Trigonometric function with specific properties

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Discussion Overview

The discussion revolves around the existence of a trigonometric function f(theta) that satisfies the condition f(Pi-theta) = -f(theta) and has its square f(theta)^2 maximized at Pi/2. The scope includes theoretical exploration of trigonometric properties and potential functions.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if there exists a trigonometric function f(theta) such that f(Pi-theta) = -f(theta) and f(theta)^2 is maximized at Pi/2.
  • Another participant suggests that if f(Pi-theta) = -f(theta), then f(Pi/2) must equal zero, leading to the conclusion that f(theta) could be the zero function, which is deemed uninteresting.
  • A later reply proposes that if f is allowed to be infinite at Pi/2, then it could satisfy the conditions, referencing the behavior of the tangent function.
  • One participant clarifies that f cannot be infinite as it represents a wave function in their research, reinforcing the idea that f(theta) = 0 may be the only solution.
  • Another participant introduces the sine function, arguing that it satisfies the condition f(Pi-theta) = -f(theta) and has local maxima for its square at Pi/2 + nPi, questioning if they are missing something.
  • A follow-up acknowledges a mistake in the sine function's properties but does not resolve the inquiry.

Areas of Agreement / Disagreement

Participants express differing views on the existence of suitable functions, with some suggesting the zero function as a solution while others propose the sine function. The discussion remains unresolved regarding the existence of a non-trivial function that meets the criteria.

Contextual Notes

There are limitations regarding the assumptions about the function's behavior at Pi/2, particularly concerning whether it can be infinite or must remain finite. The discussion also reflects varying interpretations of trigonometric identities and properties.

Physicslad78
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Is there any trigonometric function f(theta) where theta is an angle such that

f(Pi-theta)=-f(theta) and such that its square (i.e f(theta)^2) is maximum at Pi/2?


Thanks
 
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try theta = PI/2

then f(PI-(PI/2)) = f(PI/2)

so:

if f(PI-theta)=-f(theta) for all theta, then f(PI/2) must equal zero.

and you also said that:

f(PI/2)^2 is its maximum. but f(PI/2)^2 is 0.

so f(theta) = zero, for all theta. A boring and useless function.

edit: Presuming f(theta) cannot be infinite
 
Last edited:
Hi Physicslad78! Hi Georgepowell! :smile:

(have a pi: π and a theta: θ :wink:)
Physicslad78 said:
Is there any trigonometric function f(theta) where theta is an angle such that

f(Pi-theta)=-f(theta) and such that its square (i.e f(theta)^2) is maximum at Pi/2?
Georgepowell said:
if f(PI-theta)=-f(theta) for all theta, then f(PI/2) must equal zero.

It depends exacty what the question envisages …

if f is allowed to be infinite at π/2, then you could have f(π/2)- = -f(π/2)+ = ±∞ :wink:
 
if f is allowed to be infinite at π/2, then you could have f(π/2)- = -f(π/2)+ = ±∞ :wink:

Ahh, I missed that. How about the tan() function then?
 
Thanks a lot guys..f cannot be infinite as it is a wave function in a physics system I am researching on..I just wanted to make sure that there aint any solutions except the f(theta)=0 as tiny-tim pointed out..Thanks again...
 
Wait... what about sin(x)?

sin(Pi-theta) = sin(PI)cos(theta) - sin(theta)cos(pi) = -sin(theta)

And [sin(x)]^2 has local maxima at PI/2 + nPI.

Am I missing something?
 
AUMathTutor said:
Wait... what about sin(x)?

sin(Pi-theta) = sin(PI)cos(theta) - sin(theta)cos(pi) = -sin(theta)

And [sin(x)]^2 has local maxima at PI/2 + nPI.

Am I missing something?

Yup! … cosπ = minus 1 :wink:
 
oops. lol.
 

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