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$S i n 24^{0} + \sin 32^{0} + \sin 204^{0} + \sin 212^{0} =$

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Detailed Solution

$Sin 24^{\circ} + \sin 32^{\circ} + \sin 204^{\circ} + \sin 212^{\circ} = \sin 24 ° + \sin (180 ° + 24 °) + \sin 30^{\circ} + \sin (180 ° + 32 °) = \sin 24 ° - \sin 24 ° + \sin 32 ° - \sin 32 ° = 0$

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