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E=sin(pi/3)+sin(2pi/3)+sin(3pi/3)+...+sin(2013pi/3)

Răspuns :

[tex]E=sin \frac{\pi}{3}+sin \frac{2\pi}{3}+sin \frac{3\pi}{3}+......+sin \frac{2011\pi}{3}+sin \frac{2012\pi}{3}+sin \frac{2013\pi}{3} \\ \\ E = \frac{ \sqrt{3} }{2}+ \frac{ \sqrt{3} }{2}+ 0-\frac{ \sqrt{3} }{2}- \frac{ \sqrt{3} }{2}+.....+ \frac{ \sqrt{3} }{2}+ \frac{ \sqrt{3} }{2}+ 0 [/tex]

Cum aflam cat e sin 2011pi/3 2012,2013.

Impartim pe 2011 la 3 si ne da 670 rest 1 deci: [tex]sin( 670\pi+\frac{\pi}{3}) = sin \frac{\pi}{3} = \frac{ \sqrt{3} }{2} [/tex]

2012 : 3 = 670 rest 2. [tex]sin( 670\pi+\frac{2\pi}{3}) = sin \frac{2\pi}{3} = \frac{ \sqrt{3} }{2} [/tex]

2013 : 3 = 671 rest 0. [tex]sin \frac{2013\pi}{3} =sin(671\pi) = 0[/tex]

In suma, se reduce tot, inafara de primii doi termeni ai sumei si avem:

[tex]E = \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{2} \Rightarrow E = \frac{2 \sqrt{3} }{2} \Rightarrow E = \sqrt{3} [/tex]