Question

$cosAcos2Acos4Acos8A$ = $\frac{sin16A}{16sinA}$

Answer 1

sin 2X = 2sinXcosx 

So taking X=8A 

sin 16A = 2cos8Asin8A 

Taking X=4A 

sin16A = 2cos8A(2cos4Asin4A) 

Similarly, sin 4A=2cos2Asin2A 

And sin2A=2sinAcosA 

so sin 16A = 2*2*2*2 sinA cosA cos2A cos4A cos8A 

sin 16A/sin A = 16 cosA cos 2A cos 4A cos 8A 

Answer 2

Answer:

CosAcos2Acos4Acos8A

=1/2sinA[(2sinAcosA)cos2Acos4Acos8A]

=1/2sinA(sin2Acos2Acos4Acos8A)

=1/4sinA[(2sin2Acos2A)cos4Acos8A]

=1/4sinA(sin4Acos4Acos8A)

=1/8sinA[(2sin4Acos4A)cos8A]

=1/8sinA(sin8Acos8A)

=1/16sinA(2sin8Acos8A)

=1/16sinA(sin16A)

=sin16A/16sinA (Proved)