# Trigonometric function with specific properties

## Main Question or Discussion Point

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

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:
tiny-tim
Homework Helper

(have a pi: π and a theta: θ )
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?
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)+ = ±∞

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

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?

tiny-tim
Homework Helper

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

oops. lol.