Why in calculus the angles are measured in radians? [duplicate]

Why is the formula $\lim\limits_{h\rightarrow 0}\frac{\sin h}{h}=1$ not valid when $h$ is measured in degrees?

Solution 1:

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?

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.

Solution 5:

Check out this article:

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-8-limits-of-sine-and-cosine/MIT18_01SCF10_Ses8a.pdf

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.