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Why is $\cfrac{\sin\theta}{\sin\sin\theta}$ very close to 1 radian $(57.2958°)$? I tested this out for $\theta=1^{\circ}$ and $\theta=45^{\circ}$ and various other angles, but I always seem to get about the same answer.

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  • $\begingroup$ What is $\theta$ here? $\frac{\sin(\theta)}{\sin(\sin(\theta))}$ is not a constant. (Also, the value you give is how many degrees are in a radian - it's confusing to refer to it as "1 Radian") $\endgroup$ – Milo Brandt Aug 11 '16 at 3:16
  • $\begingroup$ If I calculate with any angle value, like sin1/sin(sin1) = 57,2958 or sin45/sin(sin(45)) = 57,2972 and so on, the value is very similar of the value of 1 Radian, Is not the same, but is incredible similar with all the angles that I had try. $\endgroup$ – Ju M. Aug 11 '16 at 3:26
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Presumably, you are calculating $\sin(\theta)$ with $\theta$ given in degrees. In that case, one has the following approximation of $\sin(\theta)$ for small $\theta$: $$\sin(\theta)\approx \frac{\pi}{180}\cdot \theta$$ Since $\sin(\theta)$ always has absolute value no more than $1$, we can use this approximation to say that $\sin(\sin(\theta))\approx \frac{\pi}{180}\sin(\theta)$. Thus, one has

$$\frac{\sin(\theta)}{\sin(\sin(\theta))}\approx \frac{\sin(\theta)}{\frac{\pi}{180}\sin(\theta)}=\frac{180}{\pi}$$ which is what you are observing. Note that this expression never actually obtains equality, but stays pretty close.

It's worth noting that when you work in radians, you have that $\sin(\theta)\approx \theta$. So this term of $\frac{\pi}{180}$ comes in because, when the argument is in degrees, we need to multiply by $\frac{\pi}{180}$ to get the equivalent angle in radians. That is, the conversion factor shows up due to there being a conversion from degrees to radians implicit in the computation.

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  • $\begingroup$ As an aside, we get the small-angle approximation from the series expansion of $\sin x$, or just by looking at a graph of $\sin x$ and $x$ together at the origin. $\endgroup$ – Parcly Taxel Aug 11 '16 at 3:46
  • $\begingroup$ But we arent dealing with small angles, we are dealing with all angles. $\endgroup$ – Jacob Wakem Aug 11 '16 at 3:46
  • $\begingroup$ @Alephnull And that is where the small-angle approximation fails. The result goes to a value of $\csc 1$ radians, as I pointed out in my answer. $\endgroup$ – Parcly Taxel Aug 11 '16 at 3:55
  • $\begingroup$ @Alephnull because we take the sin of $\theta$ we are really dealing with $\frac x {\sin x}$ with $-1 \le x \le 1$. That is a reasonably small angle-$\sin 1 \approx 0.84$, which one might consider not so far from $1$ $\endgroup$ – Ross Millikan Aug 11 '16 at 4:27
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    $\begingroup$ $1°$ in radians is $\pi/180\approx 0.017$. That's quite small, indeed. $\endgroup$ – celtschk Aug 11 '16 at 5:31
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Consider $$y=\frac t{\sin(t)}$$ and, for small $t$ (using radians), by Taylor $$y=1+\frac{t^2}{6}+O\left(t^3\right)$$ Replace $t$ by $\sin(\theta)$ to get $$\frac{\sin(\theta)}{\sin(\sin(\theta)}=1+\frac 16\sin^2(\theta)+\cdots$$ which implies $$1\leq \frac{\sin(\theta)}{\sin(\sin(\theta)}\leq \frac 76$$

For the range $0\leq \theta \leq \pi$, the average value, given by $$\frac 1\pi \int_0^\pi\frac{\sin(\theta)}{\sin(\sin(\theta)}\,d\theta \approx1.09133$$ (this has been obtained using numerical integration) while $$\frac 1\pi \int_0^\pi\left(1+\frac 16\sin^2(\theta)\right)\,d\theta =\frac{13}{12}\approx 1.08333$$

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When I put the expression into Wolfram Alpha I find that the function is not a constant. It has a minimum of 1 radian, as you thought, but a maximum of around 1.19 radians.

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