Question

cos A cos 2A cos 4A cos 8A = 1/ 16

Answer 1

hey yr ans..


Take the complex number x = cosa + isina, then x^15 = 1 and 1+x+x^2+x^3+...x^13+x^14=0. Since 1/x = cosa - isina, 
cosa = (x+1/x)/2 
cos2a = (x^2+1/x^2)/2 
cos4a = (x^4+1/x^4)/2 
cos8a = (x^8+1/x^8)/2 
cos a cos 2a cos 4a cos 8a = (x+1/x)(x^2+1/x^2)(x^4+1/x^4)(x^8+1/x^8)... 

So need to show that 

P = (x+1/x)(x^2+1/x^2)(x^4+1/x^4)(x^8+1/x^8) = 1 

Clear denominators by multiplying by x^15, which is 1: 

P = (x^2+1)(x^4+1)(x^8+1)(x^16+1) 
(x+1)P = 1 + x + x^2 + x^3 + ...x^29 + x^30 + x^31 
(x+1)P = (1 + x + x^2 + x^3 + ...x^14)(1+x^15)+ x^30 + x^31 
Remember 1+x+x^2+x^3+...x^13+x^14=0, so 
(x+1)P = x^30+x^31 = (x^15)^2(x+1) = x+1 
and P = 1 

If a were chosen so that x^(2^n-1) = 1 with n>0 (n=4 here), and n factors were in the product: 
cosa cos2a ...cos(2^(n-1)a, then the ending above would be 
(x+1)P = (1+x+...x(2^n-2))(1+x^(2^n-1))+x^(2^n-2)... = (x^( 2^(n-1) ))^2(1+x) = 1+x 
Since x is not -1, P = 1 and the product is 1/2^n

please mark as brain list

Answer 2

Given cosA x cos2A x cos4A x cos8A = sin16A/16sinA
RHS
sin16A/16sinA
[2sin8A cos 8A]/16sinA
[cos8A(2sin4A cos4A)]/8sinA
[cos8A cos 4A(2 sin 2A cos 2A)]/4sinA
[cos 8A cos 4A cos 2A(sin2A)]/2 sinA
[cos8A cos 4A cos 2A(2sinA cos A)]/2
sin A cos A cos 2A cos 4A cos 8A
Hence RHS=LHS