Why Sine is an odd function and Cosine is an even function?

  • Thread starter hmm?
  • Start date
19
0
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?
 

Office_Shredder

Staff Emeritus
Science Advisor
Gold Member
3,734
99
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
 
452
0
Looking at the Taylor Series should help.
 
19
0
Office_Shredder said:
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
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.
 
998
0
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:
19
0
Data said:
You'll see this works for angles [itex]\theta > \pi[/itex] too if you think about it.
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:

Related Threads for: Why Sine is an odd function and Cosine is an even function?

Replies
2
Views
728
  • Posted
Replies
5
Views
9K
  • Posted
Replies
3
Views
5K
  • Posted
Replies
3
Views
2K
  • Posted
Replies
2
Views
1K
Replies
3
Views
34K
  • Posted
Replies
9
Views
7K
  • Posted
Replies
15
Views
19K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top