Question 1101500
<br>
Here is a graph of sin(x) (red) and (2/3)sin((1/2)x) (green):<br>
{{{graph(400,200,-2*pi,4*pi,-2,2,sin(x),(2/3)sin(x/2))}}}<br>
The graph of sin(x), as I assume you know, oscillates between 1 and -1; its amplitude is 1.  And it completes one period every 2pi (~6.28) radians.<br>
For your modified function, the 2/3 is the amplitude; so now the graph oscillates between 2/3 and -2/3.  The smaller amplitude of the green graph is easily seen.<br>
The factor (1/2) on the sine function changes the period of the function.  You know that the sine function completes a period between 0 and 2pi:
{{{0 <= x < 2pi}}}<br>
So the function sin((1/2)x) similarly completes a period between 0 and 2pi; that means it completes a period every 4pi.  You can see that by solving the inequality:
{{{0 <= (1/2)x < 2pi}}}  -->  {{{0 <= x < 4pi}}}<br>
You can see on the graph that the modified function (green) has a period of 4pi (~12.56).<br>
It is likely that while you are studying this topic, you will be looking for the period of modified trig functions frequently.  Instead of going through this kind of analysis to find the period in every example, just use the fact that the period of sin(nx) (or cos(nx)) is just equal to (2pi/n).<br>
For this example, the period of sin((1/2)x) is {{{(2pi/(1/2)) = 4pi}}}.<br>
To summarize what is involved in this example, the (2/3) changes the amplitude from 1 to (2/3)*1 = 2/3; the (1/2) changes the period from 2pi to (2pi/(1/2) = 4pi.<br>
In general, the graph of a*sin(bx) will have an amplitude of a and a period of (2pi/b).