sina+sin(120+a)+sin(240+a) = 0
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We have to prove that,
sin a + sin (120+a) + sin (240+a) = 0 .......... (1)
consider the left hand side of equation (1),
sin a + sin (120+a) + sin (240+a) = sin a + sin 120.cos a + cos 120.sina + sin 240.cosa + cos 240.sin a (∵sin(a+b) = sina.cosb +cosa.sinb )
sin a + sin (120+a) + sin (240+a) = sin a + √3/2 . cos a + (-1/2) . sin a +
(-√3/2) . cos a + (-1/2) . sin a
(∵sin120 = √3/2, sin240 = -√3/2) , (cos 120 = cos 240 = -1/2)
sin a + sin (120+a) + sin (240+a) = sin a + √3/2 . cos a - 1/2 . sin a
-√3/2 . cos a - 1/2 . sin a
sin a + sin (120+a) + sin (240+a) = sin a - [(1/2 + 1/2) . sin a]
sin a + sin (120+a) + sin (240+a) = sin a - (1) . sin a
sin a + sin (120+a) + sin (240+a) = sin a - sin a
sin a + sin (120+a) + sin (240+a) = 0
which is equal to right hand side of equation (1).
Hence proved.
Hope it will help you. Thanks.
sin a + sin (120+a) + sin (240+a) = 0 .......... (1)
consider the left hand side of equation (1),
sin a + sin (120+a) + sin (240+a) = sin a + sin 120.cos a + cos 120.sina + sin 240.cosa + cos 240.sin a (∵sin(a+b) = sina.cosb +cosa.sinb )
sin a + sin (120+a) + sin (240+a) = sin a + √3/2 . cos a + (-1/2) . sin a +
(-√3/2) . cos a + (-1/2) . sin a
(∵sin120 = √3/2, sin240 = -√3/2) , (cos 120 = cos 240 = -1/2)
sin a + sin (120+a) + sin (240+a) = sin a + √3/2 . cos a - 1/2 . sin a
-√3/2 . cos a - 1/2 . sin a
sin a + sin (120+a) + sin (240+a) = sin a - [(1/2 + 1/2) . sin a]
sin a + sin (120+a) + sin (240+a) = sin a - (1) . sin a
sin a + sin (120+a) + sin (240+a) = sin a - sin a
sin a + sin (120+a) + sin (240+a) = 0
which is equal to right hand side of equation (1).
Hence proved.
Hope it will help you. Thanks.
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