# Trig help

1. Jul 31, 2006

### hmm?

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. Jul 31, 2006

### Office_Shredder

Staff Emeritus
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

3. Aug 1, 2006

Looking at the Taylor Series should help.

4. Aug 1, 2006

### hmm?

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.

5. Aug 2, 2006

### Data

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 $\theta$ counterclockwise, without changing its length, then you get a new point (x,y) (with $x^2+y^2=1,$ 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 $\sin{\theta} = y$ and $\cos{\theta} = x$.

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

Seeing that $\cos$ 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, $\pi = 180^\circ$)

Last edited: Aug 2, 2006
6. Aug 2, 2006

### hmm?

If you could provide me a demonstration of lets say $\sin(-\frac{7\pi}{6}) = -\sin(\frac{7\pi}{6})$ I would be most greatful, Data. As you can see, this is what excites the most trouble in my understanding--when $$\theta \geq 180$$. My apologies if I come off tenacious, but not understanding this completely vexes me inside!

Ahh! It should $$\frac{7\pi}{6}$$ radians.

Thanks.

Last edited: Aug 2, 2006