Find the value of ${sin}^{2} 18^{0} - {cos}^{2} 72^{0}$

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Solution

${sin}^{2} 18^{\circ} - {cos}^{2} 72^{\circ}$
$= {sin}^{2} 18^{\circ} - {cos}^{2} (90^{\circ} - 72^{\circ})$
$= {sin}^{2} 18^{\circ} - {sin}^{2} 18^{\circ} = 0$

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