Question

4 *sin de pi supra 12 * cos de pi supra12=1

Answer 1

[tex]\it 4 sin\dfrac{\pi}{12} cos\dfrac{\pi}{12}=4\cdot \dfrac{\sqrt6 -\sqrt2}{4}\cdot\dfrac{\sqrt6+\sqrt2}{4} = \dfrac{(\sqrt6-\sqrt2)(\sqrt6 +\sqrt2)}{4}= \\\;\\ \\\;\\ =\dfrac{(\sqrt6)^2-(\sqrt2)^2}{4} =\dfrac{6-2}{4} =\dfrac{4}{4}=1[/tex]