3.339. a) $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4} ;$ b) $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$.
## Solution.

a) $\cos 36^{\circ}=\sin 54^{\circ} \Leftrightarrow \cos 2\left(18^{\circ}\right)=\sin 3\left(18^{\circ}\right)$.

## From this we get:

$$
\begin{aligned}
& 1-2 \sin ^{2} 18^{\circ}=3 \sin 18^{\circ}-4 \sin ^{3} 18^{\circ} \Leftrightarrow \\
& \Leftrightarrow 4 \sin ^{3} 18^{\circ}-2 \sin ^{2} 18^{\circ}-3 \sin 18^{\circ}+1=0 \Leftrightarrow \\
& \Leftrightarrow\left(\sin 18^{\circ}-1\right)\left(4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}-1\right)=0 \Leftrightarrow 4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}-1=0
\end{aligned}
$$

Solving this equation as a quadratic in $\sin 18^{\circ}$, we have:

1) $\sin 18^{\circ}=\frac{-1-\sqrt{5}}{4}0$;
2) $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$, then $\cos 36^{\circ}=1-2 \sin ^{2} 18^{\circ}=1-2\left(\frac{\sqrt{5}-1}{4}\right)^{2}=$ $=1-\frac{6-2 \sqrt{5}}{8}=\frac{2 \sqrt{5}+2}{8}=\frac{\sqrt{5}+1}{4}$.

The equality is valid.

b) $\sin 36^{\circ}=\cos 54^{\circ} \Leftrightarrow 2 \sin 18^{\circ} \cos 18^{\circ}=4 \cos ^{3} 18^{\circ}-3 \cos 18^{\circ} \Leftrightarrow$

$\Leftrightarrow 2 \sin 18^{\circ}=4 \cos ^{2} 18^{\circ}-3,2 \sin 18^{\circ}=4\left(1-\sin ^{2} 18^{\circ}\right)-3 \Leftrightarrow$
$\Leftrightarrow 4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}=1 \Leftrightarrow 4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}+\frac{1}{4}=\frac{5}{4}$,

$\left.\left(2 \sin 18^{\circ}+\frac{1}{2}\right)^{2}=\frac{5}{4} \Rightarrow 1\right) 2 \sin 18^{\circ}+\frac{1}{2}=-\frac{\sqrt{5}}{2}, \sin 18^{\circ}=\frac{-1-\sqrt{5}}{4}, \varnothing ;$

2) $2 \sin 18^{\circ}+\frac{1}{2}=\frac{\sqrt{5}}{2}, \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$.

The equality is valid.