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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Jul 2007 00:44:13 IST
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This is a special class of problems. I have given the solution in details. Try to follow each of the steps.
Let 2  / 7 = x
sin 2/7 + sin 4/7 + sin 8/7
= sinx + sin2x + sin4x ................(1)
Let P = (sinx + sin2x + sin4x)2 + (cosx + cos2x + cos4x)2
= 3 + 2 [ (sinxsin2x + cosxcos2x) + (sin2xsin4x + cos2xcos4x) + (sin4xsinx+cos4xcosx) ]
[ expanding, then arranging and grouping the terms ]
= 3 + 2 (cosx + cos2x + cos3x)
= 3 + 2( cosx + cos2x + cos4x)
[ since 7x = 2 , 3x = 2 - 4x, cos3x = cos(2 - 4x) = cos4x ]
Now, let us calculate the value of cosx + cos2x + cos4x
Let Q = cosx + cos 2x + cos4x
Therefore, 2Q sinx = 2cosxsinx + 2cos2xsinx + 2cos4xsinx
= sin2x + sin3x - sinx + sin 5x - sin 3x = - sinx
[ since , sin 5x = sin (7x - 2x) = sin(2 - 2x) = - sin2x ]
Therefore, 2Q sinx = - sinx i.e Q = - 1 / 2 ............ (2)
Finally,
P = (sinx + sin2x + sin4x)2 + Q2
= 3 + 2Q
 (sinx + sin2x + sin4x)2 + 1 / 4 = 3 + 2 (-1/2) = 2 [since , Q = - 1/2 ]
 (sinx + sin2x + sin4x)2 = 2 - 1/4 = 7 / 4
(sinx + sin 2x + sin4x) =   7 / 2
sin 2 / 7 + sin 4 / 7 + sin 8 / 7 =  7 / 2
Now, we have to reject the negative value i.e - 7 / 2
sin 2 / 7 > sin / 7 ( 2 / 7 = 51 deg and / 7 = 25 deg approx. )
i.e sin 2 / 7 - sin / 7 > 0
i.e sin 2 / 7 + sin 8 / 7 > 0 [ since, sin 8 / 7 = - sin / 7 ]
i.e sin 2 / 7 + sin 4 / 7 + sin 8 / 7 > 0 [ since, sin 4/7 > 0 as 0 <4/7 < ]
Therefore, negative value for (sin 2 / 7 + sin 4 / 7 + sin 8 / 7) is inadmissible.
Hence, sin 2 / 7 + sin 4 / 7 + sin 8 / 7 = 7 / 2
Cheers !!
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You never know what is enough till you know what is more than enough.
Titun |
this reply: 10 points
(with 2 
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