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(I'm not sure this rubric is the right one for this question, but since it didn't seem to fit any of the other rubrics, I put it here.)

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Radians. Thank you.In summary, the conversation discusses the use of Euler's formula, which can be proven using power series and the limit of sin(x)/x. The conversation also touches on the concept of radians and degrees, and it is explained that the derivative formulas for trigonometric functions only hold true when the arguments are in radians. This is because the limit used to derive the derivative is based on the arc length formula, which is expressed in radians. However, the derivatives can still be used for degrees by converting the values to radians and back.

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(I'm not sure this rubric is the right one for this question, but since it didn't seem to fit any of the other rubrics, I put it here.)

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It would work fine with another scaling, except it would have another scaling.

given small x we have approximately that

exp(i x)~1+i x

cos(x)+i sin(x)~1+i x

which combined with the addition laws gives us the identity

if we had for some reason wanted to use other scaling we have

otherexp(i x/a)=exp(i x)~1+i x

othercos(x/b)+i othersin(x/b)=cos(x)+i sin(x)~1+i x

we would have the identity

otherexp(i x/a)=othercos(x/b)+i othersin(x/b)

for suitable a and b

in calculus we would say

a=otherexp'(0)

b=othersin'(0)

and of course when a=b=1 we have the usual functions

given small x we have approximately that

exp(i x)~1+i x

cos(x)+i sin(x)~1+i x

which combined with the addition laws gives us the identity

if we had for some reason wanted to use other scaling we have

otherexp(i x/a)=exp(i x)~1+i x

othercos(x/b)+i othersin(x/b)=cos(x)+i sin(x)~1+i x

we would have the identity

otherexp(i x/a)=othercos(x/b)+i othersin(x/b)

for suitable a and b

in calculus we would say

a=otherexp'(0)

b=othersin'(0)

and of course when a=b=1 we have the usual functions

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Ah, that makes sense. Thank you both, D H and lurflurf

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[tex]

\lim_{x \rightarrow 0}{\frac{\sin{x}}{x}} = 1

[/tex]

which works only in radians.

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Not true. Here is the series for sine(x) whennomadreid said:Ah, one detail. In doing the MacLauren series for sin and cos, in the expansion one only needs that sin(0) = 0 and cos(0) = 1.

[tex]\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]

This is not valid when

[tex]\frac{d}{dx}\sin x^{\circ} = \frac{\pi}{180} \cos x^{\circ}[/tex]

and thus the Maclaurin series for [itex]\sin x^{\circ}[/itex] is

[tex]\sin x^{\circ} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{180}\right)^{2n+1}x^{2n+1}[/tex]

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nomadreid said:

(I'm not sure this rubric is the right one for this question, but since it didn't seem to fit any of the other rubrics, I put it here.)

All of those terms are circular functions. Radians are based on pi. 360 Degrees is arbitrary.

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You may be thinking that, because (sin x)'= cos x, (sin x)''= -sin x, (sin x)'''= -cos x, etc. and similarly for the derivatives of cos x, in order to find the MacLauren series, we only need to evaluate those at x= 0.nomadreid said:

But the point is that those derivative formula are only true if x is in

In particular, the derivative of sin x is given by

[tex]\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}[/tex]

Using the trig identity [itex]sin(x+h)= sin(x)cos(h)+ cos(x)sin(h)[/itex] that becomes

[tex]\lim_{h\to 0}\frac{sin(x)cos(h)- cos(x)sin(h)- sin(x)}{h}[/tex]

[tex]= \lim_{h\to 0} sin(x)\frac{cos(h)- 1}{h}+ cos(x)\frac{sin(h)}{h}[/tex]

so to have the derivative of sin(x) be cos(x) we must have

[tex]\lim_{h\to 0}\frac{cos(h)- 1}{h}= 0[/tex]

and

[tex]\lim_{h\to 0}\frac{sin(h)}{h}= 1[/tex]

and, as Dickfore noted, those are only true if h is in

In fact, if you are going to use the trig functions as general

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(1) First, nowhere in your proof is it explicitly shown that x and h need to be in radians. I presume it is implicit in the use of the limit of sin(h)/h = 1 which is derived from using the arc formula arc length = rx, where x is in radians, and r is the radius. Right?

(2) Assuming this to be the case, although the proof that sin'(x) = cos x and so forth rests on x being in Radians, nonetheless sin'(x) = cos x works perfectly well for x in degrees, since one just substitutes from the radians to the equivalent number of degrees and back again. The derivative of sin x at x=60 degrees is indeed cos(60 degrees). Therefore I would appreciate some clarification to your statement that these derivatives are only valid if x is in Radians.

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sin'(0)=pi/Arccos(-1)

As I said above the sine and cosine in another scaling are easily expressible in terms of the usual one, but that does not make them equal. Some confusion might be resulting from thinking of degrees as a way of expressing a number versus as being a different number. Like if we said

60 degrees=π/3

or

degrees=π/180

by sine in degrees we mean a function where

(60,√3/2)

not

(π/3,√3/2)

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