simplify:-
${cos}^{2} 72^{\circ} - {sin}^{2} 54^{\circ} =$

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Solution

Simplify $c o s^{2} 72 - s i n^{2} 54$

$c o s^{2} 72 = c o s^{2} (90 - 18) = s i n^{2} 18 = (\frac{\sqrt{5} - 1}{4})^{2}$

$s i n^{2} 54 = s i n^{2} (90 - 36) = c o s^{2} 36 = (\frac{\sqrt{5} + 1}{4})^{2}$

$\Rightarrow (\frac{\sqrt{5} - 1}{4})^{2} - (\frac{\sqrt{5} + 1}{4})^{2}$

$= \frac{1}{16} [(\sqrt{5} - 1)^{2} - (\sqrt{5} + 1)^{2}] = \frac{1}{16} [(\sqrt{5} - 1 - \sqrt{5} - 1) (\sqrt{5} - 1 + \sqrt{5} + 1)]$

$= \frac{1}{16} [(- 2) \cdot 2 \sqrt{5}] = \frac{- \sqrt{5}}{4}$

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