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Why is the sine and cosine always between $-1$ and $1$? If I would have circle with a radius other than $1$, then it wouldn't be between $-1$ and $1$ anymore, would it?

This also ties in with another thing I'm getting confused about: the $x$ and $y$-coordinates of a point on a circle are defined by $\cos\theta$ and $\sin\theta$. But what if the radius would be defined as anything other than having a length of $1$ unit? Would you still be able to find the $(x, y)$ coordinate of the point on the circle?

5xum
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Garth Marenghi
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    This is all caused by the definition of sin and cos. They are defined by unit circle. If you have a circle with different radius, you'll have other functions, which are sin and cos scalled by the ratio of the radius and the unit length. – V-X Oct 21 '14 at 12:17
  • The sin and cosine of real numbers. – Enredanrestos Sep 01 '15 at 17:20

2 Answers2

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Short answer: The sine function takes values from $-1$ to $1$ because that's just the nature of the function.

Longer answer:

The elementary definition of the sine function for angles from $0$ to $\pi/2$ is that it is the ratio of the length of one (opposite to the corner of which the sine is calculated) catete of a right angle triangle to the hypothenuse of the triangle, and since the hypothenuse is always the longest side of a right angle triangle, the sine of an angle will always be smaller than $1$ (and, for this definition, larger than $0$).

The more general definition of the function is that the sine of an angle is equal to the height of the point on the unit circle given the correct angle, and since the unit circle reaches only heights between $-1$ and $1$, so does the sine function.

For your other question, the answer is fairly simple. The unit circle can be defined as the set $$S=\{(x,y)| x^2+y^2 = 1\}$$

Using the definition of the sine and cosine functions, it is simple to show that this set is equal to the set $$\{(\cos \theta, \sin\theta)|\theta\in[0,2\pi)\}.$$

The circle centered around $0$ with a radius different than $1$, say the radius $R\neq 1$, is defined as $$S_R=\{(x,y)| x^2+y^2 = R^2\}$$

and can be shown to be equal to the set $$\{(R\cos \theta, R\sin\theta)|\theta\in[0,2\pi)\}.$$

5xum
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  • I'm still a little bit confused. Why have the unit circle and have all these things based off it, when it doesn't seem applicable to circles with different radii? – Garth Marenghi Oct 21 '14 at 13:02
  • @GarthMarenghi We base things off the unit circle because that is the natural expansionof the elementary definition. And in what way are these things "not applicable" to other circles? – 5xum Oct 21 '14 at 13:06
  • "Not applicable" was an unfortunate use of words on my part. What I meant was that trying to learn trig, you (almost) exclusively use the unit circle. You're urged to learn the angles and the coordinates by heart, but what about other circles with different radii? $\frac{cos \theta}{1}= cos \theta$ doesn't apply then anymore for example. And there are a lot of circles with radii $\neq 1$. – Garth Marenghi Oct 21 '14 at 13:14
  • @GarthMarenghi $\frac{\cos \theta}{1}=\cos\theta$ is true no matter what circle you look at. I don't understand what you are trying to say here. In a larger circle, the $x$ coordinate will be $R\cos \theta$, and the radius will be $R$, so the ration between the $x$ coordinate and the radius will still be $\frac{R\cos\theta}{R} = \cos \theta$. – 5xum Oct 21 '14 at 13:16
  • That's okay :-) Just some internal struggles of coming to terms with it all. Thanks for you though, it did help in understanding. – Garth Marenghi Oct 21 '14 at 13:34
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If the radius was not $1$ then you could not say that the x and y-coordinates of a point on a circle are described by $\cos \theta$ and $\sin \theta$, since $\cos^2 \theta +\sin^2 \theta =1$.

So you could say that the sine (or cosine) of real angles is always between negative one and positive one because the hypotenuse is the longest side of a right angled triangle.

Henry
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  • This seems odd; this way you seem to not able to find coordinates of circles with radii $\neq 1$. – Garth Marenghi Oct 21 '14 at 13:03
  • @Garth Marenghi: not at all. If the centre is at $(0,0)$ and the radius is $r$ then the co-ordinates can be described as $(r\cos \theta, r\sin\theta)$ and you will then have $(r\cos \theta)^2+(r\sin\theta)^2=r^2$, meaning that $r$ will be greater (or equal) in magnitude to each of $r\cos \theta$ and $r\sin\theta$. – Henry Oct 21 '14 at 17:38