What is there in the derivation of the Taylor/Maclaurin series for #sin(x)# that determines if the series assumes #x# is radians or degrees?

Answer 1
If the angle is not measured in radians, then the differentiation rule #d/(dx)(sinx)=cosx# is false.
In establishing (proving) that rule we use #lim_(hrarr0)sinh/h = 1#
That is also not true if #h# is taken to be the measure of an angle in units other than radians.
And that's because proving #lim_(hrarr0)sinh/h = 1# uses "arc length = central angle in radians times radius"
If the central angle #theta# is measured in degrees, the arc length, #s# may be found by using the proportion:
#s/C = theta/360# or #s/(2 pi r) = theta/360#
With #theta# measured in radians, the #360# is replaced by #2 pi#.
If #theta# measures #m# in degrees then measures #m pi / 180# radians. Instead of the simple formulas:
#s=r theta# and #d/(d theta)(sin theta) = cos theta #

We have:

#s = (pi r m )/ 180# and the derivative is #pi/180 cos m# or maybe it's #180 / pi cos m#. (My brain hurts when I try to use degrees to do calculus.)
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2
You could also just say that the idea of the MacLaurin series for #sin(x)# is so you can say that #d/dx(sin(x)) = cos (x) = 1# and #sin(x)# is approximately equal to #x# near #x = 0#.
When you do #sin(0.02)#, in radians it does return you an answer of 0.02 with negligible error (you get 0.01999867). When you get that, you know you're in the right units. In degrees, you get 0.9110879, which is not a 1:1 relation of the argument with the answer.

But this is just a quick check. The actual proof might not be like this; I don't remember how it actually went.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

The derivation of the Taylor/Maclaurin series for sin(x) assumes that x is in radians. This is because the Taylor/Maclaurin series formulas are based on derivatives, and the derivatives of trigonometric functions like sin(x) are defined in terms of radians. Therefore, when using these series to approximate trigonometric functions, it's essential to ensure that the input values are in radians. If the input values are in degrees, they must first be converted to radians before using the series.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7