Question

A tough series question
$cos (a) cos (2a)cos (4a)cos (8a)......cos (2^{n}a)=?$
please help me!!!

Answer 1

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]

Answer 2

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]