62cant think of a simple soln right away.. but seems interesting..
can ewe try to introduce complex nos?
eiφ + eiθ = a+ib b+ai (edited)
I am not sure.. but can this help? just thinking!
1thanx..i'll try to solve by ur method...
1actually i got another easier way!!
take cosθ=x and cosφ=y
then x+y =
also
tanθ/2tanφ/2=c
hence
(sinθ/2sinφ/2)/(cosφ/2cosθ/2)=c
multiplying both numerator and denominator with 4cosθ/2cosφ/2
we get
(sinθsinφ)/(2cos2θ/2)*(2cos2φ/2)=c
replacing 2cos2θ/2 by cosθ+1 and similarly the other
we get the expression (i)
now square the first equation and the second equation and add them
you will get the relation between
sinθsinφ and cosθcosφ
thus you can write expression (i) in terms of cosθ and cosφ
thus you have two equations and two variables!!
1nishant i suppose it should be b+ia
62
Lokesh Verma
·2008-12-01 06:33:28
yes.. u are right dear.. (bhawna ko samjho :) :D
62
Lokesh Verma
·2008-12-01 06:44:17
but yeah.. thanks for poiting out.. i corrected it :)
33haan bhavnao ko samajhna chahiye...
are ye to maine kaha tha to Vyom :P