Răspuns:
[tex]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}[/tex]
Calculam separat [tex]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} =[/tex]
[tex]=\frac{\sqrt{6} }{4}-\frac{\sqrt{2} }{4}[/tex]
Inlocuim:
[tex]\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}[/tex]
