Question

вычислите sin^3 15°+cos^3 15°​

Answer 1

Ответ:     $sin^315^\circ +cos^315^\circ =\dfrac{3\sqrt6}{8}$   .

$sin15^\circ =sin(45^\circ -30^\circ )=sin45^\circ \cdot cos30^\circ -sin30^\circ \cdot cos45^\circ =\\\\=\dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2}-\dfrac{1}{2}\cdot \dfrac{\sqrt2}{2}=\dfrac{\sqrt2\, (\sqrt3-1)}{4}=\dfrac{\sqrt6-\sqrt2}{4}\\\\\\cos15^\circ =cos(45^\circ -30^\circ )=cos45^\circ \cdot cos30^\circ +sin45^\circ \cdot sin30^\circ =\\\\=\dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt2\, (\sqrt3+1)}{4}=\dfrac{\sqrt6+\sqrt2}{4}$

$sin^315^\circ +cos^315^\circ =\Big(\dfrac{\sqrt6-\sqrt2}{4}\Big)^3+\Big(\dfrac{\sqrt6+\sqrt2}{4}\Big)^3=\\\\\\=\dfrac{\sqrt{6^3}-3\cdot 6\cdot \sqrt2+3\cdot \sqrt6\cdot 2-\sqrt{2^3}}{64}+\dfrac{\sqrt{6^3}+3\cdot 6\cdot \sqrt2+3\cdot \sqrt6\cdot 2+\sqrt{2^3}}{64}=\\\\\\=\dfrac{2\cdot \sqrt{6^3}+2\cdot 3\cdot \sqrt6\cdot 2}{64}=\dfrac{12\sqrt6+12\sqrt6}{64}=\dfrac{24\sqrt6}{64}=\dfrac{3\sqrt6}{8}$