Question

Doar 1.a) raspuns pe foaie

Answer 1

Răspuns:

$sin(\frac{5\pi }{24})*cos(\frac{\pi }{8} )=\frac{sin(\frac{5\pi }{24}+\frac{\pi }{8} )+sin(\frac{5\pi }{24}-\frac{\pi }{8} )}{2} =\frac{sin\frac{8\pi }{24}+sin\frac{2\pi }{24} }{2} =\frac{sin\frac{\pi }{3}+sin\frac{\pi }{12} }{2}$

Calculam separat $sin\frac{\pi }{12}=sin(\frac{\pi }{3}-\frac{\pi }{4} ) =sin\frac{\pi }{3}*cos\frac{\pi }{4}-sin\frac{\pi }{4}*cos\frac{\pi }{3}=\frac{\sqrt{3} }{2} *\frac{\sqrt{2} }{2}-\frac{\sqrt{2} }{2} *\frac{1 }{2} =$

$=\frac{\sqrt{6} }{4}-\frac{\sqrt{2} }{4}$

Inlocuim:

$\frac{\frac{\sqrt{3} }{2}+\frac{\sqrt{6}-\sqrt{2} }{4} }{2} =\frac{\frac{2\sqrt{3}+\sqrt{6}-\sqrt{2} }{4} }{2} =\frac{\sqrt{3}(2+\sqrt{2})-\sqrt{2} }{8}$