Question

решите выражения
1)$2\sqrt{2}sin\frac{\pi }8} cos\frac{\pi }8}$
2)$\sqrt{48} cos^{2} \frac{7\pi }{12} -\sqrt{48}sin^{2}\frac{7\pi }{12}$[/tex]
3)$\sqrt{48} cos^{2} \frac{\pi }{12} -\sqrt{12}$

Answer 1

Ответ:

Объяснение:

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Answer 2

$1) \: 2\sqrt{2}sin\frac{\pi}{8}cos\frac{\pi}{8}=sin\frac{\pi}{4}\cdot \sqrt{2}=\frac{\sqrt{2}}{2}\cdot \sqrt{2}=\frac{2}{2}=1$

$2) \: \sqrt{48}cos^2\frac{7\pi}{12}-\sqrt{48}sin^2\frac{7\pi}{12}=4\sqrt{3}\cdot \frac{1+cos\frac{7\pi}{6}}{2}-4\sqrt{3}\cdot \frac{1-cos\frac{7\pi}{6}}{2}=2\sqrt{3}\cdot (1-\frac{\sqrt{3}}{2})-2\sqrt{3}\cdot (1-(-\frac{\sqrt{3}}{2}))=2\sqrt{3}-3-2\sqrt{3}\cdot (1+\frac{\sqrt{3}}{2})=2\sqrt{3}-3-2\sqrt{3}-3=-6$

$3) \: \sqrt{48} cos^{2} \frac{\pi}{12} -\sqrt{12}=4\sqrt{3}\cdot \frac{1+cos\frac{\pi}{6}}{2}-2\sqrt{3}=2\sqrt{3}\cdot (1+\frac{\sqrt{3}}{2})-2\sqrt{3}=2\sqrt{3}+3-2\sqrt{3}=3$