# In the expresion $4\sin (\theta) \frac{d \theta}{d t}=\frac{dy}{dt}$ Which is the correct unit of $\theta$ ¿ radian or degrees? [closed]

If I change units of an angle for radians to degrees in the next expresion $$4\sin (\theta) \frac{d \theta}{d t}=\frac{dy}{dt}$$ the value of $$\frac{dy}{dt}$$ changes.

For example at a rate of change of $\frac{d\theta}{dt} = 30deg , \qquad$ and $\frac{d\theta}{dt}=\frac{\pi }{6}rad$ the rate of change is the same, but the final expresion is not.

So which is the correct unit? and mathematicaly why is the reason?.

I already know that the correct unit are radians, Im looking for a more formal and deeper explanation of why this units are the correct.

Thanks

## closed as unclear what you're asking by Ben Crowell, Brandon Enright, JamalS, Neuneck, ACuriousMind♦Nov 10 '14 at 13:36

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• Related: physics.stackexchange.com/q/33542/2451 and links therein. – Qmechanic Nov 9 '14 at 18:23
• Both are units of angle. Why should one be more correct than the other? (Note that the sine is defined in a way that the function takes radians input, though) – ACuriousMind Nov 10 '14 at 13:36
• @ACuriousMind: One must be more correct, becouse if you use one or the other, the final expression of $\frac{dy}{dt}$ will change his value. – JuanMuñoz Nov 11 '14 at 17:30

For dimensional consistency, firstly, I would expect the number '4' to be dimensionfull. Additionally, whether the angle is to be taken in radians or in degrees depends on where this equation came from. If I venture a guess, then I suppose at some point you differentiated a relation between $y$ and $\theta$. In that case, the relation you started of with will specify the units for $\theta$ after which the differentiation would have to carried out accordingly. Remember, $$\dfrac{\mathrm{d}}{\mathrm{d}\theta}\cos{\theta}=-\sin\theta\\ \dfrac{\mathrm{d}}{\mathrm{d}\theta}\cos{\theta^{\circ}}=-\dfrac{\pi}{180}\sin\theta^{\circ}\\ \dfrac{\mathrm{d}}{\mathrm{d}\theta^{\circ}}\cos{\theta^{\circ}}=-\sin\theta^{\circ}\\$$
• Where @surajshankar: did you get that only if $\theta$ is in radians then the derivative is $-\sin (\theta)$. I know that the demonstration if this rule, is a limit and never uses the hipotesis that $\theta$ is in radians – JuanMuñoz Nov 9 '14 at 20:54