Answers

2014-05-05T11:02:15+05:30
Dividing and multiplying by
[cos(a)cos(2a).................cos(2^na)*(sin(a)sin(2a)..................sin(2^na)*2^n)]/[sina sin2a...................sin2^na * 2^n]
now 2*sina*cosa=sin2a
so
[sin2a*sin4a*sin8a*...............sin2^na*sin2^n+1a]/[(sina*sin2a*sin4a*sin8a*............sin2^na)*2^n]
so
[sin2^n+1a]/[sina*2^n]

1 1 1
2014-05-08T18:12:59+05:30
Dividing and multiplying by
[cos(a)cos(2a).................cos(2^na)*(sin(a)sin(2a)..................sin(2^na)*2^n)]/[sina sin2a...................sin2^na * 2^n]
now 2*sina*cosa=sin2a
so
[sin2a*sin4a*sin8a*...............sin2^na*sin2^n+1a]/[(sina*sin2a*sin4a*sin8a*............sin2^na)*2^n]
so
[sin2^n+1a]/[sina*2^n]
2 5 2