Prove that: sin2 72∘−sin2 60∘=√5−18

1.	Prove that: sin2 72∘−sin2 60∘=√5−18
Answer» Prove that: $s i n^{2} 72^{\circ} - s i n^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}$

Answer»

Prove that:

$s i n^{2} 72^{\circ} - s i n^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}$

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