Problem :
Find a geometric construction or a proof by hand to show :
$$\sin\left(1+\frac{1+\sqrt{5}}{2}\right)<1/2$$
As attempt I introduce the inequality :
$$\sin\left(1+\frac{1+\sqrt{5}}{2}\right)-1+\frac{1+\sqrt{5}}{2}-\frac{559}{500}<0$$
Then I introduce :
$$f(x)=\sin\left(1+x\right)-1+x-\frac{559}{500}$$
Or :
$$g(x)=\sin\left(x\right)+x-2-\frac{559}{500}$$
Then we can use power series around $x=610/233$ but it's tedious by hand .
How to solve the problem ?
Since $\dfrac{1}{2}=\sin\dfrac{5\pi}{6}$ and $y=\sin x$ is strictly decreasing for $x\in\left(\dfrac{\pi}{2}, \pi\right)$, it suffices to prove that
$$ 1+\frac{1+\sqrt{5}}{2}>\frac{5\pi}{6} $$
Since $2.236^2=4.999696<5$, we know that $2.236<\sqrt{5}$, and hence $1+\dfrac{1+\sqrt{5}}{2}>2.618$. Hence, we have
$$\begin{align} 1+\frac{1+\sqrt{5}}{2} &>2.618 \\ &>\frac{1775}{678} & \text{ since }2.618=\frac{1775.004}{678}\\ &=\frac{5}{6}\cdot\frac{355}{113} \\ &>\frac{5\pi}{6} \end{align}$$
Yes there are some steps which have to be verified by multiplication, but definitely doable by hand.
Note that $\phi^2=1+\phi$. It suffices to prove that $$ \phi^2>\frac{5\pi}{6}\implies \frac{6}{5}\phi^2>\pi. $$
By applying the cosine law to the golden triangle, we find $$ \frac{6}{5}\phi^2=\frac{3}{5(1-\cos(\frac{\pi}{5}))} =\frac{3}{5\left(1-\frac{\sqrt5+1}{4}\right)} =\frac{3}{5}(3+\sqrt5). $$ Different proofs that $\cos(\pi/5)=\frac12\phi$ can be found here.
By taking $\sqrt5$ to four decimal places ($\sqrt{5}\approx2.2360$, rounding down) and doing a little unpleasant computation by hand, it can be shown that $$ \frac{3}{5}(3+\sqrt5)>\frac{3}{5}(3+2.2360)=3.1416>\pi. $$ Hope this helps!
This answer is just an extension of my answer to this question, which asks why $\frac{6}{5}\phi^2\sim\pi$.
First, notice that $${3(3+2.236)} = {5\cdot 3.1416}$$
Now, recall we need to prove $$\sin\left(1+\frac{1+\sqrt5}{2}\right)<\sin\frac{5\pi}{6}$$ which as Michael Rozenberg and others pointed out, is logically equivalent to showing $$1+\frac{1+\sqrt5}{2}>\frac{5\pi}{6}$$ Multiplying both sides by $6/5\pi$ and simplifying is equivalent to $$\frac{3(3+\sqrt5)}{5\pi}>1$$
Since $\sqrt 5 > 2.236$ and $\pi < 3.1416$, we have $$\frac{3(3+\sqrt5)}{5\pi} >\frac{3(3+2.236)}{5\cdot3.1416} = 1\quad\quad \blacksquare$$
Here is another that avoids $\pi$ approximations other than $3 < \pi < 4$ and takes liberty with the notion of calculating by hand.
For $x \ge 0$ we have $\sin x \le u(x)=\sum_{k=1}^7 (-1)^{n+1} {x^{2k+1} \over (2k+1)!}$, so in principle we can upper bound $\sin$ when calculating with rationals.
Using $2 < \sqrt{5} < 3$ we have ${\pi \over 2} < 2 <{5 \over 2} < 1+ \phi < 3 < \pi$ and $\sin$ is decreasing on the interval $({\pi \over 2}, \pi)$.
If we can find a rational ${p \over q} \in (2,3)$ such that ${p_n \over q_n} \le 1+\phi$ and we have $u({p_n \over q_n}) < {1 \over 2}$ then we are finished.
Noting that $q(x)= x^2-x-1$ is increasing and non negative for $x \ge 0$, $q(\phi) = 0$ and $q({1618 \over 1000}) < 0$, we see that ${2618 \over 1000}$ satisfies the criteria and $u({2618 \over 1000}) = {4132642497822341133247468087 \over 9278984069824218750000000000} < {1 \over 2}$.
I assume we know
$\sqrt{5}>2.236$
$\pi<3.141593$
$\sin\frac{\pi}{6}=\frac{1}{2}$
$\sin$ is strictly increasing on $\left[0,\pi/2\right]$
Now, $\sin\left(1+\frac{1+\sqrt{5}}{2}\right)=\sin\left(\pi-\left(1+\frac{1+\sqrt{5}}{2}\right)\right)$.
We can verify by hand that $\pi-\left(1+\frac{1+\sqrt{5}}{2}\right)\approx0.5\in\left[0,\pi/2\right]$.
So, by (4), it suffices to show that $$\pi-\left(1+\frac{1+\sqrt{5}}{2}\right)<\frac{\pi}{6}\text{ or }\frac{5}{6}\pi-\left(1+\frac{1+\sqrt{5}}{2}\right)<0.$$ But this last inequality holds because
$$\frac{5}{6}\pi-\left(1+\frac{1+\sqrt{5}}{2}\right)<\frac{5}{6}\left(3.141593\right)-\left(1+\frac{1+2.236}{2}\right)=-0.0000058\dot{3}.$$
We need to prove that $$\sin\left(1+\frac{1+\sqrt5}{2}\right)<\sin\frac{5\pi}{6}$$ or $$1+\frac{1+\sqrt5}{2}>\frac{5\pi}{6}.$$ Can you solve the question from here?
