geometrygolden ratiotrigonometry
I noticed that $\phi+1\approx 1.000015\left(\frac{5\pi}{6}\right)$, where $\phi=\frac{1+\sqrt5}{2}$, the golden ratio.
So I wonder, is there a clever way to show that $\sin (\phi+1)<\frac12$, without a calculator?
Using Maclaurin series, we have $\sin (\phi+1)<\sum\limits_{k=1}^7 (-1)^{k-1}\frac{(\phi+1)^{2k-1}}{(2k-1)!}$, which, in principle, could be shown to be less than $\frac12$ without a calculator. But is there a better way?
The following is exact. :-)
$$\phi=\frac{\frac{\pi}{\pi}+\sqrt{\frac{e+e+e+e+e}{e}}}{\frac{e}{e}+\frac{\pi}{\pi}}$$
Take a few well-known constants such as $\gamma,1,\phi,2,e,3,\pi$, a few functions $\log,\exp,\sin,\cos$ and the six basic operations $+,-,\times,\div,a^b,\sqrt[b]a$. Form as many simple expressions as you can by combining them. For example, there are more than $6000$ expressions of the form $f(x)\text{ op }g(y)$ many of which fall in a small range, say $[0.1,10]$. Hence you can expect a few pseudo-coincidences.
Below a plot of the values of $2717$ such expressions with a value in the above range.
This exhaustive search reveals the interesting
$$\frac{e^\gamma}{e^e}\approx\frac{e}{e^\pi}$$ $$0.117529\approx0.117467$$
which is nothing but $$\gamma-e+\pi\approx 1.$$ (Actually $1.00052649003$.)
Best Answer
Based on the comment by @Travis Willse:
$5\times 339^2>758^2$
$\sqrt5>\frac{758}{339}=\left(\frac{355}{113}\right)\left(\frac{5}{3}\right)-3$
$\phi+1=\frac{\sqrt5+3}{2}>\frac{5}{6}\left(\frac{355}{113}\right)>\frac{5}{6}\pi$ because $\frac{355}{113}>\pi$
$\pi>3>\phi+1$
$\therefore \sin (\phi+1)<\frac12$