of $y=\cos 3 x$ is $2 \pi / 3 .$ Use the same method to show that the period of $y=A \cos B x,$ with $B>0,$ is $2 \pi / B$
So we have equation. Why is equal to a co sign? BX? So we know that the graph starts or one period will start. We're X is equal to zero and then one p aed stops where X is equal to to buy or, in other words, tax the point where P X or three X is equal to to buy. So this would that mean that X is equal to two pi over be so, therefore for dysfunction. What we can say is that period is equal to two high over B.