this is because the graph repeats every pi units on the x-axis.
the graph of cot(x) is shown below:
the period of cot(4x) would be pi/4.
this is because the graph repeats every pi/4 units on the x-axis.
the graph of cot(4x) is shown below:
period for sine and cosine and secant and cosecant function = 2pi / frequency
frequency for sine and cosine and secant and cosecant function = 2pi / period.
period for tangent and cotangent function = pi / frequency.
frequency for tangent and cotangent function = pi / period.
in your probolem, the frequency of the cotangent function is 4.
this makes the period pi/4.
if you were given the period as pi/4, then the freq8uency would be calculated as pi / (pi/4) which would then become 4.
the general formula for repeating trigonometric functions is
y = a*sin(b*(x-c))+d
a is the amplitude.
b is the frequency
c is the horizontal displacement.
d is the vertical displacement.
your equation would be shown as:
y = a*cot(b*(x-c)+d
in your equation:
a = 1
b = 4
c = 0
d = 7
your amplitude is 1.
your frequency is 4
your horizontal displacement is 0.
your vertical displacement is 7.
vertical displacement means that your center line for the graph will be 7 units above the x-axis.
here's the graph of y = cot(4x).
here's the graph of y = cot(4x) + 7
you can see that the centerline for the grpah has been vertically displaced by 7 units in a positive direction.
here's a couple of references for you to ponder.
http://www.regentsprep.org/regents/math/algtrig/att7/sincostan.htm
http://www.regentsprep.org/regents/math/algtrig/att7/othergraphs.htm
note that the normal period for sine and cosine is shown as 2pi and the normal period for tangent and cotangent is shown as pi.