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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?

nmasanta
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Sudhanshu Bharadwaj
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  • You can take $\sin(\sin(x))$ degrees if you want. There is nothing stopping you. The question is, how is the result meaningful? – SlipEternal Jun 26 '19 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 '19 at 15:32
  • See https://math.stackexchange.com/q/1797756/30222 – tomasz Jun 27 '19 at 16:31

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$\sin(\sin x)$ could be evaluated if $x$ is in radian or in degrees.

You will get different answers so we need to clarify which unit is used.

For example in radian $$\sin(\sin(90)) \approx 0.7795$$ but in degrees $$\sin(\sin(90))\approx0.017$$

The trig formulas for differentiation or integration are only valid in radians, but the trig identities are valid in any unit.

For instance $$ \sin^2 (x) + \cos^2(x) =1$$ is valid in any unit as well as $$\tan (x) = \frac {\sin (x)}{\cos(x)}$$

Mohammad Riazi-Kermani
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