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$$\lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1$$

Could anyone explain why $\theta$ is in radians? Is it because if $\theta$ is not in radians then $\frac{\sin{\theta}}{\theta} \neq 1$? Is there a detailed explanation?

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You've seen this? –  J. M. Aug 20 '12 at 12:35
    
That would depend on how you interpret it. If you interpret $n^\circ$ as another way to say $\frac {n}{180}\pi$, then it doesn't really matter. If, however, you scale sine instead, so that your $\sin (x)$ is actually $\sin (\frac {x}{180}\pi)$, then it does matter. –  tomasz Aug 20 '12 at 12:40
    
So will it be more convenient to have θ in radians because my textbook often have θ in radians? I rarely see people have it in angles, actually is it just convenient to solve algebraically? –  z_z Aug 20 '12 at 12:57
    
You might want to consider units as well. You can't plug units into arbitrary functions: $2^{10 \text{m}}$ doesn't mean anything. Radians, since they're defined as the ratio of two lengths, are dimensionless and so can be safely plugged into the sine function without issues. –  Robert Mastragostino Aug 20 '12 at 15:54
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It is also necessary to use radians to get $\sin'=\cos$. If degrees were used, you'd have $\sin'=(\pi/180)\cos$. No matter what units are used, you have $\sin'=\text{constant}\cdot\cos$. Only when radians are used is the constant equal to $1$. –  Michael Hardy Aug 20 '12 at 16:29

7 Answers 7

up vote 3 down vote accepted

$$ \lim_{x\to0}\frac{\sin x^\circ}{x} = \lim_{x\to0}\frac{\sin_\text{(radians)}\left(\frac{\pi x}{180}\right)}{x}. $$

Now let $u=\dfrac{\pi x}{180}$. Notice that as $x\to0$, we have $u\to0$. And $x=\dfrac{180u}{\pi}$. So the limit becomes

$$ \lim_{u\to0}\frac{\sin_\text{(radians)}\left( u \right)}{ \left( \frac{180u}{\pi} \right) } = \lim_{u\to0} \frac{\pi}{180} \cdot \frac{\sin_\text{(radians)} u}{u} = \frac{\pi}{180} \cdot \lim_{u\to0} \frac{\sin_\text{(radians)} u}{u} = \frac{\pi}{180}\cdot 1= \frac{\pi}{180}. $$

Therefore $$ \lim_{x\to0}\frac{\sin x^\circ}{x} = \frac{\pi}{180}. $$

Only when radians are used is the limit $1$.

Likewise, only when radians are used is $\sin'$ equal to $\cos$. If degrees are used then $\sin' = \frac{\pi}{180}\cdot\cos$.

This is similar to the situation with derivatives of exponential functions: $$ \frac{d}{dx} a^x = \left(a^x\cdot\text{constant}\right). $$ Only when $a=e$ is the constant equal to $1$. That's what's "natural" about $e$.

Later note: I'd add a graphic to back this up if I could conveniently draw it and upload it. But here I'll describe it. First look at this: http://upload.wikimedia.org/wikipedia/commons/4/45/Unitcircledefs.svg Then imagine what happens if $\theta$ is infinitely small but positive. In that case, the curved arc you're looking at looks like a straight line, and the vertical line that defines the sine is the same straight line. So the ratio of their lengths is $1$. But if you use some units besides radians, so that the length of the arc is not measured in the same units as the length of the vertical line, then their lengths are two different numbers, even though their lengths are the same, so the ratio of those numbers is not $1$.

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Million thanks! I understand now! –  z_z Aug 21 '12 at 2:36

In the xy-plane, draw a circle with radius 1 around the origin.
Given a point $(x,y)$ on the circle, the angle between the positive $x$-axis and the vector from the origin to $(x,y)$ in radians is just the length of the part of the circle from the point $(1,0)$ counterclockwise to $(x,y)$. If the angle is small, then $x$ and the length of the part of the circle are almost the same.
This is my sloppy way of saying that $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$. This limit is only $1$ if $\theta$ is given in radians. If the angle $\theta$ is given in degrees, then in radians this is $\frac{2\pi}{360}\theta$. By the usual limit rules, in case $\theta$ is given in degrees, the limit is $\frac{\pi}{180}$.

