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$$ \sin( \pi /2) = 1 $$ $$ \sin(\phi ) = 0.99889 \approx 1 $$ Can we say that: $$ \pi/2 \approx \phi $$ and hence, $$ \pi \approx 2 \phi $$

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    Well, numerically, they are decently close... – 2017-02-27
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    Sine is a $2\pi$-periodic function, hence there are other possibilities of $\phi$. E.g $\phi \approx \pi/2 + 2n\pi$ with $n\in\mathbb{Z}$. – 2017-02-27
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    @Eff true, though that doesn't seem to be what the OP is getting at. Restricting the interval to $[0,\pi]$ is natural here. Really this is all coincidence of course, simply because $\pi/2$ and $\phi$ have an absolute difference of about $0.047$ – 2017-02-27

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Can we say that: $$ \pi/2 \approx \phi $$ and hence, $$ \pi \approx 2 \phi $$

Yes, but not because $\sin \phi \approx \sin (\pi/2)$. Since there is no other context here (and therefore no way for me to know which methods are allowed or not), I will say that we can simply see $\pi/2 \approx \phi$ from a calculator.

But in general, $\sin x \approx \sin y$ does not mean that $x \approx y$. This is because sine is periodic. For example, in radians, $\sin 0 \approx \sin(13815119.1366)$, but certainly $0$ is not approximately equal to $13815119.1366.$

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    If we restrict ourselves to the interval $[0,\pi]$ then we have $\sin x \approx \sin y \implies x \approx y$, and this domain restriction is pretty natural. – 2017-02-27
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    @BrevanEllefsen: No we don't -- for example, $\sin 2 \approx \sin 1.141593$ but $2 \not\approx 1.141593$. – 2017-02-27
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    @HenningMakholm oh yeah, bijection is required. Alright, then $[0,\pi/2]$ – 2017-02-27
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Aside from the fact that $\sin$ is a periodic function, the value of $\sin(x)$ changes very slowly when $x$ is near $\frac\pi2.$ Hence the fact that $\sin(x) \approx 1$ and $0 < x < \pi$ gives us some idea that $x$ is close to $\frac\pi2,$ but it does not give us a very good indication of how close.

In this case, $\sin(\phi)$ differs from $\sin\left(\frac\pi2\right)$ by less than $0.12\%,$ but $\phi - \frac\pi2 \approx 0.0472,$ so $\phi$ is more than $3\%$ greater than $\frac\pi2.$ That's an approximation of some sort, but not nearly as good an approximation as the approximation $0.99889 \approx 1 .$

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Note that the derivative of $\sin x$ is $0$ at $\pi/2$, and so a small change in $x$ won't affect the result a ton. Since $\pi/2$ and $\phi$ have an absolute difference of $0.047...$, this can be considered a small change in $x$, and will yield a small change in the output.

Note that the derivative being zero at $\pi/2$
is important, because it means $\sin(x) $ is roughly flat in small neighborhood around $\pi/2$. If we chose a function with non-zero derivative we could get much more different result; for example, $e^{\pi/2} - e^\phi = - 0.23...$