Even and odd Trig functions...my ti-92 says cos is an odd function...I think

[math_knight]

The book points out how cos(t) and sec(t) are even functions (how f(-x)= f(x)), and then it shows the graph of two angles in Quadrant I and IV. t= 1/6 pi and t = - 1/6 pi.

The book sorta glosses over the topic and then goes on to give examples of angles that are conveniently on Quadrant I or IV, where x is positive.

But what about an angle of -2/3 pi, both x, and y are negative. A negative/negative = positive but what about angel t = 2/3 pi?

a.) one way which the book taught was just to think of cos (t) as the x- coordinate (if using a unit circle), in which case x would be negative for t=-2/3 pi and 2/3 pi

AND

b.) my ti-92 says cos(-2/3 pi) = -.5 --> cos(-x) = -cos (x) --> f(-x) = -f(x) --> odd function

AND

c.) my ti-92 says cos (2/3 pi) = -.5 also


d.) The "All Student Take Calculus mnemonic" says cos (t) would only be positive in the I and IV Quadrant...cos(2/3 pi) is in Quadrant II

A bit confusing when you introduce odd/even concept. I must not have a clear conceptualization of f(-x) = f(x)?

EVEN function means ---> function(negative input)=output still positive?

I hope I didn't belabor the topic...any help would be appreciated.

 

[Dr.Peterson]

But what about an angle of -2/3 pi, both x, and y are negative. A negative/negative = positive but what about angel t = 2/3 pi?

The fact that -/- = + tells you that the tangent is positive. But you aren't talking about the tangent! So I'm not sure what your point is here.

a.) one way which the book taught was just to think of cos (t) as the x- coordinate (if using a unit circle), in which case x would be negative for t=-2/3 pi and 2/3 pi

Yes, these angles are in quadrants III and II, where x is negative, so the cosines are both negative; and since they have the same reference angle, you have shown that cos(-2/3 pi) = cos(2/3 pi). That supports the claim that the cosine is an even function. (The fact that this happens for any angle is what makes the function even.)

b.) my ti-92 says cos(-2/3 pi) = -.5 --> cos(-x) = -cos (x) --> f(-x) = -f(x) --> odd function

AND

c.) my ti-92 says cos (2/3 pi) = -.5 also

You don't seem to be paying any attention to what x means here.

Your calculator is showing exactly the same fact that reasoning about quadrants did: the cosines of -2/3 pi and 2/3 pi are the same.

That is, if we take x as 2/3 pi, the cosines of x and -x are the same: cos(-x) = cos(x). So it is an even function.

Can you explain why you say "cos(-x) = -cos (x)"?

d.) The "All Student Take Calculus mnemonic" says cos (t) would only be positive in the I and IV Quadrant...cos(2/3 pi) is in Quadrant II

A bit confusing when you introduce odd/even concept. I must not have a clear conceptualization of f(-x) = f(x)?

EVEN function means ---> function(negative input)=output still positive?

I hope I didn't belabor the topic...any help would be appreciated.

You're right, you don't understand what it means to be an even function.

An even function is one that doesn't care about the sign of the input. If x is some positive number, or its opposite (-x), you get the same output from the function. That is, f(x) and f(-x) are always the same number.

This isn't primarily about signs at all, as far as y is concerned. That is probably distracting you. You need to know what the sign of the trig functions is in order to find their values correctly, but that's a side issue.

Let's get rid of trig for the moment. Consider the classic even function, f(x) = x^2. It's an even function because if you replace x with -x, it doesn't change the output: f(-x) = (-x)^2 = x^2 = f(x).

Now consider the classic odd function, g(x) = x^3. That's an odd function because if you replace x with -x, it changes the sign of the output (without changing the absolute value): g(-x) = (-x)^3 = -x^3 = -g(x).

We call a function that behaves in this respect like even powers an even function, and one that behaves like odd powers an odd function.

 

[JeffM]

The book points out how cos(t) and sec(t) are even functions (how f(-x)= f(x)), and then it shows the graph of two angles in Quadrant I and IV. t= 1/6 pi and t = - 1/6 pi.

The book sorta glosses over the topic and then goes on to give examples of angles that are conveniently on Quadrant I or IV, where x is positive.

But what about an angle of -2/3 pi, both x, and y are negative. A negative/negative = positive but what about angel t = 2/3 pi?

a.) one way which the book taught was just to think of cos (t) as the x- coordinate (if using a unit circle), in which case x would be negative for t=-2/3 pi and 2/3 pi

AND

b.) my ti-92 says cos(-2/3 pi) = -.5 --> cos(-x) = -cos (x) --> f(-x) = -f(x) --> odd function

AND

c.) my ti-92 says cos (2/3 pi) = -.5 also


d.) The "All Student Take Calculus mnemonic" says cos (t) would only be positive in the I and IV Quadrant...cos(2/3 pi) is in Quadrant II

A bit confusing when you introduce odd/even concept. I must not have a clear conceptualization of f(-x) = f(x)?

EVEN function means ---> function(negative input)=output still positive?

I hope I didn't belabor the topic...any help would be appreciated.

You have, I suspect, got confused by what is an admittedly confusing notation.

