Period changes for sine and cosine

Preliminaries

 

We define the $\cos$cos and $\sin$sin functions as the horizontal and vertical coordinates of a point that moves on the unit circle. In the diagrams below, this is shown for an angle $\alpha$α in the first and second quadrants.

If we imagine the point moving anticlockwise on the unit circle so that the radius from the point makes an ever-increasing angle with the positive horizontal axis, eventually the angle exceeds $360^\circ$360°; but the values of the $\cos$cos and $\sin$sin functions repeat the values of the coordinates from the angle $360^\circ$360° smaller. We say $\sin$sin and $\cos$cos are periodic functions with period $360^\circ$360°.

Thus, for any angle $\alpha$α, there is a sequence of angles with the same value of $\sin\alpha$sinα.

$...,\alpha-720^\circ,\alpha-360^\circ,\alpha,\alpha+360^\circ,\alpha+720^\circ,\alpha+1080^\circ,...$...,α−720°,α−360°,α,α+360°,α+720°,α+1080°,...

 

Transformations involving period

Again, consider the angle $\alpha$α made by the point moving around the unit circle. If a new angle $\alpha'$α′ is defined by $\alpha'=k\alpha$α′=kα, We know that $\sin\alpha'$sinα′ has period $360^\circ$360°, but we see that $\alpha'$α′ reaches $360^\circ$360°  when $\alpha=\frac{360^\circ}{k}$α=360°k​. So, $\sin k\alpha$sinkα and $\cos k\alpha$coskα must have period $\frac{360^\circ}{k}$360°k​ with respect to $\alpha$α.

Example 1

The function $\sin2x$sin2x begins to repeat when $2x=360^\circ$2x=360°. That is, when $x=180^\circ$x=180°. So, $\sin2x$sin2x has period $180^\circ$180°. The period is multiplied by $\frac{1}{2}$12​ when $x$x is multiplied by $2$2.

 

 

 

Thus, we see that for functions $\sin kx$sinkx and $\cos kx$coskx where $k$k is a constant, the period of the function with respect to $kx$kx is $360^\circ$360° but the period with respect to $x$x is $\frac{360^\circ}{k}$360°k​.

We can use these ideas to deduce the formula for a sine or cosine function from a graph.

Example 2

 

This graph looks like the graph of a cosine function since it has the value $1$1 at $0$0. However, the period is $288^\circ$288°.

We know that $\cos k\alpha^\circ$coskα° has period $\frac{360^\circ}{k}$360°k​ and, in this case, $\frac{360^\circ}{k}=288^\circ$360°k​=288°. Therefore, $k=\frac{360^\circ}{288}=1.25$k=360°288​=1.25.

The graph must belong to the function given by $\cos\left(1.25\alpha^\circ\right)$cos(1.25α°).

 

Worked examples

Question 1

Question 2

Question 3