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Trig help

  1. Jul 31, 2006 #1
    Hello,

    I'm curious if anyone can shed some light on my seemingly opaque brain as to why Sine is an odd function and Cosine is an even function?
     
  2. jcsd
  3. Jul 31, 2006 #2

    Office_Shredder

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    Try graphing it, that should help. Sin(x) is odd because sin(-x)=-sin(x), whereas cos(x) is even because cos(-x)=cos(x). When looking at the definitions of sin and cos on the unit circle it should be obvious.... if you go backwards x radians instead of forwards, you end up on the opposite side of the x-axis, but the same side of the y-axis. This is because you start at x=0 on the x-axis
     
  4. Aug 1, 2006 #3
    Looking at the Taylor Series should help.
     
  5. Aug 1, 2006 #4
    From what I gather, I think your explanation and the books is similiar. So as far as I can understand Cos(-60)=1/2 and Sin(-30)=-1/2 which satisfies the Sin(-x)=-Sinx and Cos(-x)=Cosx--this is not hard for me to comprehend, but I was thinking, what about Cos(-120)? Does this not equal -1/2, or Sin(-210)=1/2--does it vary with in certain quadrants? Anyways, thanks Officer shredder for pointing me in the right direction.
     
  6. Aug 2, 2006 #5
    Look at the "unit circle" definition for the functions. As Office_Shredder said, the parity of the functions should be pretty obvious.

    In case you don't know what I'm talking about:

    Consider the point (1,0). If you rotate that point around the origin by an angle [itex]\theta[/itex] counterclockwise, without changing its length, then you get a new point (x,y) (with [itex]x^2+y^2=1,[/itex] which is why this is referred to as the "unit circle" defition - you're just rotating around a circle of radius 1 centered at the origin). We define [itex]\sin{\theta} = y[/itex] and [itex]\cos{\theta} = x[/itex].

    So if we rotate 0 degrees, you get [itex]\sin{\theta} = 0[/itex] since that just leaves you with (x,y)=(1,0). If you rotate around counterclockwise (ie. [itex]\theta[/itex] is positive) with an angle [itex]\leq \pi[/itex], you see that [itex]y \geq 0[/itex], so [itex]\sin{\theta} = y \geq 0[/itex]. If you then rotate by [itex]-\theta[/itex] (ie. clockwise by the same angle) then you find that [itex]y[/itex] is now of opposite sign but the same magnitude, ie. [itex]\sin{\theta} = -\sin{(-\theta)}[/itex]. You'll see this works for angles [itex]\theta > \pi[/itex] too if you think about it.

    Seeing that [itex]\cos[/itex] is even is equally easy with that definition.

    (note i'm using radians for angles here which you may or may not be familiar with - in the above, [itex]\pi = 180^\circ[/itex])
     
    Last edited: Aug 2, 2006
  7. Aug 2, 2006 #6
    If you could provide me a demonstration of lets say [itex]\sin(-\frac{7\pi}{6}) = -\sin(\frac{7\pi}{6}) [/itex] I would be most greatful, Data. As you can see, this is what excites the most trouble in my understanding--when [tex]\theta \geq 180 [/tex]. My apologies if I come off tenacious, but not understanding this completely vexes me inside!

    Ahh! It should [tex]\frac{7\pi}{6} [/tex] radians.

    Thanks.
     
    Last edited: Aug 2, 2006
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