sin(x) derivative in degrees

So the small angle approximation for sin(x) is just x, but this only works in radians, but I never thought too hard about what this means about the gradient of the function in degrees.

so if youre working in radians (_r), the derivative of sin(x)_r is cos(x)_r, and so the gradient at x= 0, is just one, but in degrees (_d), at x = 0, the gradient of sin(x)_d is pi/180 (which is why the small angle approx doesnt work in degrees).

But it gets weird if you work in degrees, as the derivative of sin(x)_d = pi/180* cos(x)_d -> (which I dont really know how to prove, I thought it would be simple chain rule, but that gave me the wrong answer, so my only logic is the degrees graph is more stretched out (by factor of 180/pi, so it has a 180/pi more stretched out gradient).

Also, I have never really seen this being asked before in tests or covered in my syllabus for maths -> if a question asks you to give the derivative of sin(x), and theyre working in degrees, does that mean you give cos(x) or the other one?

(edited 1 month ago)

Reply 1

davidfor

Original post by mosaurlodon

So the small angle approximation for sin(x) is just x, but this only works in radians, but I never thought too hard about what this means about the gradient of the function in degrees.
so if youre working in radians (_r), the derivative of sin(x)_r is cos(x)_r, and so the gradient at x= 0, is just one, but in degrees (_d), at x = 0, the gradient of sin(x)_d is pi/180 (which is why the small angle approx doesnt work in degrees).
But it gets weird if you work in degrees, as the derivative of sin(x)_d = pi/180* cos(x)_d -> (which I dont really know how to prove, I thought it would be simple chain rule, but that gave me the wrong answer, so my only logic is the degrees graph is more stretched out (by factor of 180/pi, so it has a 180/pi more stretched out gradient).
Also, I have never really seen this being asked before in tests or covered in my syllabus for maths -> if a question asks you to give the derivative of sin(x), and theyre working in degrees, does that mean you give cos(x) or the other one?

I don't think that they'll ask for the derivative of sin(x) when working in degrees in an exam.
I don't think you're wrong for the most part, except for "sin(x)_d = pi/180* cos(x)_d" where "x" should be "x * pi/180" to convert from degrees to radians for the sake of differentiating.

Reply 2

mqb2766

Original post by mosaurlodon

So the small angle approximation for sin(x) is just x, but this only works in radians, but I never thought too hard about what this means about the gradient of the function in degrees.
so if youre working in radians (_r), the derivative of sin(x)_r is cos(x)_r, and so the gradient at x= 0, is just one, but in degrees (_d), at x = 0, the gradient of sin(x)_d is pi/180 (which is why the small angle approx doesnt work in degrees).
But it gets weird if you work in degrees, as the derivative of sin(x)_d = pi/180* cos(x)_d -> (which I dont really know how to prove, I thought it would be simple chain rule, but that gave me the wrong answer, so my only logic is the degrees graph is more stretched out (by factor of 180/pi, so it has a 180/pi more stretched out gradient).
Also, I have never really seen this being asked before in tests or covered in my syllabus for maths -> if a question asks you to give the derivative of sin(x), and theyre working in degrees, does that mean you give cos(x) or the other one?

To add to the previous reply you have
y = kx
where x is in radians and y in degrees so k=180/pi. Then
sin_d(y) = sin_r(x)
where the subscript denotes the units of the argument and
dy = k dx
so
d sin_d(y) / dy = 1/k d sin_r(x) / dx = 1/k cos_r(x) = pi/180 cos_d(y)

It makes sense because if you twiddle y in sin_d(y) by 1 degree, the change in the function is small (scaled by pi/180) compared to twiddling x in sin_r(x) by 1 radian

(edited 1 month ago)

Reply 3

DFranklin

Original post by mosaurlodon

So the small angle approximation for sin(x) is just x, but this only works in radians, but I never thought too hard about what this means about the gradient of the function in degrees.
so if youre working in radians (_r), the derivative of sin(x)_r is cos(x)_r, and so the gradient at x= 0, is just one, but in degrees (_d), at x = 0, the gradient of sin(x)_d is pi/180 (which is why the small angle approx doesnt work in degrees).
But it gets weird if you work in degrees, as the derivative of sin(x)_d = pi/180* cos(x)_d -> (which I dont really know how to prove, I thought it would be simple chain rule, but that gave me the wrong answer, so my only logic is the degrees graph is more stretched out (by factor of 180/pi, so it has a 180/pi more stretched out gradient).
Also, I have never really seen this being asked before in tests or covered in my syllabus for maths -> if a question asks you to give the derivative of sin(x), and theyre working in degrees, does that mean you give cos(x) or the other one?

To add to the other comments, the chain rule does work.

\sin x^o = \sin(\frac{x\pi}{180})

(where the RHS is in the "natural" radians unit).
Writing

y = \frac{x\pi}{180}

, we have

\dfrac{d}{dx} \sin x^o = \dfrac{d}{dx} \sin y = \dfrac{dy}{dx} \dfrac{d}{dy} \cos y = \dfrac{\pi}{180} \cos y = \dfrac{\pi}{180} \cos x^o

.

In general, the trig differentiation rules only work when you're measuring in radians. If you're measuring in degrees there's an implicit scaling going on (the pi/180 factor) - the way to proceed is to make that scaling explicit so you are actually working in radians.

Reply 4

mosaurlodon

Thanks for all the replies - ill try to keep this rule in mind, looks like only working in radians keeps things much simpler.

There was this one vague memory I had about 3b1b's proof of sin's derivative being cos, so I rewatched it and I was quite confused about how it implied we were working in radians, but then I realised his nudge of "dtheta" was really just pi/180 dtheta and then everything checks out.

https://www.youtube.com/watch?v=S0_qX4VJhMQ around 16:00