# Why do both trig functions have the same Macluarin series?

Both the degree version and the radian version of the trig functions have the same Maclaurin series, yet they are different. How is this possible? How can two different functions have the same Maclaurin series?

• Please clarify, I'm not sure what you're talking about. – Raskolnikov Sep 17 '14 at 21:32
• They don't have the same series. Why do you think they do? – MPW Sep 17 '14 at 21:32
• @Raskolnikov The degree version of sine is not the same as the radian version. For instance sin(pi degrees) isn't equal to sin (pi radians) – dfg Sep 17 '14 at 21:35
• Almost certainly the assumption is that the angle is measured in radians for such series. – MPW Sep 17 '14 at 21:39
• Not so. The derivative of $\sin$ in degrees at $0$ is $\pi/180$. You may be assuming, incorrectly, that the derivative of $\sin$ is $\cos$ regardless of the unit of the variable. – Kevin Carlson Sep 17 '14 at 21:49

## 1 Answer

Practically nobody does calculus using the "degrees" version of trigonometric functions, and this is one of the reasons why. The function that gives the sine of a number of radians is not the function that gives the sine of a number of degrees, and you cannot just start taking derivatives of both functions and expecting that the only thing you need to do to reconcile them is to use the "correct" units of input.

For clarity, I'll write $\sin_r(x)$ to mean "sine of the value $x$ when $x$ is given in radians," and $\sin_d(x)$ to mean "sine of the value $x$ when $x$ is given in degrees." Similarly, I'll wrote $\cos_r(x)$ and $\cos_d(x)$ to distinguish the two cosine functions. Then $$\frac{d}{dx} \sin_r(x) = \cos_r(x).$$ But $$\frac{d}{dx} \sin_d(x) \ne \cos_d(x).$$ In fact, the two sides of that inequality are not even close, as you can easily confirm by integrating $\cos_d(x)$ from $60$ degrees to $90$ degrees. Since $\cos_d(x) \ge \frac 12$ over that entire range, the answer cannot be less than $\frac 12 \cdot(90-60) = 15,$ which plainly is much greater than the difference between $\sin_d(90)$ and $\sin_d(60).$

In fact, since $\sin_d(x) = \sin_r\left(\frac{\pi}{180} x\right),$ we find that $$\begin{eqnarray} \frac{d}{dx} \sin_d(x) &=& \frac{d}{dx} \sin_r\left(\frac{\pi}{180} x\right) \\ &=& \frac{\pi}{180} \cos_r\left(\frac{\pi}{180} x\right) \\ &=& \frac{\pi}{180} \cos_d(x). \end{eqnarray}$$ If you construct your MacLaurin series using correctly calculated derivatives, you will find it looks much different than the usual MacLaurin series, but you should find that your new MacLaurin series is what you would get if you were to substitute $\frac{\pi}{180} x$ for $x$ in the usual MacLaurin series. (There would be a lot of powers of $\pi$ and $180$ occurring in the terms of the series.)