Question 554415
not sure if i know the answer to this, but, in general, the formula for a repeating trigonometric function is:
y = a(b(x-c)+d
a is the amplitude.
b is the frequency
c is the horizontal displacement
d is the vertical displacement.
the function you are describing looks like the reciprocal of the cosine function.
as a cosine function, the formula would be:
y = cos(x-pi/2)
the frequency is 1
the horizontal displacement is pi/2
what this means is:
when x = 0, you are taking the cosine of (-pi/2)
when x = pi/2, you are taking the cosine of 0
etc.
the graph that peaks at x = 0 is the cosine(x) graph.
the graph that peaks at x = pi/2 is the cosine(x-pi/2) graph.
this is because:
when x = 0, cosine(x) = cosine(0) = 1
and:
when x = pi/2, cosine(x-pi/2) = cosine(0) = 1
the graph of y = sec(x-pi/2) is the reciprocal of the graph of cos(x-pi/2)
the graph of the cosine function of x would look like this:
{{{graph(600,600,-2pi,2pi,-1,1,cos(x))}}}
the graph of the cosine function of (x-pi/2) would look like this:
{{{graph(600,600,-2pi,2pi,-1,1,cos(x-pi/2))}}}
if you superimpose the second graph on the first, you will see that there is a shift of pi/2 as shown below:
{{{graph(600,600,-2pi,2pi,-1,1,cos(x),cos(x-pi/2))}}}
since sec(x) = 1/cos(x), this means that the graph of:
y = sec(x-pi/2) can be graphed as:
y = 1/cos(x-pi/2)
that graph is shown below:
{{{graph(600,600,-2pi,2pi,-5,5,1/cos(x-pi/2))}}}
when you superimpose that graph on the graph of cos(x-pi/2) you get the following:
{{{graph(600,600,-2pi,2pi,-5,5,cos(x-pi/2),1/cos(x-pi/2))}}}
i'm not exactly sure what you mean by dilation factor.
i couldn't see where it would apply in this problem.
if you are talkin about a change in frequency which would result in a change in period, that doesn't apply to this problem as far as i can see.