" (ii) "cos x*cos2x*cos4x...cos(2^(n-1)x)=(sin2^(n)x)/(2^(n)sin x)AA n in N

Prove that cos x+cos3x+cos5x+cos(2n-1)x=(sin2nx)/(2)sin x

If (x)=(cos x+i sin x)(cos3x+i sin3x)...(cos(2n-1)x+i sin(2n-1)x) then f(x) is

The general solution of the equation 8cos x cos2x cos4x=(sin6x)/(sin x) is x=((n pi)/(7))+((pi)/(21)),AA n in Zx=((2 pi)/(7))+((pi)/(14)),AA n in Zx=((n pi)/(7))+((pi)/(14)),AA n in Zx=(n pi)+((pi)/(14)),AA n in Z

If cos((x)/(2))cos((x)/(2^(2)))cos((x)/(2^(square)))......cos((x)/(2^(n)))=(sin x)/(sin((x)/(2^(n)))) prove (1)/(2)tan((x)/(2))+(1)/(4)tan((x)/(4))...(.1)/(2^(2n))tan((x)/(2^(n)))=(1)/(2^(n))cot((x)/(2^(n)))-cot x

cos ^(3) x sin 2x= sum _(x=0) ^(n) a _(r) sin (rx) AA x in R, then