- #1

- 47

- 0

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

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Physicslad78
- Start date

- #1

- 47

- 0

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

Thanks

- #2

- 179

- 0

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

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:

- #3

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

(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)

- #4

- 179

- 0

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?

- #5

- 47

- 0

- #6

- 492

- 0

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?

- #7

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

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 =

- #8

- 492

- 0

oops. lol.

Share: