(Daneză)
Euclid showed in Book XIII, Proposition 10 of his Elements that the area of the square on the side of a regular pentagon inscribed in a circle is equal to the sum of the areas of the squares on the sides of the regular hexagon and the regular decagon inscribed in the same circle. In the language of modern trigonometry, this says:
Ptolemy used this proposition to compute some angles in his table of chords in Book I, chapter 11 of Almagest.
Solution 1 CiP
We know the values $$sin 18^\circ=\frac{\sqrt{5}-1}{4},\;sin 30^\circ=\frac{1}{2},\;sin 36^\circ=\frac{\sqrt{10-2\sqrt{5}}}{4}.$$
The calculation is simple: $sin^2 18^\circ +sin^2 30^\circ =\left (\frac{\sqrt{5}-1}{4} \right )^2+\left (\frac{1}{2} \right )^2=\frac{5-2\sqrt{5}+1}{16}+\frac{1}{4}=\frac{6-2\sqrt{5}+4}{16}=\frac{10-2\sqrt{5}}{16}=sin^2 36^\circ .$
$\blacksquare$
Solution 2 CiP
First we show that
$cos 36^\circ -cos 72^\circ=\frac{1}{2}. \tag{1}$
For this, we denote $A:=cos 36^\circ -cos 72^\circ.$ We will calculate, with double-angle formula and product-to-sum identity
$$2sin 36^\circ \cdot A=2sin 36^\circ \cdot cos 36^\circ-2sin 36^\circ \cdot cos 72^\circ=sin (2\cdot 36^\circ)-(sin 108^\circ-sin36^\circ)=$$
$$=(sin 72^\circ-sin 108^\circ)+sin 36^\circ=sin 36^\circ,$$
(because of reflection in $90^\circ$ formulae, $sin108^\circ=sin(180^\circ-108^\circ)=sin 72^\circ)$, hence (1) follow.
Now we calculate with power-reduction formula
$sin^2 36^\circ-sin^2 18^\circ=\frac{1-cos (2\cdot 36^\circ)}{2}-\frac{1-cos(2\cdot 18^\circ)}{2}=\frac{1}{2}(cos 36^\circ-cos 72^\circ)\;\underset{(1)}{==}\;\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}=sin^2 30^\circ.$
$\blacksquare$
Remark CiP We also know $cos 36^\circ =\frac{\sqrt{5}+1}{4},\;cos 72^\circ =\frac {\sqrt{5}-1}{4}$ and (1) results easier.
<End Rem>
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