Provesin 2 72-sin 2 60-5-1 - 8

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Provesin 2 72...

Question

Prove

sin²72° - sin²60° = √5 -1/8

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Prove that:

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

{sin}^{2} 72^{\circ} - {sin}^{2} 60^{\circ} = \frac{\sqrt{5} - 1}{8}

s i n^{2} 60^{\circ} + c o s^{2} (3 x - 9^{\circ}) = 1

{sin}^{2} θ + {sin}^{2} (60^{\circ} + θ) + s i n^{2} (60^{\circ} - θ) =

\frac{4}{{cot}^{2} 30^{\circ}} + \frac{1}{{sin}^{2} 60^{\circ}} - {cos}^{2} 45^{\circ}

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