Deriving the Period of a Tan Function - Trig Graphs

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SUMMARY

The period of the tangent function can be derived using the formula for the general tangent function, expressed as y = A tan[B(x - C)] + D. The period is calculated as π/B, where B is the coefficient of the variable in the function. For example, in the function y = tan(4t), the period is π/4. Understanding the relationship between the tangent function and the unit circle is essential, as tan(x) = sin(x)/cos(x) and the periodicity can be demonstrated using the identity tan(x + π) = tan(x).

PREREQUISITES
  • Understanding of trigonometric functions and their graphs
  • Familiarity with the unit circle and its properties
  • Knowledge of the general form of trigonometric functions: y = A sin[B(x - C)] + D
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Learn how to derive the period of various trigonometric functions, including sine and cosine
  • Study the properties of the unit circle and its application in trigonometry
  • Explore the concept of phase shifts in trigonometric functions
  • Practice graphing tangent functions with different coefficients and translations
USEFUL FOR

Students studying trigonometry, educators teaching math concepts, and anyone looking to deepen their understanding of the tangent function and its graphical representation.

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

Also, how did you define the tangent function?
 
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?
 
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?
 
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

\tan(x) = \frac{\sin(x)}{\cos(x)}

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

\tan(x+\pi) = \tan(x)

To do this, do you know some formulas for ##\sin(x+\pi)## and ##\cos(x+\pi)##?
 
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 \sin(x+\pi) and \cos(x+\pi) help me get there.

Thanks
 
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.
 
graph tan(4t)

ok. here is one I missed. graph y= tan(4t)
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 y= A tan [B (x-C)] + D. A is amp, ∏/B is the period, C will give the phase shift, and D 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 4t? This is what I missed.
Thanks,
 
##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##
 

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