The definition \(\displaystyle f(x) \text { is an even function } \iff f(x) = f(-\ x)\)

is not talking about the sign of x. The x may be a negative number, in which case -x is a positive number. The notation refers to additive inverses, not positive and negative values.

So a slightly different but logically equivalent way to define an even function is

\(\displaystyle f(x) \text { is an even function } \iff f(|x|) = f(-\ |x|)\)

That may make it clearer that if f(x) is an even function, x is not being restricted to non-negative values.

 

[math_knight]

Thank you guys both...I still to stare at all this for it to make sense of it...

Your calculator is showing exactly the same fact that reasoning about quadrants did: the cosines of -2/3 pi and 2/3 pi are the same.

That is, if we take x as 2/3 pi, the cosines of x and -x are the same: cos(-x) = cos(x). So it is an even function.

heh...yeah I see that now...still doesn't want to fit in my brain for some reason

Can you explain why you say "cos(-x) = -cos (x)"?

I'm just thrown off a bit...the x^2, x^3 examples you give are the one's the book gives but, I was thinking in terms of input/output (which, granted, should turn out the same),

but for example f(x)=x^2 ; let x (the input)= 2, -2 means y (the output) = 4, 4 , always positive

but cos(-2/3 pi) = -.5

The input was negative, and the output was ALSO negative...that's not what happens with X^2. It's like X^2 CONVERTS a negative input, to a POSITIVE output. But like you both said, I don't understand what it means for a function to be even/odd. I think x^2, x^3 are bad examples, or at least bad examples to FINISH (as the book did). Maybe other examples should be given in the book.

You have, I suspect, got confused by what is an admittedly confusing notation.

The definition \(\displaystyle f(x) \text { is an even function } \iff f(x) = f(-\ x)\)

is not talking about the sign of x. The x may be a negative number, in which case -x is a positive number. The notation refers to additive inverses, not positive and negative values.

So a slightly different but logically equivalent way to define an even function is

\(\displaystyle f(x) \text { is an even function } \iff f(|x|) = f(-\ |x|)\)

That may make it clearer that if f(x) is an even function, x is not being restricted to non-negative values.

Yes, I think I am beginning to see...I just have to stare at that for a moment. I never saw the absolute value notation used like that to explain even/odd functions.

 

[Dr.Peterson]

I'm just thrown off a bit...the x^2, x^3 examples you give are the one's the book gives but, I was thinking in terms of input/output (which, granted, should turn out the same),

but for example f(x)=x^2 ; let x (the input)= 2, -2 means y (the output) = 4, 4 , always positive

but cos(-2/3 pi) = -.5

The input was negative, and the output was ALSO negative...that's not what happens with X^2. It's like X^2 CONVERTS a negative input, to a POSITIVE output. But like you both said, I don't understand what it means for a function to be even/odd. I think x^2, x^3 are bad examples, or at least bad examples to FINISH (as the book did). Maybe other examples should be given in the book.

Actually, x^2 and x^3 are the motivating examples that are often given first; they certainly shouldn't be the last! What I mean is that this is why we use the words even and odd -- because even powers are even functions, and odd powers are odd functions.

You seem to be stuck on the mere idea of what the sign of the result is, rather than on the main idea, which is symmetry of the graph. I hope your book pointed this out. It is not whether the result is always positive. It is whether ...

• the graph is the reflected from one side to the other (so you could fold along the y-axis, and the two halves will match) -- an even function has this "line symmetry" about the y-axis; or whether,
• if you put a pin at the origin and spin the graph 180 degrees around, it matches the original -- an odd function is one that has this "point symmetry" about the origin.

See here, or here for more about symmetry.

 

[JeffM]

Thank you guys both...I still to stare at all this for it to make sense of it...



heh...yeah I see that now...still doesn't want to fit in my brain for some reason




I'm just thrown off a bit...the x^2, x^3 examples you give are the one's the book gives but, I was thinking in terms of input/output (which, granted, should turn out the same),

but for example f(x)=x^2 ; let x (the input)= 2, -2 means y (the output) = 4, 4 , always positive

but cos(-2/3 pi) = -.5

The input was negative, and the output was ALSO negative...that's not what happens with X^2. It's like X^2 CONVERTS a negative input, to a POSITIVE output. But like you both said, I don't understand what it means for a function to be even/odd. I think x^2, x^3 are bad examples, or at least bad examples to FINISH (as the book did). Maybe other examples should be given in the book.



Yes, I think I am beginning to see...I just have to stare at that for a moment. I never saw the absolute value notation used like that to explain even/odd functions.

Whoa. There is no implication that if f(x) is an even function that it is a positive function.

\(\displaystyle \text {Given: } f(x) = -\ (x^2 + 1) < 0 \text { for all real } x \implies\)

\(\displaystyle f(x) < 0 \text { for all real } x \text { and}\)

\(\displaystyle f(a) = -\ a^2 - 1 = -\ (-\ a)^2 - 1 = f(-\ a).\)

The function given above is an even function that is negative for all x.

The graph of an even function is symmetric with respect to the vertical axis. It may be positive everywhere or negative everywhere or positive somewhere and negative somewhere, but the graphs on either side of the vertical axis will be mirror images of each other.