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How can sin(-θ)=-sinθ?

  1. May 20, 2014 #1
    I'm having a hard time understanding this concept when cos(-θ)=cosθ . It doesn't seem to make sense.
     
  2. jcsd
  3. May 20, 2014 #2

    Matterwave

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    Take a look at the graph. It should be pretty clear from looking at the graphs that sin is an odd function, while cos is an even function.

    Alternatively, think of the unit circle, and what it means to have a (small i.e. <90 degrees)) ##-\theta##. This means you're going below the x-axis instead of above it. In this case, the opposite side (for sin) goes below the x-axis, and the y-value is negative, while the adjacent side (for cos) still goes to the right and the x-value is still positive.
     
  4. May 20, 2014 #3
    Now THAT makes sense. I was looking at it from a purely mathematical formula. I did not think of the graph that comes along with it.
     
  5. May 21, 2014 #4

    jtbell

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  6. May 21, 2014 #5

    ChrisVer

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    Well, although the answer is given- mathematically it depends on how you define the trigonometric functions sin and cos....
    The result comes from the geometric meaning of cos and sin being the adjacent and opposite respectively over the hypotenuse... when you have θ and -θ, the adjacent remains the same, while the opposite changes sign- thus the cosine remains the same, while sin will get a minus sign (the hypotenuse has the opposite and adjacent squared, so their sign doesn't play a role)...
    [itex] cos (\theta) = \frac{adjacent}{hypotenuse}[/itex]
    [itex] cos (-\theta) = \frac{adjacent}{hypotenuse}= cos(\theta) [/itex]
    [itex] sin (\theta) = \frac{opposite}{hypotenuse}[/itex]
    [itex] sin (-\theta) = \frac{-opposite}{hypotenuse}=-\frac{opposite}{hypotenuse}=-sin (\theta)[/itex]


    Another way to see it, is by their definition through exponentials with imaginary powers....

    [itex] sin(\theta)= \frac{e^{i \theta} - e^{-i \theta}}{2i}[/itex]

    [itex] sin(-\theta)= \frac{e^{-i \theta} - e^{i \theta}}{2i}=- \frac{e^{i \theta} - e^{-i \theta}}{2i}=-sin(\theta)[/itex]

    [itex] cos(\theta)= \frac{e^{i \theta} + e^{-i \theta}}{2}[/itex]

    [itex] cos(-\theta)= \frac{e^{-i \theta} + e^{i \theta}}{2}= \frac{e^{i \theta} + e^{-i \theta}}{2}=cos(\theta)[/itex]


    Another way is through defining them with as sums (taylor expansion)

    [itex] sin(\theta)= \sum_{n=1}^{∞} (-1)^{n+1} \frac{\theta^{2n-1}}{(2n-1)\factorial}[/itex]

    changing θ to -θ you will get a minus overall because [itex](-1)^{odd} =-1[/itex]

    For cos you have even powers, so it will bring about +1, remaining the same

    Another way of seeing the Taylor series, is looking at the function of sin and cos as a general odd or even functions, which under the change of sign of their arguments will give an overall - or + (respectively) sign...
    [itex] f(-x)=-f(x) , f(x) odd[/itex]
    [itex] f(-x)=f(x) , f(x) even[/itex]
    This becomes obvious for cos and sin when you look at their graphs...

    Also things can be seen by the unit circle which sin and cos "draw" on a plane (that is closely associated with the exponentials I gave above, because of Euler's formula).
    you have that [itex] sin^{2} \theta + cos^{2} \theta=1[/itex]
    this is closely related to having the module of a vector equal to unity, thus the vector "draws" a unit circle. Making this assumption, you can set [itex] x= cos \theta[/itex] and [itex]y= sin \theta [/itex] and you will have:
    [itex]x^{2} +y^{2}=1[/itex]
    if you put x,y as vector components: [itex] \vec{R}= x \vec{e}_{x} + y \vec{e}_{y}[/itex]
    you have:
    [itex] |\vec{R}|^2 = 1[/itex]
    and R's x component represents the cos and y component represents the sin...
    Going around the circle, you can see that for [itex]\theta[/itex] angle over the x axis and for [itex]-\theta[/itex] angle which is under the x-axis, the vector R (starting from origin and reaching the unit circle) has the same x component but opposite y components....
     
    Last edited: May 21, 2014
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