Why is the formula $\lim\limits_{h\rightarrow 0}\frac{\sin h}{h}=1$ not valid when $h$ is measured in degrees?
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10A better question would be why angles are measured in anything other than radians. – zhw. Oct 02 '16 at 19:27
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1@zhw I think, you are pointing to the fact that radian is the natural unit for angles! – SRS Oct 02 '16 at 19:30
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Yes, that would my opinion. – zhw. Oct 02 '16 at 19:49
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4Type "radians" in the search box, press Enter and ... http://math.stackexchange.com/questions/1339540/why-does-the-derivative-of-sine-only-work-for-radians http://math.stackexchange.com/questions/720924/why-do-we-require-radians-in-calculus http://math.stackexchange.com/questions/526043/radian-an-arbitrary-unit-too and so on... – leonbloy Oct 02 '16 at 19:53
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I would suggest you to have an understanding of measurement of angles. Normally students are introduced to the use of protractor (with degrees marked on it) without any explanation as to why a semi circular tool is needed to measure angles. Sometime back I wrote a post (http://paramanands.blogspot.com/2014/11/measuring-angle.html) on it. – Paramanand Singh Oct 02 '16 at 19:54
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One also needs to understand the definition of $\sin h$ as a function i.e. given an $h$ how can we theoretically describe $\sin h$? When $h$ is measured in radians the definition of $\sin h$ is very natural and the use of degrees is more practical/historical (somewhat similar to the way $\log$ to the base $e$ is a natural choice once we try to define $\log$ properly and base $10$ is more due to practical usage). – Paramanand Singh Oct 02 '16 at 20:00
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https://www.youtube.com/watch?v=2HmMzhZ8zJg – Count Iblis Oct 02 '16 at 21:25
6 Answers
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Definition of 1 radian:
Angle subtended at the centre of a circle by an arc equal in length to the radius.
Definition of 1 degree:
The angle subtended by one three-hundred-and-sixtieth of the circumference of a circle.
Definition of 1 gradian:
The gradian is a unit of measurement of an angle, equivalent to $1 \over 400$ of a turn, $9\over 10$ of a degree or $\pi \over 200$ of a radian.
Clearly, the radian has the most concrete definition: a ratio of two distances. It doesn't require any other thing for its support.
In the definition of degrees, you have to know the definition of an angle (to measure circumference) before you can divide it by $360$.
And clearly, gradians are helpless without degrees or radians.
So, isn't the choice obvious?
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The comment above : "degree" is the same as "real numbers", this is how I interpret it. – Peter Oct 02 '16 at 19:47
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1Grads don't need degrees any more than degrees need grads: i.e. both are defined as rational parts of the (angle corresponding to) full circle. Both can of course be defined in terms of the other. – ilkkachu Oct 02 '16 at 20:45
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The definition of a radian is precisely the same as the definition of a degree, differing only in the specific ratio used. They're both defined as angles subtended by an arc on a circle when the ratio of the arclength to the radius / circumference is something specific. The natural (i.e., integer) definition for degree uses a circumference:arc length ratio of 360, while the natural definition for radian uses a radius:arc length ratio of 1, but both can be expressed just as easily using the other ratio. 180/$\pi$ R:AL for degrees, and 2$\pi$ C:AL for radians. – MichaelS Oct 02 '16 at 21:44
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Look at the proof of the limit you are asking. And no, using L'Hospital is not a proof. You will see that the proof fails if you use degrees instead of radians, and the main point is that you can compare the numerical value of radians with length in a very direct way, just by the definition of radians. So you can say something like $\sin x \leq x$ when $x$ is in radians, but not when it in degrees.
Theo
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Good approach. For more details on the last couple of lines check my answer. – Qwerty Oct 02 '16 at 19:50
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No. Definitely $\sin(x^\circ)\leq x^\circ$. Sine of sth. is always less or equal that sth. – Michael Hoppe Oct 02 '16 at 20:38
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You seem to be under the mistaken impression perpetuated by far too many math teachers that radians are defined as a ratio of two lengths. They aren't. Radians are a unit of angle just as much as degrees. The sine function accepts units of angle and outputs a ratio of two lengths, regardless of whether the input angle is radians, degrees, or whatever other angular unit you enjoy using. $sin(x$ $radians)$ is just as incompatible with $x$ $radians$ as degrees are. – MichaelS Oct 02 '16 at 21:53
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Yes, but because of their definition, their numerical value acts directly as a unit of length. If I write $\sin(x)\leq x$ there is no meaningful distinction between $x$ as radian measure of an angle and $x$ as a real number. You don't even need radians or angle measures to define the trig functions, you can do it directly on real numbers. If you write the same inequality with degrees it only makes sense after you introduce an awkward scaling factor that is jot even inherent to the definition of degrees. – Theo Oct 02 '16 at 22:20
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Check out this article:
The (intuitional) arguments used in this article really on radians and the fact that only for radians we have:
$$r\theta=\text{arclength}$$
If it was the case that we used degrees instead, we would of had:
$$\text{arclength}=(\frac{\theta}{360})2\pi r=\frac{\pi}{180}\theta r$$
And the whole limit would be scaled by this factor of $\frac{\pi}{180}$ as a result.
