Ok, please ignore my silliness.
So, why do we use radians in calculus and why is it considered more scientific than degrees. And how did mathematicians know or prove that radians would work for all integration, differentiation in other fields or is there a method to prove it?
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Aditya Agarwal
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2Radians measure lengths of arcs on the unit circle. This is a much more natural way of measuring angles than arbitrarily defining a unit that is $\frac{1}{360}$th of a full rotation. What exactly do you mean by "prove that radians would work"? Explain what that means. – anon May 28 '15 at 03:37
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1Tangentially related (no pun intended): an open letter to students – Juan Sebastian Lozano May 28 '15 at 03:39
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Because we don't want factors of $\frac {\pi}{180}$ all over our equations. – May 28 '15 at 03:39
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Taylor series for trigonometric functions are useful in radian measure. $$\sin x = x - \frac{x^3}{1\cdot 2\cdot 3} + \frac{x^5}{1\cdot 2 \cdot 3\cdot 4\cdot 5} - \ldots$$ Then limit $\displaystyle\lim_{x\to 0} \frac{\sin x}{x}$ is $1$, not $\pi/180^\circ$. – Oleg567 May 28 '15 at 03:41
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I believe the "right" answer is that, for small $x$, we have an approximate proportionality $\sin x \approx kx$, where the constant $k$ depends on how $x$ Is measured (one $360$th of a circle (degrees), one $2\pi$th of a circle (radians), one $5$th of a circle, etc). If you choose your standard of angle measure wisely, you get $k=1$. Much like $e$ as the base of the natural exponential function. – pjs36 May 28 '15 at 03:55
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This is a good question, and it is unfortunate that its simple answer is not emphasized more than it is. Of course, in order to prove that the "simple" answer is true, one needs to go through some details that are not quite as simple. See my answer below for the simple answer and some hints at the somewhat less simple parts of the reasoning. – Michael Hardy May 28 '15 at 03:56
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@Michael_Hardy I saw your answer. But still I dont think I got my answer. Because is there a way to prove without reasoning, just a rigorous proof? – Aditya Agarwal May 28 '15 at 04:01
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If degrees are used then $$ \frac d{dx} \sin x = \frac \pi {180} \cos x. $$
If some units other than degrees are used then $$ \frac d{dx} \sin x = \left(\text{some constant}\cdot\cos x\right). $$
If the units are radians, then the "constant" is $1$, so $\dfrac d{dx}\sin x = \cos x$.
As for proving this, go back to the proof that $$ \lim_{x\to0} \frac{\sin x} x = \text{something}. $$ The "something" is $\dfrac\pi{180}$ if degrees are used, and $1$ if radians are used, and something else if some other units are used. Read that proof, and read the proof that $\sin'=\cos$, and notice how the proof of the former proposition is used in proving the latter proposition.
It's really the same as what's "natural" about the number $e$. $$ \frac d{dx} 2^x = (\text{some constant}\cdot2^x). $$ $$ \frac d{dx} 10^x = (\text{some other constant}\cdot10^x). $$ $$ \frac d{dx} e^x = (\text{some constant}\cdot e^x). $$ Only when the base is $e$ is the "constant" equal to $1$.
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Okay, but what about in integrals? We use radians because the constant simplifies to 1. But how did mathematicians know that making an angle in which arc length is equal to the radius of the circle will help simplify the constant in differentiation? And why there is $\frac{\pi}{180}$ before $\cos$? – Aditya Agarwal May 28 '15 at 03:57
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The $\frac{\pi}{180}$ is there because of the chain rule. When we express $sin(x)$ in terms of degrees, we are actually talking about a completely different function; if we want it to be the same function, we must write $sin(\frac{\pi}{180}x)$. Differentiating this requires the chain rule. – Jonathan Hebert May 28 '15 at 04:00
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Okay, I know about the exponential function, but again, why we must write $\sin(\frac{\pi}{180}\theta)$? There must be a solid reason for it? – Aditya Agarwal May 28 '15 at 04:05
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That is again, a way of writing radians, so again, my question seems unquenched here. – Aditya Agarwal May 28 '15 at 04:05
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$1\text{ radian}= \dfrac{180}\pi\text{ degrees}$. So $\dfrac{d}{dx}\Big(\text{sine of }\Big(x\text{ degrees}\Big)\Big)$ $= \dfrac d{dx} \Big(\text{sine of }\Big(x\cdot\pi/180\text{ radians}\Big)\Big)$ $= \Big(\text{cosine of }\Big(x\cdot\pi/180\text{ radians}\Big)\cdot \pi/180\Big)$ $=\text{sin of }\Big(x\text{ degrees}\Big)\cdot\pi/180$. ${}\qquad{}$ – Michael Hardy May 28 '15 at 04:33
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Radians make many formulas much simpler.
For example, the length of an arc of a circle subtended by an angle of $\theta^\circ$
$$s=2\pi r\theta/360^\circ.$$
If $\theta$ is measured in radians: we get $s=r\theta$.
Or, the area of a sector is $$A=\pi r^2\theta/360^\circ,\quad\theta\text{ in degrees}$$ but $$A=r^2\theta/2\quad\theta\text{ in radians}.$$
It makes sense, that the formulas for derivatives, and antiderivatives would be much nicer in radians.
Tim Raczkowski
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Okay, but how did mathematicians find out that using radians only would simplify rlthe constant? And what would be the derivative of sine function in degrees and why? – Aditya Agarwal May 28 '15 at 04:00
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@AdityaAgarwal : The short answer to "How did they find out?" depends on going back to the proof that $\lim\limits_{x\to0}\dfrac{\sin x}x = 1$. See my posted answer here. – Michael Hardy May 28 '15 at 04:02