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In radians the derivative of sine is the cosine. But why isn't this the same in degrees? According to this I'd first have to convert it to radians before this works. But when you look at the graph of sine whether the x is in degrees or radians, the graph stays the same... I don't understand.

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    $\begingroup$ Why is the derivative of $f(x)=x$ equal to $1$ at the origin but not of $g(x)=2x$? $\endgroup$ Aug 13, 2013 at 1:41
  • $\begingroup$ @marianoSuárez-Alvarez because the slope of g(x) is > 1? $\endgroup$ Aug 13, 2013 at 1:47

3 Answers 3

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Here's a graph of $y = \sin x$.

$y = \sin(x)$

The red line is $\sin x$ if $x$ is interpreted as an angle in degrees, and the blue line is $\sin x$ if $x$ is interpreted as an angle in radians.  See the difference?

Here's a zoomed-in version

zoomed in version

What is $\sin(1.57\text{ rad}),$ and what is $\sin(1.57˚)?$

Intuitively, you can think of $\sin(x˚)$ as a "stretched out" version of the original graph. As a result, the slope of the tangent flattens, changing the derivative.

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    $\begingroup$ @user1534664: the best way to think about sine, in the long term, is as a function from numbers to numbers. That function can be interpreted as taking an angle in radians and spitting out a ratio that relates to that angle. If you have an angle measured in degrees, you need to convert it to radians before you feed it to the sine function. $\sin(n^\circ)$ means $\sin\left(\frac\pi{180}n\right)$. Even better, you should always think of the $\phantom)^\circ$ symbol as meaning "times $\frac\pi{180}$". $\endgroup$
    – dfeuer
    Aug 13, 2013 at 1:59
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    $\begingroup$ A good resource for teachers is The $\tau$ Manifesto (which also has cool art), but it's probably not the best for students. $\endgroup$
    – dfeuer
    Aug 13, 2013 at 2:03
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    $\begingroup$ @dfeuer so basically I should just accept that the sine function only accepts radians as its domain? and if I want the sin of a degree I need to convert it first? $\endgroup$ Aug 13, 2013 at 2:06
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    $\begingroup$ I think the place to start is to accept that unless you specify otherwise, sin accepts radians as its domain. Radians, as it ends up, has a much more "natural" definition, to the point that can be thought of as not really being a unit, unlike degrees. So yes. The mathematical sine takes only radians. No degrees allowed, unless you change them into radians first. $\endgroup$ Aug 13, 2013 at 2:09
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    $\begingroup$ @user1534664: to put one last spin on it, "degrees" is of the general nature as "dozen", "score", and "moles". You wouldn't expect the smell of three fish to equal the smell of three dozen fish, so you shouldn't expect the sine of three to equal the sine of three degrees. $\endgroup$
    – dfeuer
    Aug 13, 2013 at 2:54
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I'm gonna assume that both system of measure exist, and that we all know that $$ \frac{\text{measure in degree}}{360}=\frac{\text{measure in rad}}{2\pi}. $$ I will also take for granted the classic derivative formulas for $\sin(x),\cos(x)\ldots$

It is often useful to think in terms of a new function to see this. There is no particular reason as to why they are not the same except that because that is the way the maths work. Nothing better to understand why something isn't true than to show it is not.

Let's define the degree sine function and call it $S(x)$ and the degree cosine function and call it $C(X)$. These are the functions that take $x$ degrees and return the value of $\sin(x°),\cos(x°)$. $\,S(x)$ corresponds to the red line in Omnomnomnom's answer. Now let us compute the derivative of $S(x)$, based on what we know. $$ \begin{align} \frac{d}{dx}S(x)&=\lim_{h\to 0}\frac{S(x+h)-S(x)}{h}&\\ & = \lim_{h\to 0} \frac{\sin(\frac{2\pi}{360}(x+h))-\sin(\frac{2\pi}{360}x)}{h} &\text{from measure to measure relation}\\ &=\frac{\pi}{180}\cos{\frac{\pi}{180}}=\frac{\pi}{180}C(x).& \text{from classic derivative formula} \end{align} $$ In a similar way, we'd find $C^{'}(x)=-\frac{\pi}{180}S(x)$.

Some not directly related but similar reading about radians vs degrees can be found here and also here.

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    $\begingroup$ Why muck with those limits if you're just going to use the chain rule and known derivatives in the end? $\endgroup$
    – dfeuer
    Aug 13, 2013 at 4:55
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    $\begingroup$ @dfeuer This was tagged as pre-calculus, I had no idea what was known to the op. He knew derivative, so I went with the basic definition. $\endgroup$ Aug 13, 2013 at 5:03
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this is because of the contraction of the functions as the domain changes. in degrees 0 to90 corresponds to 0 to 3.14 radians hence suitably as the general derivative is made with radians as domains we have convert other units into the general.

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    $\begingroup$ I'm pretty sure this is wrong, 0 to Pi in radians is 0 to 180 in degrees. Your writing is also very confusing to me. $\endgroup$ Nov 21, 2015 at 7:42

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