Deriving the Period of a Tan Function - Trig Graphs

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In summary: So to find the period of the function, simply plug in ##4t## into the general formula and solve for ##t##.In summary, the tan function has a period of ∏/b. How is this derived? I know it has to do with tan = y/x right? but I just don't understand how to derive the period when you're graphing a tan function.
  • #1
datafiend
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I know the general equation for trig functions and how to manipulate them:
y=A sin [B (x-c)] + D
howver , the tan function has a period of ∏/b. how is this derived? I know it has to do with tan = y/x right? but I just don't understand how to derive the period when you're graphing a tan function.

Thanks,
Randy
 
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  • #2
You're asking about the function ##f(x) = \tan(x)##? Then what is ##b##?

Also, how did you define the tangent function?
 
  • #3
datafiend said:
I know the general equation for trig functions and how to manipulate them:
y=A sin [B (x-c)] + D
howver , the tan function has a period of ∏/b. how is this derived? I know it has to do with tan = y/x right? but I just don't understand how to derive the period when you're graphing a tan function.

Thanks,
Randy

In addition to what micromass said, "tan = y/x" is not correct. There's an angle involved that you don't show. Tangent of what? It's a little like saying "√ = 4". Square root of what?
 
  • #4
hmmm...maybe I'm not being clear. If using the "unit circle" the sin function is y/r, cos is x/r, tan is y/x, where r=1. Is this not how you plot a sin/cos/tan function on the x/y plane?
 
  • #5
datafiend said:
hmmm...maybe I'm not being clear. If using the "unit circle" the sin function is y/r, cos is x/r, tan is y/x, where r=1. Is this not how you plot a sin/cos/tan function on the x/y plane?

Well first of all, if you take a unit circle definition, then I don't see why you bother to write the ##r##.

Second, you ignored the post by Mark. There is no such thing as a ##\tan##. You need to take the ##\tan## of some angle.

Anyway, let's move on to you question. You know that

[tex]\tan(x) = \frac{\sin(x)}{\cos(x)}[/tex]

holds for all ##x## for which the fraction on the right is defined.
Can you try to show that

[tex]\tan(x+\pi) = \tan(x)[/tex]

To do this, do you know some formulas for ##\sin(x+\pi)## and ##\cos(x+\pi)##?
 
  • #6
tan(x) formula

I'm sorry, but I don't see how this is germane to a tan function. In a standard problem that asks to A:graph a tangent function B:show the period of the function C: show the asymptotes D: give the domain and range. I don't see how the formulas [tex]\sin(x+\pi)[/tex] and [tex]\cos(x+\pi)[/tex] help me get there.

Thanks
 
  • #7
Seeing as this is a standard problem, I moved it to the homework forums. Now, please provide an attempt at solving the problem before we can continue.
 
  • #8
graph tan(4t)

ok. here is one I missed. graph [tex] y= tan(4t) [/tex]
A: find the period.
B: find the phase shift
c: give the domain/range
d: find the asymptotes
the general formula for the tan/cot funcit is [tex] y= A tan [B (x-C)] + D [/tex]. A is amp, [tex] ∏/B [/tex] is the period, [tex] C [/tex] will give the phase shift, and [tex] D [/tex] is the vertical translation.
I know that at the points on the unit circle (0,1) and (0,-1) the function is UNDEFINED, so this is the asymptote. Now WHAT IS THE [tex] 4t [/tex]? This is what I missed.
Thanks,
 
  • #9
##4t## is your independent variable, instead of an ##x## or ##\theta## that you might usually see. You can use this 4t to find the period of this particular function. ##tan\theta## contains a period of ##\pi## but since you're dealing w/ ##tan(n\theta##), your period will be ##\frac{\pi}{n}## which will give you a ratio of ##\pi## relative to your function, in other words n(##\frac{\pi}{n}##) = ##\pi##
 

What is the period of a tan function?

The period of a tan function is the length of one complete cycle of the function. In other words, it is the distance between two consecutive peaks or troughs on the graph.

How do you derive the period of a tan function?

The period of a tan function can be derived by using the formula T=2π/b, where b is the coefficient of x in the function. This means that the period is equal to 2π divided by the value of b.

What is the relationship between the period and the coefficient of a tan function?

The coefficient of a tan function is directly proportional to the period. This means that as the coefficient increases, the period of the function also increases. This can be seen in the graph of the function, where a larger coefficient results in a wider graph with a longer period.

Can the period of a tan function be negative?

No, the period of a tan function cannot be negative. The period is a measure of distance and cannot have a negative value. It is always either positive or zero.

How does the period of a tan function relate to its frequency?

The period and frequency of a tan function are inversely related. This means that as the period increases, the frequency decreases and vice versa. The frequency is equal to 1 divided by the period.

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