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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Mar 2008 17:04:03 IST
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Prove that (without xpanding)
1 cos(b-a) cos(c-a)
cos(a-b) 1 cos(c-b) = 0
cos(a-c) cos(b-c) 1
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17 Mar 2008 17:07:30 IST
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plz anybody...........
come with da soln. early.......
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17 Mar 2008 17:12:08 IST
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Put a=b=c. So, cos in all places becomes 1, and the determinant becomes zero... I'll tell u another way too, that's more satisfactory...
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17 Mar 2008 17:13:23 IST
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How could you put a=b=c? I mean I can't see the logic.
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'Your eyes show the strength of your soul.' |
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17 Mar 2008 17:23:20 IST
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@securitas2311
there is no logic yaar in puttuin a=b=c......
plz tell da other way............
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17 Mar 2008 17:30:09 IST
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write it as a product of two determinants yaar....
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"I a universe of atoms.......an atom in the universe" |
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cos A sin A 0 cos A sin A 0
cos B sin B 0 X cos B sin B 0
cos C sin C 0 cos C sin C 0
I havent checked this...just giving you the idea.....My ans might be wrong also
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"I a universe of atoms.......an atom in the universe" |
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17 Mar 2008 17:59:38 IST
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thanks .....anand.......for
it ll b like this.......
cos a sin a 0 cos a cos b cos c
cos b sin b 0 * sin a sin b sin c
cos c sin c 0 0 0 0
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17 Mar 2008 18:00:41 IST
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thanks .....anand......
it checkd it.......it ll b like this.......
cos a sin a 0 cos a cos b cos c
cos b sin b 0 * sin a sin b sin c
cos c sin c 0 0 0 0
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17 Mar 2008 19:19:49 IST
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your determinant is the same as mine......
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"I a universe of atoms.......an atom in the universe" |
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