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The length $L$ of an arc of a circle with radius $r$ and subentending angle $\theta $ is $$\begin{equation*} L=r\theta \end{equation*}$$ if and only if $\theta $ is measured in radians$^1$. Under this assumption and for $0<\theta<\pi/2$ radians $$\begin{equation*} \sin \theta <\frac{L}{r}=\theta <\tan \theta \end{equation*}$$

and $$\begin{equation*} 1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }, \end{equation*}$$

which implies that $$\begin{equation*} \lim_{\theta \rightarrow 0}\frac{\theta }{\sin \theta }=1\Leftrightarrow \lim_{\theta \rightarrow 0}\frac{\sin \theta }{\theta }=1, \end{equation*}$$

because $\frac{\sin \theta }{\theta }$ is an even function.

$^1$If $\theta $ is measured in degrees, then \begin{equation*} L=\frac{\pi r}{180}\theta . \end{equation*}

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As pointed out in one comment, everything depends on your definition of the sine function. If you fix one sine function, then nothing changes, since it is just a change of variables by dilation. But radians are pure numbers, while degrees are numbers with a unit, and we should probably remark that $90^\circ$ is not a real number. If we denote by $\sin$ the sine function defined on $\mathbb{R}$, the expression $\sin 90^\circ$ must be interpreted as $$ \sin \left(\frac{90^\circ}{180^\circ} \pi \right). $$

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The apparent confusion stems from an overloading of $\sin$, the function: the function that gives the sine of an angle when measured in degrees is different from the function that gives the sine of an angle when measured in radians. They are related by a scale in the input and so the limit changes with that scale.

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What is the difference? I really don't get it... –  z_z Aug 20 '12 at 13:00
    
@z_z, the value of $\sin 30$ depends on how you interpret the number $30$. If it's degrees, then the value is $0.5$. If it's radians, then the value is $-0.988\ldots$. –  lhf Aug 20 '12 at 13:05

It doesn't matter whether theta is in radians/degrees/what-so-ever. The result will remains the same (as long as the the units you use are linear to radians. meaning, theta units could not be radians squared, because then the limit will change (and will be 0 in that case).

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2  
This is not correct. If you use degrees, the limit is $\pi/180$. –  Michael Hardy Aug 20 '12 at 17:44
    
Oh, I thought you will change both theta to the new units. –  DiGMi Aug 21 '12 at 9:23
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You can't change anything to the new units when there's no trigonometric function involved. Sine-in-radians is a function; sine-in-degrees is a different function. That's where the difference between radians and degrees is. –  Michael Hardy Aug 21 '12 at 15:29

"Angular distance" is not a real number. However, they are closely related: given a particular choice of angular distance unit (e.g. "radian"), one can convert a real number $x$ into an angular distance by writing "$x$ radians". Conversely, given any angular distance, we can divide out the unit to obtain a real number.

"Angular displacement" is another related concept. Every angular distance gives an angular displacement. The difference is, e.g., that the displacements given by "0 degrees" and "360 degrees" are the same displacement, despite being different distances.

Now, the confusion you're observing comes from two different uses of the term $\sin$. Geometrically, we define $\sin$ as a function of angular displacement, e.g. by the usual "opposite / hypotenuse" definition and its variants. This can be directly viewed as a function of angular distances as well.

But don't we usually write $\sin$ as a function of real numbers? Well, if we have canonically chosen a unit of angular distance, then we can convert between real numbers and angular distance, and so we can define $\sin x$ as, say, $\sin(x\ \text{radians})$.

But the function so defined is really a function both of $x$ and the chosen unit of angular distance! If you change the unit, you change the function.

There are many good non-geometric reasons why "$\sin(x\ \text{radians})$ is a useful function of real numbers. So, today, mathematicians use the notation $\sin x$ to be that function, rather than the geometric one.

So if you want to use $\sin$ to find "opposite / hypotenuse", then you have to convert angular distance to radians to get the number to plug into $\sin$.

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