1cos δθ=l.(l+δl) +m.(m+δm) +n.(n+δn)
=l2+m2+n2+l.δl+m.δm+n.δn
=1+l.δl+m.δm+n.δn
cos δθ-1= l.δl+m.δm+n.δn
1-cos δθ=-(l.δl+m.δm+n.δn)
we know 1-cos δθ=2sin2 δθ/2 .........for small δθ.....1-cos δθ=2(δθ/2)2
(δθ)2/2= -(l.δl+m.δm+n.δn)...............................................i
now we know that l2+m2+n2=1......................ii
(l+δl)2+(m+δm)2+(n+δn)2=1..........................iii
now just subtract eq 3 from 2 and substitute 1.......u ill get teh req ans
11Hint :- Expansion of cosx.
I cud not get much from joker's soln.
1
Che
·2010-01-06 09:03:12
l2+m2+n2+2(l.δl+m.δm+n.δn)+(δl)2+(δm)2+(δn)2 =1.................iii
l2+m2+n2=1........................ii
iii-ii
2(l.δl+m.δm+n.δn)+(δl)2+(δm)2+(δn)2 =0
from i
2[-(δθ)2/2] + (δl)2+(δm)2+(δn)2=0
(δθ)2= (δl)2+(δm)2+(δn)2
hp.....
1
Che
·2010-01-06 09:03:52
btw wat u din get soumik in my soln [2]
11
Devil
·2010-01-06 09:17:57
No its ok.....I used expansions.
24tell ur method also soumik..
and thx thejoker[1]
