Cos x cos 2x cos 4x.......cos(2^(n-1) x)=Sin2^n x÷2^n sin x

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Use identity cosx = Sin(2x)/2sinx So.... Cos(2x) = Sin(4x) / 2Sin(2x).... ...------. 1 .....Notice ...power of 2 in LHS is 1 and in RHS is (1 +1) in equation 1 So.... Where power of 2 in LHS is n-1, in RHS it must be n.... {Sin2x/2Sinx} {Sin4x/2Sin2x} .. .. . . {Sin(2^n)x/2Sin(2^n-1)} = Sin(2^nx)/2^n Sinx

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Cos x cos 2x cos 4x.......cos(2^(n-1) x)=Sin2^n x÷2^n sin x

Cos x cos 2x cos 4x.......cos(2^(n-1) x)=Sin2^n x÷2^n sin x

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