- #1

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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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- Thread starter Physicslad78
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- #1

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

Thanks

- #2

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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

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

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(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

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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

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- #6

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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

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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

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oops. lol.

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