# In $\sin(\sin(x))$ Why should I Calculate the $\sin$ of $(\sin(x))$ radians not the $\sin$ of $\sin(x)$ degrees? [duplicate]

Why is radians the only unit for which I can use take $$\sin(\sin(x))$$ (Why is it the the default unit of trigonometry). This does not work if change the definition of a radian. But works if we change the definition of $$\pi$$ itself

Is this due to how we derive the Taylor Polynomial of $$\sin(x)$$ and $$\cos(x)$$ ? or How the differential of $$\sin x$$ in radians is $$\cos x$$ ?(i.e because $$\frac{\sin x}{x}$$ converges to $$1$$ as $$x\to 0$$)

Or is it because of some convention or assumption somewhere?

## marked as duplicate by tomasz, YuiTo Cheng, воитель, postmortes, Thomas ShelbyJun 28 at 5:53

• You can take $\sin(\sin(x))$ degrees if you want. There is nothing stopping you. The question is, how is the result meaningful? – InterstellarProbe Jun 26 at 15:28
• "How the differential of sinx in radians is cosx" That's the reason why everyone uses radians. You can use degrees if you want, but calculus will be messier. – D_S Jun 26 at 15:32
• – tomasz Jun 27 at 16:31

$$\sin(\sin x)$$ could be evaluated if $$x$$ is in radian or in degrees.
For example in radian $$\sin(\sin(90)) \approx 0.7795$$ but in degrees $$\sin(\sin(90))\approx0.017$$
For instance $$\sin^2 (x) + \cos^2(x) =1$$ is valid in any unit as well as $$\tan (x) = \frac {\sin (x)}{\cos(x)}$$