Rigorous arguments (that would be considered proofs unlike this one), using squeeze theorem, involve the similar fact that for radians:
$$\frac{1}{2}r^2\theta=\text{sector area}$$
The reason we have: $r\theta=\text{arclength}$ for radians is because by definition there are $2\pi$ radians in a circle and hence the proportion of the circle given by $\theta$ radians is $\frac{\theta}{2\pi}$. If this is the same proportion of the arc-length to the circumference we must have $\frac{\theta}{2\pi}{2\pi r}=\text{arclength}=r\theta$.
A question that might come to mind is why l'hopitals rule may seem to not work. In order to use l'hopitals rule, you need to compute:
$$\frac{d}{dx}\sin x=\lim_{h \to 0} \frac{\sin(x+h)-\sin(x)}{h}$$
Which through angle addition formulas eventually results in having to compute the limit in your question. We have seen that computing this limit is dependent on wether you use radians or degrees. Hence the derivative will very depending on wether you use radians or degrees. From the above we see that l'hopitals rule still works, yet it is circular.
Ahmed S. Attaalla
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1This answer simply restates the question in the form of properties of derivatives. I doubt it answers OP's question. – Ittay Weiss Oct 02 '16 at 19:37
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One answer is that sin and cos are functions on the points of the unit circle, and the most natural way to parametrize the unit circle is with the arclength parameter.
Vik78
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the basis is Euler's formula $e^{it}=cos(t)+isin(t)$ which is valid only if $t$ is in radians. – hamam_Abdallah Oct 02 '16 at 21:56
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It's very simple. Recall that the percent sign “%” is the abbreviation for the sequence of symbols and numbers “${}\cdot\frac1{100}$”, so $42\%=42\cdot\frac{1}{100}=0.42$. Similar the sign “${}^\circ$” is the abbreviation for “${}\cdot\frac{\pi}{180}$”. From here $\sin(h^{\circ})\neq\sin(h)$. But nevertheless $$\lim_{h^\circ\to0}\frac{\sin(h^\circ)}{h^\circ}=1.$$ So it is true that the limit is $1$ in both cases. It will be ever the case that the limit is $1$ if you divide $\sin$ by its argument and let the argument tend to zero.
Michael Hoppe
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When you use degrees for angles and the definition of trig functions, expressions like $\frac{\sin{h^{\circ}}}{h^{\circ}}$ make no sense. How do you divide a real number by degrees? Do you mean dividing by the numerical value of the degree? That's why $\frac{\pi}{180^{\circ}}$ makes no sense, while $(\frac{\pi}{180})^{\circ}$ does. – Theo Oct 02 '16 at 20:57
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When you use radians to define trig function, you in fact define trig functions for any real value, thus $\sin{h}/h$ makes perfect sense, just divide two real numbers. – Theo Oct 02 '16 at 20:59
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As $30%$ is the number $0.30$, so is $30^\circ$ the number $30\cdot\frac{\pi}{180}=\frac{\pi}{6}$. The point is: degrees are real numbers as percents are. – Michael Hoppe Oct 02 '16 at 21:20
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The percentage was defined that way. The degree were not, they are not some shorthand notation for real numbers. They become real numbers after you transform them, and, in hindsight, you already have knowledge of radians. That's the beauty of radians, you don't have to transform anything. You can also easily define trig functions of real numbers without mentioning angles. – Theo Oct 02 '16 at 21:30
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Surely degrees weren't defined historically as numbers, but mathematically they are: how do you else derivate $\sin(x^\circ)$? – Michael Hoppe Oct 02 '16 at 21:34
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Depends what you mean. Do you mean the function $\sin$ defined on the space of angular measures taking real values? In this case you can't do much without abusing notation, or, if you want to get formal, introducing additional structure on the space of angular measures. But if you mean the function on the real numbers taking real values, defined as $f(x)=\sin(x^{\circ}) $, then its derivative is not $\cos(x^{\circ}) $. – Theo Oct 02 '16 at 22:38
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It is usual to use degrees for triangle calculation and radians for calculus. If the angle tends to $0°$, we have
$$\lim_{x\rightarrow 0} \frac{\sin(\frac{\pi x}{180})}{x}=\lim_{x\rightarrow 0}\frac{\pi}{180}\cdot \frac{\sin(\frac{\pi x}{180})}{\frac{\pi}{180}x}=\frac{\pi}{180}\ne 1$$
Peter
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Historically ? It makes sense to use degrees for triangles (because you are interested in the angle in degrees). But the inputs for trigonometric functions are usually real numbers, so you have to convert into radians. – Peter Oct 02 '16 at 19:30
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