Why does the power series of sin(x) require radian?

I learnt the maclaurin series of sin x and then i realised it cant work for both degrees and radians. Of course it should work in radians but i cant seem to understand why. In the proof of the serie, which is x - x^3/3! + x^5/5! - x^7/7! + ....(cba with latex) , you only require sin 0 = 0 and cos 0 = 1. this is true for both degree and radian so that cant be the reason why degree dont work. now i understand that you need sinx/x -> 1 as x -> 0, but this seem to be true for degree as well, because as an angle x in degree get smaller, so does sin x, so they both -> 0. so sinx/x -> 1 for degree too (surely?)

Then i found a proof that prove sinx/x = 1 without referring to degree or radian again. link

**Direct Proof from Definition of Sine**

This proof works directly from the definition of the sine function:

| | | | sin*x* | = | | ∑*n*=0∞(−1)*n**x*2*n*+1(2*n*+1)! | | | By the definition of the sine function | |

| | | | | = | | (−1)0*x*2⋅0+1(2⋅0+1)!+∑*n*=1∞(−1)*n**x*2*n*+1(2*n*+1)! | | | | |

| | | | | = | | *x*+∑*n*=1∞(−1)*n**x*2*n*+1(2*n*+1)! | | | | |

now this confuse me even more cos now we have a proof of sinx/x without referring to deg or rad, which suggest that both will work.

i know deg cant work for the power serie, cos then you could divide the 360deg angle into any number of parts and call that a measurement and it would work. but why radian? its just dividing the 360 angle into 2pi parts

sorry if anything is unclear. please ask. this is probably a noob question with a obvious answer but its bugging me for ages. i just cant find the answer or understand it

Re: Why does the power series of sin(x) require radian?

The issue is that the Maclauren Series uses the derivatives of a function to develop the series:

$\displaystyle f(a) = f(0) + af'(0) + \frac {a^2}{2!}f''(0) + \frac {a^3} {3!} f'''(0) + ...$

For the case of f(x) = sin(x) this yields:

$\displaystyle \sin(x) = \sin (0)+ x \cos (0) + \frac {x^2}{2!} (- \sin (0)) + \frac {x^3}{3!}(- \cos (0)) + ... $ which becomes the series you are familiar with:

$\displaystyle \sin (x) = x - \frac { x^3}{3!} + \frac {x^5}{5!} + ...$

This formula works because if x is in radians then the derivative of sin(x) is cos(x), the second derivative of sin(x) is -sin(x), etc. If x is in units of degrees then the derivative would be $\displaystyle \frac {d(\sin (x) )}{dx} = ( \frac {\pi}{\180} ) \cos(x)$. The second derivative is $\displaystyle \frac {d^2(\sin (x) )}{dx^2} = - (\frac {\pi}{\180})^2 \sin(x)$, etc.

The infinite series for x in degrees then becomes:

$\displaystyle sin(x) = ( \frac { \pi}{180} ) x - (\frac {\pi}{180})^3 \frac {x^3}{3!} + (\frac { \pi}{180})^5 \frac {x^5}{5!} -.... $

Re: Why does the power series of sin(x) require radian?

oh thanks!

thats it, but i cant remember that you need radian for differentiation, where does it say that in the proof of the derivative of sin, i vaguely remember that there was an arc and a circle involved

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Re: Why does the power series of sin(x) require radian?

It comes about because despite what you wrote in post #1 if x is in degrees $\displaystyle \lim_{x \to 0}\frac {\sin (x)}x = \frac {\pi} {180}$, not 1. This is because the arc length for angle $\displaystyle \theta $ if theta is in degrees is $\displaystyle s=R \theta \frac {\pi} {180} $, not $\displaystyle R \theta$.

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Re: Why does the power series of sin(x) require radian?

The degree is an extremely arbitrary unit of measure, which disrupts all the subtle connections between the trigonometric functions. Radians are natural because they take the arc length and give you the x and y coordinates.

Re: Why does the power series of sin(x) require radian?

thank you so much guys!

one last thing, is it true that only in radian sinx/x->1 as x->0? is this not true for degree? if so why?

thanks

Re: Why does the power series of sin(x) require radian?

It cannot possibly be true for both of them...the function which takes x in degrees and gives the sin of the angle is $\displaystyle \sin(\frac{180x}{\pi})$, where the inner sin function works in radians.

If we try to apply the same limit to this function, we get:

$\displaystyle lim_{n \to 0} \frac{\sin(\frac{180x}{\pi})}{x} = \frac{180}{\pi}$

So no, it certainly does not hold, and there's no reason why it would.