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Even odd trig functions

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    I dont know how other people learned trig, but in my book, the first 2 chapters have nothing to do with angles, but are mostly circles and trig functions of real numbers. Does this make sense to anyone? I am in the next chapter already and I know there are other, easier ways of calculating trig, but in these chapters, I cant do that. I know how to show sin and cos are odd and even with a random number and a calculator, but I have to show it the way presented below. Ive googled and my book only shows the properties with answers, no worked probs. Can anyone explain to me if I am doing this right or what I am doing wrong? thanks!

    a few examples
    1. f(x)= x2 sin x
    2. f(x)= x3+ cos x
    3. f(x)= xsin3x
    4. f(x)= cos(sin x)

    2. Relevant equations
    even odd properties that have all trig functions odd except cos and sec

    3. The attempt at a solution
    1. f(-x)= (-x)2 sin (-x) = x2 sin -x = -f(x)= odd.
    the answer says -x2 sin x = -f(x) = odd. why am i wrong?

    2. f(-x)= (-x)3 + cos(-x) = -x3+ cos x = neither

    3. (-x)sin3(-x)
    frankly i have no clue how to do that one.

    4. cos(sin(-x)) would be cos (-x) = x = f(x) = even?????
  2. jcsd
  3. Oct 30, 2008 #2


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    Right angle triangles are probably the first everyday context in which the trigonometric functions sine and cosine arise, and as a result, these two functions are often first introduced IN that context. That having been said, the definitions of trigonometric functions are best (i.e. most generally) treated in terms of the unit circle. The properties of trigonometric functions (not to mention their myriad uses and their ubiquity) is a rich area of mathematics. You should stop thinking of trigonometric functions as merely "useful tools in geometry problems" and start thinking of them as mathematical functions that just happen to exist (independently of any sort of application) and that have unique properties worth investigating. So, try to keep an open mind.

    This definitely makes no sense. Oddness and evenness in this context are symmetry properties of functions and have nothing to do with oddness or evenness of specific values of those functions for specific arguments.

    You're not wrong. It is odd. You just didn't complete the last step, which is to show that since sin(-x) = -sin(x) (i.e. since the sine function is, itself, odd), the whole thing becomes -x^2 sin(x), which is -f(x).

    I'm pretty sure that odd powers of odd functions are odd. Which means that both functions in the product are odd, which means that the product is even.

    I don't know what you're doing here. It's baffling. Where did the sine go? How did cos(-x) become x?


    = cos(-sin(x)) (using the oddness of sine)

    = cos(sin(x)) (using the evenness of cosine)

    = f(x)
    Last edited: Oct 30, 2008
  4. Oct 30, 2008 #3
    1. You are not wrong. You have x^2sin(-x). But sin(-x)=-sin(x), and so x^2sin(-x)=-x^2sin(x).

    2. f(-x)= (-x)3 + cos(-x) = -x3+ cos x = neither

    3. (-x)sin3(-x)
    frankly i have no clue how to do that one.

    Again, sin(-x)=-sin(x).

    4. cos(sin(-x)) would be cos (-x) = x = f(x) = even?????

    cos(sin(-x))= cos(-sin(x))
    Cos(-a)=cos(a); cos(-sin(x))=cos(sin(x))=f(x).
    Even. So yes, you were absolutely correct in your assumption.

    This is all supposing that you are allowed to know that sin(x) is odd and cos(x) is even, but generally you are allowed to know the properties of standard trig functions by heart. At least we are at my school.
    Good luck with the rest of the problems!
  5. Oct 30, 2008 #4


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    Some advice:

    Here you've been given mostly combinations (sum and products) of functions (and one composition of functions). So you could prove general properties like:

    (odd function)*(odd function) = even function

    (odd function)*(even function) = odd function


    then things would go much faster. It's not even really that hard.

    let f(x) be odd. Let g(x) be even.

    Let h(x) = f(x)g(x)

    Is h(x) odd or even?

    h(-x) = f(-x)g(-x) = -f(x)g(x) = -h(x)

    ==> h(x) is odd.

    Since this applies generally for any two functions, we can say that:

    "The product of an odd function and an even function will be odd."

    There, I did one for you ;-)

    Knowing this would have allowed you to do the first problem by inspection. You can prove other properties that apply to problems 2-4
  6. Oct 30, 2008 #5
    thank you both so much. as for the cos(sin(-x)) , i meant to say the cos of the negative sin would be even due to the cos, but i completely left out the sin on that one.

    and thanks for the explanations. even though i already understood even and odd functions, bringing sin and cos into them confused me for no real reason except that I am not used to trig. and your explanations are so thorough compared to my book, which just lists which functions are even and odd. i will be printing this page out. thanks again.
  7. Nov 2, 2008 #6
    One more question about even and odd trig functions. would a graph of a function that is symmetrical about the y axis be even and one symmetric about the x axis be odd?
  8. Nov 2, 2008 #7


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    Not quite. Your first statement is correct. *Any* time you talk about symmetry, what you're really saying is that you can apply some sort of "transformation" and still end up with exactly what you started with. In the case of an even function, the transformation in question is a reflection across the y-axis.

    In the case of an odd function, the transformation is NOT a reflection (not even about the x-axis). Instead, the transformation is a rotation about the origin. If you rotate the graph about the origin by 180 degrees, you'll get back what you started with. These functions have symmetry about a point (or point symmetry).

    To convince yourself this is not just a reflection about the x-axis, try actually doing that to the sine function and see what you get. A more obvious example is y=x.
  9. Nov 2, 2008 #8
    ahhh, thank you for making clear what otherwise i would have ignored.
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